NOTES oil THE STRENGTH OF MATERIALS AND THE STABILITY OF STRUCTURES. BY W. M. GILLESPIE, LL. D. PLTED FOR TE CVL ENGINEERING CLASSES rN NION COLE. SCHENECTADY, N. Y. WISEMAN, RUSSELL & Co., PARER & STEREoTYPm 1870. WilAM. DAVMS, IOHN E. BUSSELL 1EGOBRAm. TxPOGaMP.EB. ENGINEERING STATICS, OR STRENGTH AND STABILITY. 1. THIS treats of the strength of materials and the stability of structures, used in Engineering and Architecture; as frames, walls, roofs, arches, bridges, &c., &o. It is divided thus: PART I. The forces which are to be resisted. PART H. The strength of cohesion, or the resistance of materials to breaking. PART III. The stability of position, or the resistae of bodies and structures to overturning. PART IV. The stability of friction, or the resistance of bodies to sliding. PART I. 2. The forces which are to be resisted are the formes which tend to produce motion and so to destroy equilibnrum. These may be statical or dynamical. 4 ENGINEERING STATICS. STATICAL FORCES. 3. Table. Weight in lbs. Material. per cubic foot. Air..08 Water - - o - 62.50 Sand - - -..... 95. Common mould - -. 90. Sandy loam - 100. Clay - - - - 110. Gravelly sand - - - 110. Gravelly clay - - - - 120. Land gravel -. 125. Water gravel - - - - - 145. Sandstone - - - - 155. Limestone - - - - 160. Granite - - - 170. Gneiss.. - - 155. Brick - - - - - 90 to 135. Masonry of brick - - 100. Masonry of sandstone - - 130. Masonry of granite or limestone - 145. Pine - - - - - 30 to 54. Hemlock - - - 40 to 48. Oak - - - - - 40 to 60. Cast iron - - - - - 450. Bar iron - - - - 480. A bar of iron one inch square and one yard long will weigh ten pounds; one bushel of wheat sixty pounds; one bushel of rye or corn fifty-six pounds; one bushel of oats thirty-six pounds; one barrel of flour one hundred and ninety-six pounds, net, or two hundred and twelve, gross; one barrel of hydraulic cement three hundred pounds; of hay eight to twelve cubic yards make one ton gross; piled wood.56 of its solid weight; a crowd of ENGINEERING STATICS. 5. men weighs seventy pounds per square foot of surface covered by them, each averaging one hundred and forty pounds and covering two square feet. An ordinary crowd weighs from fifty to sixty pounds per square foot. When so closely packed that they cannot move they weigh ninety pounds. A crowd of men is the greatest possible strain that can come upon a common road bridge. A drove of cattle weighs forty pounds per square foot of surface covered, or about half as much as men. A double row of the heaviest loaded wagons, used on the road, with teams, weighs six hundred pounds per running foot, or about fifty pounds per square foot. A railroad freight train averages half a ton, gross, per running foot. A train of the heaviest locomotives, with tenders, weighs one ton, gross, per running foot. Empty freight cars average fifteen thousand pounds each, and the load is about the same. Passenger cars weigh nineteen thousand pounds. Baggage cars fourteen thousand pounds. Thirteen passengers, with their baggage, average one ton, gross. Roofs weigh as folows: A frame truss averages 5.2 pounds per square foot of ground plan. The roof, beams and plank five pounds per square foot of superficial surface; slating five to nine pounds; shingles one and a half to three pounds; zinc one and a quarter to two pounds; lead four to seven pounds; copper one to one and a half pounds; ceiling nine pounds; tin five-eighths to. one and a quarter pounds. 6 ENGINEERING STATICS. FLUID PRESSURE. 4. This equals the product of the surface pressed, mulb tiplied by the depth of its centre of gravity, and multiplied by the weight of the fluid per unit of measure. The centre of pressure is the point where a single pressure would exactly counterbalance all the pressures acting in a contrary direction. On a vertical rectangular surface it is at one-third the height from the bottom. The pressure of the atmosphere is about fifteen pounds per square inch. SEMI-FLUIDS. 5. This name is applied to loose earth, grain, shot, &c. A B C D is a cross section of a wall. B E is the natural D C /F slope which the earth would take if free. Let a equal the angle E B C which this line makes with the vertical, and p the angle it makes with the horizon. There will be another line B F somewhere beA a tween B E and B C, such that the triangular prism of earth, whose section is'F B C, shall exercise the maximum thrust against the surface B C. Let x = angle C B F which this line makes with the vertical. We have now to find the weight and thrust of this prism. We will consider one foot in length of it. Let w = weight of a cubic foot of earth; then the weight of one foot of the prism equals I B C x C F x 1 x w = jBC xBC x; tan. x w=- BCtan. x xw. To find the horizontal thrust: Let G be its centre of gravity, G H ='weight of prism, G( I = the unknownjorizontal thrust, and the resultant force acting on the sur ENGINEERING STATICS. 7 face B F equal to G K. We shall then have, horizontal thrust = G I = wt. x tan. H G K. Now when a body is just about to slide on a surface, the angle which the direction of the force tending to produce the motion, makes with the perpendicular to the surface, equals the angle of repose for that surface, i. e., Therefore, drawing G L perpendicular to the surface B F, we have, K G L = p; andL G H = B F C (900 — x).. HIGK = -LGH-LGK = 900- x - = 90(x +'). the horizontal thrust = weight of prism, multiplied by tan. H G K = wt. x tan. [90~ — (x+ q)]. = wt. x cotan. (x+'). 2A x w x tan. x x cot. (x+ ). This is-a maximum when the factors containing x are a maximum. Now tan. x x cot. (x + A) = sin. (2 x+) - sin. q 2 sin. sin. (2 x+?p) + sin.'p sin. (2 x + p) +sin.'p. This is a maximum when sin. (2 x+ +A) is a maximum, (i. e. = 1 and then 2 x+ p = 90, or x = -(90 -') =- a); i. e., the line B F bisects the angle a, which the natural slope of the earth makes with the vertical. Cot. (x + p) = cot. (900 - ) = tan. 2 a. Then the maximum horizontal thrust = 2 B 02 x w x tan.. a x tan. aa = B -2 x, tan.2 a X w. The pressure of a perfect fluid would be = B C x 1 x B.C x wt. of fluid. =. B d X wt. of fluid. Consequently the pressure of a semi-fluid, as earth, &c., 8 ENMOINEERING STATICS. against a; surfacej equals the pressure of a fluid whose weight equals the weight of the semi-fluid, multiplied by the square of the tan. of I the angle which the natural slope of the earth makes with the verticaL This force acts horizontally against the back of the wall at the centre of pressure, i. e. ~ the height from the bottom. It is, therefore, equivalent to XB a-s x (tan. v a) 2 x w; acting against the top of the wall. If earth be exposed to saturation with water, it will act like a1 fluid of its actual weight, i. e. one and a half to two times that of water. The thrust may be resisted not only by a wall, but by timbers, or piling, or struts, &c.; and hence needs to be calculated independently. When the earth is higher than the top of the wall, we may use its entire height in the formula for the pressure, when the height is not greater than twice the height of the wall. i Natural Tan. of its Weight of Wt. of cubic ft. MA.TERIAL. slope of angle with the cubic ft. multiplied by earth. vertical. in lba tau.2 X a. Fine dry sand.......... 300.577 94 31 Common dry earth.... 43"'.435 94 18 Common moist~ earth... 54.325 106 11 Very compact earth.... 55.315 125 12.5 Loose broken stone.... 500.364 81 11 HEAT. 6. For 1~ Fah. wrought iron expands T of its length, and cast iron, with a force equal to about 15,000 pounds per square inch' of cross section, or frbm; sr- to- nie todns ENGINEERING STATICS. 9 Freezing of Water. 7. The force of this is very great. It has burst a bombshell which could resist a pressure of 18,000 pounds per square inch. Animal Force. 8. The average force of a man drawing horizontally is seventy pounds, for a short time, and thrusting horizontally thirty pounds. A horse is equal to about six men, or about four hundred and twenty pounds at a dead pull. A nominal "horse-power" is raising 33,000 pounds one foot high in one minute, which is equivalent to raising one hundred and fifty pounds, at the rate of two and a half miles per hour; eight hours of such labor is a day's work. DYNAMICAL FORCES. Solids in Motion. 9. The effects of weights passing over surfaces at high speed are greater than at rest. The most important case is that of railroad trains passing over iron bridges. It is greatest when the bridge is short and therefore light. The dynamical deflection produced by a weight in motion varies as the square of the velocity, and inversely as the square of the length of the bridge. The extra stress is caused by the centrifugal force. In an experiment upon a light model the dynamical deflection was double the statical depression; this is in an extreme case. For a bridge twenty feet long the dynamical deflection may be half more than the statical, while in a bridge fifty feet long It fifty miles per hour it was only one-quarter more. 2 10 ENGINERNG STATICS. The quantity of motion in a body is the product of its mass by its velocity = rn v. This is often called its momentum. The living force or " vis va,' of a body equals its: mas x by the square of its velocity = m v2. Some call it "living power," and half of this product "living force." The work done by a moving body equals its mass or force multiplied into the distance through which the body moves, or force acts. The product of a weight in pounds, multiplied by the distance in feet through which it falls, gives its work in "foot.-oundE." Other forces are expressed similarly. The accumulated work in a body is also called its "potential energy." The "moment of inertia" equals 2. r'. m. (See Jackson's Dynamics, Art. 45). Impact. 10. Impact cannot be compared with pressure, i. e, the force with which a moving body strikes another can not be theoretically expressed in the units of the weight of the moving body. The effects of an impact and a pressure may however be compared in certain cases. By recent experiments one pound falling one foot, produced a pressure on a spring balance of nineteen pounds. Hence was discovered -this general formula for the effect of a falling weight, viz.; W (1 2+ 2 x v) = W (1+ 18 V h). e. g. one thousand pounds faug nine feet produces a pressure 1,000 (1 + 18 f 9) = -55,000. ENGINKEERING STATICS. 11 This result might vary with every different spring used. The impact or force of ice in motion is greater in appearance than in reality. Pile Driving 11.'Inpile driving, letting &= weight of ram, and w' = weight of pile, h = height of fall, v = velocity of ram when it strikes the pile. Then u = common velocity of WV ram and pile after fall = w + w. Their "vis viva" = ++w W v w a 9 (W + wI9: The "potential energy" of the ram and pile, or the work which this "vis viva" is capable of, equals half of it tO' dv w h L Putting this under the formula w h 2 W+ wh we infer; 1st. That the effect is directly proportional 1 + W' to the fall; 2d. That for different falls and weights, such that w h remains the same, the effect will be greater the heavier the ram with a corresponding smaller fal e. g., letting w = 2 tons, w' = ton, h = 10 feet, the effect, = 20 i+~ = 17.7777 +. Let w=1 ton,w'= ton, h= 20 feet, effect =- + — -16* For the resistance which the friction of ground opposes to the sining of,a pile under a weight, see Part IV. 12 ENGINEERING STATICS. Water in motion, or Waves. 12. The greatest force observed in the Atlantic, on the west coast of Scotland, was_ six thousand pounds per square foot. The average pressure was in summer six hundred pounds, and in winter two thousand pounds. Air in motion, or Wind. 13. Experiments give the pressure in pounds, on a plane surface perpendicular to the wind, as, =,.0028 A. v.' = a little more than A x A. v.' v being velocity of wind in feet, per second, and A the area of surface in square feet. If v = velocity in miles per hour then pressure =.0049 A. v.2. Velocity in Velocity in Pressure in THE WIND. ft. per sec'd. mls per hour lbs. per sq.ft. Fresh wind............... 20 13 0.95 High wind............. 40 26 4.00 Violent wind..............i 60 40 8.00 Tempest.................. 80 54 16.00 Violent tempest........... 100 68 24.00 Hurricane................. 120 80 32.00 Greatest hurricane........ 150 102 55.00 It is safe in temperate climates to take forty pounds per square foot as the greatest force to be guarded against. Some take thirty pounds. The force operates horizontally, and its resultant passes through the centre of gravity of the surface. The pressure against a cylinder is from two-thirds to one-half the force corresponding to its diametrical section. The effect of air in motion against a body: the resistance experienced by the body moving through the air with the same velocity.:: 1.4: 1. ENGINEERING STATICS 13 PART II. STRENGTH OF COHESION, OR THE RESISTANCE OF MATERIUAI TO PARTING. Preliminary Princi1es. 14. Stress is the force which acts upon a body. Its intensity is usually expressed as so many pounds per square inch. In French measures as kilogrammes per square millimetre, which equals 1A422 pounds per square inch. The effect of a stress upon a body is its Strain, i. e., its alteration of volume and figure. When the strain is so great as to separate a solid body into parts, this constitutes its fracture. Bodies resist stress by their elasticity, which is the property they possess of resisting any change in their form or volume. They may be regarded as composed of molecules which attract and repel each other with forces which in the normal state of the body are equal. Consequently every effort to separate these molecules calls into play an attractive reaction, and every effort to press them together calls forth a repulsive reaction which is proportional to the stzess. Within certain limits the elongations, or contractions, produced by any force, are nearly proportional to those forces; beyond those limits proportionality ceases, and the elongations and contractions increase more rapidly than the forces which produce them. This limitingpoint is called the limit of elasticity. W'iet the limits of proportionality, when the force cease*. acting, the effect also ceases and the body returns to its original shape and 14 ENGINMERIG STATICS. dimensions, or nearly so. Beyond those limits it does not so return, but remains permanently elongated, contracted or bent. It is then said to have its elasticity impaired or injured, or to have taken a " set." A long continued stress produces much more effect in impairing the elasticity of bodies than does a much greater force applied for a short time. A body should never be subjected to an effort capable of seriously injuring its elasticity, for that is an approach toward fracture. The allowable strain should be very much less than this, on account of accidental forces and chemical changes to which a body is exposed.' Usually the stress intentionally applied to a body is from one-third to one-quarter of that which would cause fracture, or onehalf of that which corresponds to its limits of elasticity. The following are the corresponding stress, strains and modes of fracture: Semss. SArIN,. F Ac-rlU. Pull. Extension............ Tearing. Shearing............. Distortion........... Shearing. Thrust............ Compression. Crushing. Bending............ Bending.............. Breaking across. Twisting....... Torsion......... Wrenching. A ull is a longitudinal stress which tends to lengthen a body, A thrust is a longitudinal stress which tends to shorten a body. Usually a pull has a positive algebraical sign and a thrust a negative. ENGINEERING STATICS. 15 The other stresses are more or less transverse, and tend to distort the body. For shearing stress see Art. 30. For bending stress see Art. 36. For twisting stress see Art. 84. RESISTANCE OF BODIES TO EXTENSION. General Principes. 15. The resistance of bodies to extension is sometimes called their tensile, their direct, their absolute or their positive strength. Let a prismatic body of a length 1, and a cross section A, be subjected to a stress or pull P. It is elongated by a length 6 1. Let the ratio Z equal i, and be constant within certain limits. Within those limits i is proportional to i. e., to the stress per unit of cross section of A, the body; so that is to i, or is a constant quantity. It is called the coefficient or modulus of elasticity, and it is expressed by E, so thatT = E. If we suppose the cross section o the body to be equal to the unit of surface, usually a square inch, and suppose it is possible for the elongation to become equal to the unit of length without injury to the elasticity, the body thus continuing to elongate in a constant ratio, we should have A i = 1, since A i =A x-l=l x 1., and P = E would then be the stress per unit of surfade, capable of producing per -16 MQMFRING STATICS. rtmii bf length, an elongation equal to that of the length. This may be taken as a definition for " modulus of ekcwi ticity." Thus if 20,000 pounds elongated a bar of wrought iron, whose section is the unit of surface, TTV of its length, then for this iron E = 20,000 x 1,500 = -30,000,000 P P pounds. From - E, we obtain where A = 1, E=A4 i. Experiments then on any material will give E for the unit of cross section. Knowing this we can calculate for a prismatic body of a given cross section A, the stress which will produce a given -elongation, or the elongation which a given stress will produce, since the resistance of bodies to extension is directly proportional to their cross section, i. e., their section at right angles to the direction of the stress. Such experiments are best represented graphically, by constructing a curve of which the ordinate are the forces applied, and the elongations the abscissas. Figure 2 refers to a bar of iron elongated by successive Fig. 2. weights. From the inclinations of the curve at various successive points we can infer the laws of the action of the strain; whether it be uniform or varied, increasing or decreasing, &c. As The total resistance of a body to elongation is equal to the work done in producing that elongation, and equal to the sum of the products of the stresses exerted, multiplied by the elongations produced by them. The area 3 B C x A B, of the triangle A B C, is formed by the abscissa and ordinate, -and the first straight line ENGINEERING STATICS. 17 portion of the curve; therefore expresses the work done by the stress applied. The mean stress up to that point of strain = I B C x A B - A B - 3 B C - I final stress. So to, the quadrature or area of the curve, up to any point, gives the work done in producing the elongation corresponding to that point. Thus the quadrature up to the so called limit of elasticity, gives the work developed up to that point, by the elasticity of the body. It is called The Living Resistance of Elasticity, and it is written T,. The quadrature, up to the point of rupture, gives the work done in producing rupture. It is called The Living Resistance of Rupture, and is written Tr. To resist with safety a very sudden pull a bar or rod requires twice the strength that is necessary to resist the gradual application and steady action of the same pull. For the work done by a constant force I P, acting through a given space, equals the work done by the action through the same space of a force increasing uniformly from 0 to P. SPECIAL MATERIALLS. WFrought Iron. 16. The following table shows the relative weights and elongations of a rod of good wrought iron: 18 GFINRING STATIOS. Load per square inch in Elongation per unit of Permanent elongalbs. length tion per unit of length. 5,300.00018.000000 12,300.00047.000004 21,300.00076,000010 26,700.00101.000083 29,900.00128.000263 32,000.00229.001130 34,000 A 00429 Z003071 40,000.01049.009102 60,700.03493.03280 52,300 Rupture. Rupture. Representing this table by a curve, as in Fig. 2, we infer: 1st. Since the first portion of A is a straight line, the ratio of elongation to the load is so far constant. It continues so up to a stress of about 17,000 pounds per square inch. 2d. For a stress greater than this the line tends towards the axis of abscissas. This shows that the elongations are increasing faster than the loads. The point where this change takes place is called "the limit of elasticity." 3d. Beyond the stress of about 30,000 pounds, we have nearly a straight line again, but much more inclined toward the axis of abscissas. This shows that the elongations have again become proportional to the loads, but have a much greater ratio to them than at first, showing that the elasticity of the material is injured. The permanent elongations follow similar laws, they are very small up to a pull of 21,000 pounds, being then only.00001, but BNGINEERINO STATICS. 19 then increase very rapidly, much more so than the temporary elongations. In comparing different irons we find that the weights which alter the elasticity of soft bars do not do so for soft wire; also, that this wire breaks with a greater load but less elongation; that hard wire breaks with somewhat more weight than the soft, but with much less elongation, and that it gives way suddenly'and without warning; we also find that the hard iron has much more Te than the soft iron, i. e., that greater stresses may be applied before the limit of elasticity is reached; but soft iron has much more Tr than the hard iron, i. e., much more work must be done to break them; therefore they should be chosen for exposure to sudden shocks. Example. In soft iron, if the limit of elasticity be 22,800 pounds, and the elongation be.00086, we have Te =- x 22,800 x.00086 = 9.8 pounds per foot-of length. Suppose a bar be twenty feet long and one inch square, its Te = 20 x 9.8 = 196 foot pounds. Now if a body W falls from a height h the work of that body = a m va = 2 W xv Wh. Suppose the body to be suspended from the bar, this work is to be destroyed by the Living Elasticity Te, which is here one hundred and ninety-six pounds, which amount must not be exceeded. It would be equal to the work of a body of ten pounds falling 19.6 feet. Beyond that the elasticity would be altered. The total quadrature Tr = 6,400. Then the above bar would be broken by one hundred pounds falling from a height equal to j x 6QB0o0 x 20 x 1 = 640 feet. 20 ENGNEEMG STATICS. In like manner soft annealed wire gives Te = 9.4 and Tr = 710. For hard unannealed wire Te = 8.3 and Tr =970. These are for one inch square. Very extensible materials resist shocks much better than rigid ones. The strength of wrought iron varies greatly with its chemical composition, mode of manufacturing, size of bar, &c. Cast iron; Wrought iron; ELONGATION. stress per stress required.'Set" "Set" sq.. in., in to produce the cast iron wrt. iron Ratio lbs. ~ same elongat'n Oust iron Wrt. iron 512 1,254.00004 same 0 0. 2,486 5, 600.0002 same.00001 0. 44o o10,080.00033.same.000026.0000049 1:15 5,600 12,544.00042 same.000036.000006 1:17 6,720 15,142.00052 same.000047.0000075 1:16 8,960 20,160.00072 same.000075.0000225 1:30 12,330 26,970.00107 same.000124.0001166 1:94 13,216 29,680.00121 same.000160.0003583 1:222 33,600..... 0023.......001 Relative tensile forces to produce equal elongations in wrought and cast iron is two and one-quarter to one. Within the above limits, for equal elongations, the " set " of wrought iron is less up to strains of 12,000 pounds for cast and 27,000 for wrought iron, which produce equal elongations. Beyond that the "set" for wrought iron is much the greater. For a stress of about two-thirds rupturing stress of each, (e. g. a stress of five tons for cast and fifteen tons for wrought iron,) the elongation for the wrought iron is two and a-half times that of cast iron, and its "set" ten times that of cast iron, e. g.: ENGLEE-KG STATIC. 21 ~ ruptur.. ELONaTIONS. S.ETS. Ratio of ing wt. Ratio. set to stress. I elongaCast. Wro't. Ratio. Cast. Wro't. tion. 11,120.00095... 1:24.00011. i:10 1:9 33,600.0023......0011.... 1:2 Wrought iron is also affected by temperature. Its strength increases up to 300~ Fah., and then decreases, being only about half as strong at 1000~ as at 3000. Cast Iron. 17. The resistance of cast iron to breaking by extension is much less than that of wrought iron, but follows similar laws. Other Metale. 18. For "Table of Resistance to Extension" see following table: 22 ENGINEERING STATICS. TABRi OF oRE sITAN TO EXTENsION. Stress in Elonga- E - coe ffi R= stress inlbs. per sq Stress in I Elonga- = coeffi-inch causing rupture. lbs. per sq. tions at cient or inch casing rupture METAL. in. corres- limit of modulus of ponding to elastic'y. elasticity. Safe means limit of Extreme. for sound elasticity. material. Iron. Bars 17,000 1:200 29,000,000 35,0 48,000 1:a 200 85,000 46, 000 Boiler plate.. 26,000..........000 Annealed wire 20,000 1:1250 25,000,000 130,000........ 130,000 Steel....... 35,000 1:830 29,000,000 70,000 90,000 94,000 1:450 42,000,000 130,000 Copper. Cast....................... 19,000 Bolt.................................... 36,000 Sheet...................... 30,000 Wire............16,000,000....... 60,000 Brass. Cast........ 7,000 1:1300 9,000,000....... 18,000 Wire..... 19,000 1:740 14,000,000....... 49,000 Bronze or gun metal...... 6,000 1:1600 10,000,000.30,000 Tin (cast).. 4,500,000........ 4,600 Zinc.............. 13,500,000 7,500 Lead. Cast........ 1,400 1:500 700,000....... 1,400 Wire........ 570 1:1800 1,000,000....... Sheet...... 700,000 3,300 Cast iron....8,500 14,000000 00 16,600 Cast iron 14,000......,000,000 29,000 Corning's semi-steel....... 73,000 89,500........ * Small wires are much stronger per square inch than larger ones. For wire 20 feet to a pound, strength = 1,300 pounds = 90,000 pounds per square inch. Wire one-eighth in diameter breaks at 1,600 pounds =130,000 pounds per square inch. ENGNEERING STATICS. 23 The safe stress for extension of metals is from one-sixth to one-third of that causing rupture, depending upon their exposure to shocks, weather, &c. Some recommend taking half the stress corresponding to the limit of elasticity. These two methods do not vary much in their results. When they do adopt the safer one. Pure cobalt is' the most ductible and tenacious of metals, a wire of it is twice as strong as an iron wire. Copper (rolled) is weakened by heat above the freezing point. Aluminum has tenacity of 18,600 pounds, and hence is between zinc and copper. An alloy of ninety parts copper and ten aluminum has a tenacity of 80,000 pounds, hence between soft iron and steel. Captain MEIGs, at the capitol, allowed for greatest safe strain for iron as follows: For important parts, cast iron 1,800, and wrought iron 8,000. For secondary parts, cast iron 3,000, and wrought 9,500. From 8,000 to 10,000 is usually allowed for wrought iron in bridges. Length of bars which would break of their own weight: Steel 40,000'; wrought iron 18,000'; cast iron 5,000'; rolled copper 9,000'; lead 36Q'; cast copper 5,000'. Resistance of Wood to Extension. 19. This, as for metals, is proportional to the crosssection of the pieces. In wood even more than in metals there is strictly no limit of elasticity, but any load produces a permanent elongation, but in dry woods very much less than in wet woods. :24 E1JilrORING STA0N. T Stress corres- Elongation E = modulus R — stress AKn oF WOOD. ponding to at limit of or coefficient causing ruplimit of elas- elasticity, of elasticity. ture. ticity. Ash......... 1,800 1:855 1,600,000 17,000 Beech........ 2,300 1:570 1,300,000 11,00 Birch.................. 1,600,000 15,000 Chestnut........ 1,100,000 11,000 Elm,,......... 3300 1:414 1, 360,000 14,000 900,000!Larch.........2, 5,00 1;-520 10,000 1,300, ooo 1 Locusat ~...~.................... ~............. 16,000 Oak....... 2,850 1:600 1,700,000 10, Red pine..... 4,00 1:470 2,100,000 13,000 Yellow & White. pine1.....*... 3,100 E1 OI850 2,610,00.10,000 One-fifth to one-tenth of R is considered a safe load, depending on importance, exposure, &c. When exposed to running loads, as in bridges, one-twelfth is better; say for white pine, eight hundred pounds per square inch. The above numbers are for sound timber, free from knots and other defects. The strength of the same species of wood varies greatly according to the locality of the tree, nature of the soil, part of the tree, age of tree, its seasoning, &c, The strength of -wood, as given from the above table, supposes it to be pulled in the direction of its fibres or with the grain. At right angles to this its strength is much less. In straight grained woods like pine, this last strength is only from one-sixteenth.to one-twentieth of the former. In tough and crooked woods it is from oneseventh to one-tenth. Care must be taken when calcu ENGINEERING STATICS. 25 lating the strength of wood to resist extension, to allow for the parts which must be cut away by notches, boltholes, &c., in,order to make the connections. Other Materials. 20. Stone is rarely exposed to a strain of extension, its resistance thereto ranges from two hundred to one thousand pounds per square inch. The safe strain is onetenth of that. Brick one hundred to three hundred pounds. Glass two thousand four hundred to two thousand nine hundred pounds. The poorest mortar resists with only ten pounds per square inch. Good hydraulic mortar with from one hundred to two hundred pounds per square inch. Safe strain one-tenth in all cases. Adhesion to stone of good mortar, fifteen to thirty-three pounds. Plaster, thirty to forty-eight pounds. Best Rosendale cement, forty pounds. TABLE (RANKINE) MATERIAL. ME' "B Bricks.... 2.................. 280 to 300 Glass,....... 8,000,000 9,400 Slate............. 1,000,00 10,00to 13,000 Ordinary mortar............. 50 PARTICULAR FoiRs. Riveted Iron Plkdes. 21. The line of fracture of a riveted plate, is a line drawn on'it crossing all the lines of strain in such a manner that the section of the plate along it shall be- a mini4 26 ENGINEERING STATICS. mum. The plate must break along this line, however "zig-zag" it may be. The strength of the rivets will be considered under " Shearing stress." Wire Ropes. 22. Their breaking weight averages four thousand five hundred pounds per pound weight per fathom, and the working load one-sixth of this. TArnL oF WIe ROPE MFA{IPACTURED BY JNO A. ROEBUING, TmENT, N.J. ROPE OF 133 WIREs. ROPE OF 49 WIrS...s.I~.j.E I.. _. _! 1 6% 2X 1 20 7400 153 11 4%'54 36 00 10% 9 6 2 1 05 65 00 14X 12 4X 47 So 00 10 3 53 1% 91 54 00 13 13 8% 41 25 00 9o 4 5 1% 78 438 60 12 14 3% 35 20 00 8X 6 4% 1i 63 35 00 10% 15 8 29 16 00 7/X 6 4 3 2720 9% 16 2% 23 12 s0 6oX 7 J3% 1% 41 20 20 8 17 2% 18 880 s 8 3% 1 34 16 C0 7 18 82 15 7 60 5 9 2X % 28 11 40 6 19 % 18 6 80 4% 10 2X X 25 8 64 ~ 20 1% 11 409 4 10 2 24 5318 4% 21 1% 9 2 88 3% 10o 1% 9-16 23 4'27 4 22 X 8 13 2% 10o 1X X 22 3ta 48 3% 23 1%9 7 16 8 2% 24 1 6% 188 2% o r _f fori n25 N 6 103 2 For safe working load allow 1-6 to 1-7 of 27 % 5 0 56 1 ultimate stmeagth, 27S % 4 ^~~~~~~~7 ENGINEEEING STATICS. 27 Hempen Ro.pe 23. Breaking weight 5,500 to 12,000 pounds per squae inch of cross-section. Average of good rope 8,500 pounds. Safe strain one-half the breaking weight. Another Rule. Safe strain in pounds equals circumference in inches squared, x 200 = cross-section in inches x 2500. Another gives: Breaking weight equals one gross ton per pound weight per fathom. Chain C(able. 24. When the links are formed as in figure 3, their Fig. 3. strength is nearly equal to the strength of the bars of which the links are formC ed, i. e., equal to twice the section of the single bar. Some U. S. experiments gave a mean value of R = 41,000 pounds; extremes = 32,000 and 50,000. Thin Hollow Cylinders. 25. Let p = the interior pressure per square inch, d = diameter in inches, R = resistance to tearing per square inch, and t = thickness in inches. Then the equation for equilibrium is, pd = 2 t R. Hence, t - d For safe thickness we take t' dch R' 2f;i n which B means safe strain per square inch. The pressure is often named as so many "atmospheres," each atmosphere being fifteen pounds per square inch. 28 ElGI'EERING STATICS. Let n = number of atmospheres of pressure above that of the air, then the above formula becomes t 15 d 2 1?' For cast iron, taking R = 16,500- pounds, the proof strain for water and gas pipes may be one-third of this, and the working strain one-sixth of it. In addition to the thickness necessary to resist internal pressure a constant thickness is added to resist external shocks. Calling this additional thickness t" the formula for safe thickness becomes t' - p d + t=15 n d t 2R' R - 2 R' The Paris rule is, for cast iron take-' = 3,100 pounds and t" = 0.333 inches. For wrought iron, Paris rule makes B' = 8,500 pounds and t" 0.12 inches, or t' = 0.00086 n d + 0.12. A simpler formula for cast iron is this; t' = 0.002 n d +.4; (all in inches). For water pipes of rolled copper, t' =.00147 n d +.16, (all in inches). -For water pipes of lead,' =.00242 n d +.20, (all in inches). For water pipes of zinc, t' =.0062 n d +.16, (all in inches). For water pipes of wood, t' =.0323 n d + 1.08, (all in inches). For riiveted plate iron (steam boilers) in the first general formula, R =34,000 pound& The proof tension may be one-half of this, and the wotking tension one-eighth. For steam pipes the same; i. e. B' = 4,250 and t' 2 d +.12. 2 R' ENGINEERNG STATICS. 29 The French Government rule is, t' =.0018 n d +.12". This about corresponds to the above rule. The bursting strain on the longitudinal seams of cylindric boilers, is double the strain on the circular seams. Tubes are also liable to give way by the pressure from without, which makes them collapse. This is partly extension, partly crushing, and partly bending. It is usual to make the thickness of tubes exposed to collapsing, double of what would be required if only exposed to pressure from within. Recent experiments show that the strength of such tubes varies directly as the square of the thickness, inversely as the diameter, and inversely as the length. Mr. FAIRBAIRN'S rule, for the safe pressure in pounds per square inch, on flues of wrought iron is: multiply the constant quantity 800,000 pounds, by the square of the thickness in inches, and divide by the length in feet, and by the diameter in inches. Thick Hollow Cylinders. 26. In these the full strength of the material is not obtained, because the inner. ring may give way before scarcely any strain comes on the outer part. This is important in the case of hydraulic presses, cannons, &c. Prof. BARLow's formula is t = 2 d for bare 2 x (R -p) equilibrium. Only one-third of the bursting pressure should be used in presses. Spheres. 27. The strength of a sphere to resist bursting is pre 30 ENGNEERING STAToCB. cisely twice that of a cylinder of the same diameter. For cylinders we have p d = 2 tR,.. p For spheres 4t R A Supension Rod of Uniform Strength. 28. Let W = weight, suspended at lower end; let w = weight of rod per unit of volume; R' safe strain; x = length of rod up to any desired point; A = proper area or cross-section at the desired point. All the measurements must be in the same unit (say, inches). Formula A = e Here e is raised to a power whose numeration is w multiplied by x, and whose denomenator is R', e =- 2.71828, (the Naperian base). RESISTANCE TO SHEaNGO OR DETRmIOlN. 29. Shearing, or tangential stress, is the force which acts between two parts of a body, when each draws the other sideways, in a direction parallel to their surface of contact. It is the stress which effects rivets, trenails, notches at the ends of tie-beams, &c. A shearing stress also accompanies bending and torsion to be examined hereafter. Resistance to shearing pei square inch of cast iron is 24,000 to 42,000 pounds. Wrought iron 50,000 pounds. Safe, one-sixth of each. Chains for suspension bridges consist of long bars connected by pairs of short bars or links with bolts or pins passing through both. INIEB1TGG STATICS. 31 The best proportions for economy should be such that the bars and bolts would be equally strained, so that neither should give way before the other. This is a fundamental principle in all combinations or. parts of a structure; as in a chain all the links should be equally strong, since the strength of the chain is only that of the weakest link. To break the bolt it must be sheared across at two places at once, then the double cross-section of the bolt 11 d2 of diameter d= 2 x l4 ) x resistance to shearing per square inch, - cross-section of the bar, a x its resistance to tearing asunder. Calling these resistances equal 11 d' we have 2 x 14 = a. But these results are as 5:6. 14 Then, 2 x:cross-section bar:: 6:5. d — /_ 14 1.31 - 874. /a. 30. Rivet work for bridges, plate boilers, &c., should also be so proportioned that the section of the uncut part, multiplied by its resistance to tearing per unit of measure, should equal the section of all the rivets, multiplied by their resistance to shearing. Assuming that the riets resist shearing with about the same strength that wrought iron plates resist tearing, the sections of the rivets should about equal the sections of the iron between them.;u;: e overlap.plae: joit, single-riveted,.the sectional area of one rivet should equal the sectional area of the plate: between the holes. The diameter of the rivet is usually one and. a-half to 32 ENGINERING STATICS. two times the thickness of the plate. Calling t = thickness of the plate, d- diameter of the rivet, Z - the distance between the holes; we have Z x t - 11 d. The dis14 tance between the rivets equals 1 t, and the distance between their centres — (11 d. The overlap is about equal to the distance between their centres. In the overlap plate joint, double riveted, the sectional areas of two rivets equals the sectional area of the plate between each pair of holes in the same line. 11 d' The distance of centres -.1d + t + d. 7 The overlap is about 1.7 of this. In a plate, butt joint, with two covering plates, singleriveted, the section of the rivets which give way by being sheared across in two places at once, is as in the preceding case. Distance of centres the same 11 t + d. 7 Length of each covering plate about twice this. In a plate, butt joint, double-riveted, section of four rivets must equal that of the plate between two holes in the same line. Distance of centres = + t + d. Length of covering plate equals three and one-third to three and one-half times this. (On rivet work see Latham on wrought iron bridges, pp. 26-86.) ENGINEEING STATICS. 88 Wood. S1. When a rafter is notched-into a tie-beam, the end of the beam is liable to give way by shearing off or detrusion. Fig. 4. For greatest economy of material the resistance to thrust or compression on the surface a b ef, Fig. 4, should just equal da l- the resistance to shearing of a b c d. Calling the former resistance ten times the latter, a d should be ten times af. If the shoulder should be in the middle of the beam, as for a bolt-hole, there'would be twice the resistance to detrtusion. Resistane to detru8ion of Wood with thefibres. lbs. per sq. in. White pine.................................. 490 Ohio pine.................................... 390 Georgia pine............ 410 Spruce........................................... 470 Hemlock......................................... 540 Chestnut...................................... 690 0Ok....................... 780 Locust.... 1,180 Across thefibres. Pine................... 400 to.800 Spruoe........... -_.. 600 Lartch..................................... 1,00 to 1,700 British oak................... 2,300 Safe stress, one-quarter to one-sixth of the above. 84 ENG)INENG STATICS. RESISTANCE TO COoPIRESSION AND CRUSHING. 32. For small pressures, this resistance equals the resistance of the same material to extension, and has the same numerical modulus of elasticity, which here means the weight, which continuing to act in the same ratio would shorten the body to one-half of its length. For greater pressures the resistance to compression is very irregular. It varies directly as the. cross-section of the material. Different materials yield in different ways. Granular substances, like cast iron, stone, brick, &c., give way by oblique shearing, at a certain angle with the direction of the crushing force, which angle in cast iron is from 30~ to 40~. Sometimes a wedge-shaped piece or pyramid is forced out on two or four sides. Substances of a glassy texture give way by splitting irregularly. Tough and ductile substances, like wrought iron, yield by bulging or swelling out sideways. Fibrous substances, like wood, yield by buckling the fibres, wrinkling and splitting. When the material compressed is a certain number of times longer than it is thick, it gives way by bending or cross-breaking. The resistance varies inversely with the length (nearly). This case will be examined under Part IV. When the centre of pressure upon a surface does not coincide with its centre of figure, the maximum pressure will exceed the mean pressure. Considering the pressure to be a uniformly varying stress, the ratio of maximum to minimum pressure equals 1 + be to 1, in which b distance of centre of pressure from centre of figure, and ENGINEERNG STATICS. 35 c = a quantity depending on the form of a section of the body at right angles to the direction of the force. For a square of side h or rectangle of thickness h, in a direction perpendicular to the central line, c -. 8 For a circle of diameter h, c =. When the pressure is at one-third h from the edge, or one-sixth h from the centre, then the ratio is (1 +. h x 6) 6): 1:: 2:1, i. e., the maximum pressure equals twice the mean pressure. If the pressure be at j h from the edge, or I h from the centre, the ratio equals (1 + j h. x 6). 1::3:1. The pressure = — 0 at i h from the other side. If the pressure be at the edge, or h A from centre, then ratio is (1 + ~ h x ~:: 4:1. STONE. (Prof, Henry.) NaME or STONE. Resistance to crushing in lbs. per sq. in. Sandstone (U. S. capitol)........................ 5,200 Red sandstone (Smithsonian Institute)..........,.... 9,500 Marble.................................... 7,000 to 10,000 Malone sandstone................................. 24,000 Blue gneiss................................. 15,000 Quincy granite or sienite........................... 29,000 Brown sandstone or freestone................. 3,000 to 3,500 New Jersey freestone........................... 3,500 Connecticut freestone.............................. 3,300 Dorchester freestone....................... 3,000 Albert sandstone................................ 8,200 Caen stone..................,.......... 1,100 3u ENGtINEBING STATICB. Britania Bridge Experimens. Anglesea limestone,.......................... 7,600 Red sandstone.................................... 2,180 Brick masonry, 417 to 610 mean = 521 New brick walls of Capitol 1,333 pounds per square inch. White marble of Washington Monument 2,000 to 5,000 pounds per square inch. Briti8h Parliament Experiment. STOWs.3 Fracturing weight, Crushing weight&, in lbs. per sq. in. in lbs. per sq. in, Ind~o.n lbs. p e r, s s,. n'' " " ~6,049 7,840 ean 5,824 4,928 3,808 Magnesium limest'e 0 288 mean5,1 6,720 -8,288 mean Limestones,120 mean 1,79344 2 mean 2,91 1,568 mean 4,032 Silicious limestones. 2,912 7,168 Oolite............. 1,344 mean1568 10,568tmean Bramah's Experiments. STONES. Fracturing weight., Crushing weight. Granites........... 6 492 mean 8,288 mean 11,872 10:752 14,784 Sandstones n......,, 6,496 8,736 mea n,800 Good brick work in cement, cracked at 760, crushed at 900. Common brick crushes at from 900 to 1,900 pounds per square inch. Soft brick with from 460 to 620. ENGINEERING STATICS. 37 The strength of stone remains constant till its height equals ten or eleven times its diameter. When height is twenty-four times diameter its strength equals seventenths of original strength. At thirty times, it is T4V and at forty times, ~3. The resistance of stone to compression is about ten times that to extension. A granite column would be crushed by its own weight when about 12,000 feet high; white marble 2,000. In the greatest buildings in the world, the greatest pressure is from one-sixteenth to one-eighth the crushing weight; one-tenth is a good ratio. Common mortar crushes with from 300 to 500 pounds per square inch. Common hydraulic mortar, 1,000 pounds; Very hydraulic mortar 1,500 pounds. Concrete made from quick lime will bear safely 150 pounds per square inch. This was the foundation of the London Crystal Palace, where it bore forty pounds.' Beton' (concrete made of hydraulic cement) crushes with 600 pounds per square inch. Safe pressure, one-tenth of this. Masonry should not be loaded with more than onetwentieth the weight, which would crush the material, or one-tenth of what would crush the masonry. Cadt Iron. 33. The resistance of cast iron to crushing is from 80,000 to 140,000 pounds per square inch, the mean being 110,000. This is about six times its resistance to extension. One-fifth the crushing weight is safe. The English cast iron bridge engineers use 18,000 pounds. Captaim MEIoS makes the limit for pieces less than twelve times as long as thick, for principle parts, 38 ENGINEERING ATICS. 10,500 pounds per square inch, and for secondary 17,500. Where the length is more than three times the thickness the pieces give way by bending. It is brittle at the freezing point. Its strength varies little between 40~ and 120~, beyond 1200 it becomes weaker. Wrought Iron. 34. Its crushing weight is from 35,000 to 40,000 pounds; one-half to one-third of that of cast iron. This is about three-fourth its resistance to extension. The safe load is one-fourth of this. Captain MEIGms, for pieces less than twelve times as long as thick, uses 7,000 pounds. Great care must be taken to prevent the bending of long pieces, subjected to thrust, by supporting them at different points in their length, or by putting it in the form of tubes, &c., or by corrugating thin plates of it, or by riveting T irons to the plates. Although wrought iron is crushed with much less weight than cast iron, yet under moderate pressure (within the limit of elasticity,) it is compressed much less than cast iron. Its modulus of elasticity being almost double. Therefore it should be preferred in important structures, even for resisting compression, except where economy forbids. It was used in the Britannia tubular bridge. Crushing weight of cast brass equals 10,300 pounds per square inch. Wood. 35. Dry timber crushed along the grain yields under a pressure of from 5,000 to 10,000 pounds per square inch. Dry English oak, elm, beech, ash, 9,000 to 10,000. Pine ENGINEERING sTIC. 389 5,000 to 7,000 pounds; one-tenth of crushing weight is safe. When green its strength is very much less, sometimes one-half. For good white pine 800 to 1,000 pounds per square inch may be used for short pieces. Sheet lead or iron shoes should be interposed to prevent the fibres working into one another, where the end of one piece of wood presses against another. Where a pressure acts transversely to the fibres, if it acts on the whole side of the piece, 100 to 250 pounds is the safe limit, or say one-fourth or one-fifth of the former. When a wooden post or strut is much longer than it is thick, it yields by bending. Its strength decreases rapidly as its length increases. (See notes on bending.) ENGmNEEING RATICS. RESISTANCE TO BENDNG AND BREKING. General Prituple8 of Flexure. 36. Let A B be a body, such as a beam, fixed at one Fig. 5. end A, and having at the other end B, a weight, or some other force, applied perk G 4L pendicularly to its length. The weight of the body itself is neglected for the present. This force exerts a shearing stress equal to W, at every point of the beam, acting at right angles to it, and also a bending force along the beam. The beam will therefore assume a curved form, as shown by the dotted lines in the figure; B coming to B'. The upper fibres will be lengthened, and the lower ones shortened, while one layer of fibres will be neither lengthened nor shortened. This line is therefore called the " Neutral Axis," because along it the bending forces are neutralized, and the shearing force alone is acting. This flexure calls out a reaction of the fibres, and it will continue till the elastic forces of the body, by their reaction, come into equilibrium with the weight or other force. The fibres of a beam are lengthened or shortened in proportion to their distance from the neutral axis. Let vl - this distance for any extended fibre, and r - radius of curvature for that part Fig. 6. of the body. A Fig. 6 is an enlarged section of a,ortion of a beam. A B is the neutral axis. Thenbysimilar triangleswe have r 1: r:: " 1: vl and the elongation per unit of length,' expressed by i, - v Ty r ENGINEERING STATICS.. 41 The force P necessary to produce the elongation= Ei= E x vL, and for an elementary section d a of a body of secr tion a, it is E vL d a. This expression is the " elastic r reaction." For any compressed fibre we have in like manner, calling vus its distance from the neutral axis, its elastic reaction equals E v- d a. r The first condition of equilibrium is, that the sum of the reactions of all the extended fibres should equal that of all the compressed fibres, i. e., J Ev da a Ev2da [1] From the above we getf v d a =f d a, which shows that the neutral axis passes through the centre of gravity of the cross-section of the body. The "moment" of each fiber is the product of its elastic reaction, by its leverage, i. e., it is = E v1 d a x vl, r or vgcda x Vt. Then the moment of all the fibres equals the integral of the above = d a + E- va' d a, or calling the distance from the neutral axis v, positive or negative, according as the fibres are compressed or extended, then the expression becomes f va = 0. r 42 EKG lE-UN STATIC. This last expression is called the "moment of easticity. To find an expression for r, the radius of curvature, the equation of a curve being y =j(x), its radius of curvature r = + + dy In the case under consideration, the flexure or bending is very slight, so that-dx (which is the tangent of the slope) may be neglected when compared with unity, and we get r; and the moment of elasticity becomes d2y Ed y xJ v2 da+ a d'- xvs'd a.[2.] d x2 XC? When the piece is prismatic or cylindrical throughout its length, the above integral is constant for every section, and is its moment of inertia. Call this I, and the above moment of elasticity becomes E I d; y The second condition of equilibrium is, that the moment of elasticity of the fibres- shall equal the moment of the external forces which tend to bend the piece. This latter moment will be designated by M. BNORNG BERING STATIO8. 48 FLxmuRE aoF PIrsImATIc BwFxs When.fixed at one end. 37. When fixed at one end and loaded at the other, as Fig. 7. in Fig. 7. XA B represents the middle fibre of the beam, fixed at A, and havq ing a force at the other end, B, applied perpendicularly to its length. Its length is 1, and is supposed to be sensibly the same before and after being bent. The shearing force is constant for all points of the beam and equals W. - The bending moment of the force at any point, distant from the fixed end x units, or x' from the other, = V ( - x), = W (x'). To find the deflection y at any point distant x fom the, origin: equate the moment of elasticity of the beam, and the moment of the bending forces; i. e., E.I d Y W (- x). By integrating twice we get E I (y) = W Z )t Whence( 6 a At the end of the beam y becomes d and x becomes 1, and we have = WI3 E38] NoaE.-The value of E directly obtained by experiment on extension, and as given in the tables, is smaller than its proper value, as found from direct experitmnta on bending, owinig to tihe iameased resistance which the'cohesion of adjacent layers of fibres oppose to sliding on one another. Hence the former value of E must be modified accordingly before being used in this and the following formulas. 44- qENGINEG STATICS. When the beam is fied at one end and loaded uniform/y along its legth. Fig. 8. Fig. 8. Let w= load per unit of length, so that w 1 = whole load W. The moment of this load at any point, x from the origin is w (1 - x) i (1- x) = w (1- x) = W(X')'. Equating and integrating as in the preceeding case we get (~wQ l Q I X' + J $4) anWd 14 y~w = i0 ~ ~u - g/[E43 w Ix EM E1 8E El - - Comparing this expression for 6 with equation [3], we see that the deflection is three-eighths what it was, in the former case. When there is a uniform load acting downward as before, and also an upwardforce, P, at the end. Let xi = 1.- x. The bending moment.M w wx ( z1) -P x1 = x w e P X = wXI (XI- 2 P) 2P When xi is less than 2P, M is negative, and the curve of the beam is concave upwards.. When x1 is greater than P,'M is positive, and the curve has an infexion at the point where x = —. WI ENGEIEERING SlATIC. 45. beam upported at both ends. Fig. 9. Figs. 9 and 10. 38. Let it be loaded in the middle with -7:aW. The pressure on A and B is half W,... and the reaction of each is the same; each half of the beam is therefore in the same t~ A mmA condition as a beam fixed at one end and r loaded at the other with half W. Then for any point, x from the centre,.the moment of the bendig force =2 (2 - x). Equating as before and integrating from 0 to iI we get 6- Wi [5] 48 E [] That is,:the deflection is- one-sixteenth of that of a beam of the same length, and, having the same weight applied at its extremity. The same beam with a uniform load, Fig. 10, would give, putting wl in place of W; 6= -x. WI (1) [6.] That is, five-eighths of what it was with an equal load at the middle. This beam, compared with a beam fixed at one end and loaded uniformly, gives the ratio': I x. or 1 l x T =A 39. A beam fixed at one end and supported at the other. Fig. 11. With a load Win the middle, Fig. 1i,'the deflection at the lowest point is,.of the deflection of the same beam merely supported atboth ends. The lowest point, or point of greatest deflection, is at fiveninths of the -length from the fixed end, and the point of inflection, or reversed cur 46 ENGINBING STATICS. vature is about one-tenth the length from the fixed end. When such a beam is loaded uniformly, Fig. 12, its greatest deflection is one-third of that of the Fig. 12. same beam, or similar one of the same length, supported at both ends and loaded in the middle; or eight-fifteenths of that when loaded uniformly. The point of greatest deflection is at five-eighths the length from the fixed end, and the point of reversed curvature at oue-fourth the length from the fixed end. Fig. 13. 40. When the beam is fixed at both ends. _~: E;:When such a beam is loaded in the middle, Fig. 13, its deflection is onefourth of that of a similar beam merely supported at both ends. The point of reversed curvature is at one-quarter I from the end. If the beam i8s oaed uniformly, Fig. 14, its deflection is only one-fifth as much as when support- Fig. 14. ed at both ends and loaded uniformly, or one-eighth as much as when supported at both ends and loaded in the middle. The point of reversed curvature is sri I from the end. When a beam extends over several supports the outer portions are in the condition of Art. 39, and the inner portions is that of Art. 40. Such is the case of a bridge of several spans which are connected ENGIINEERIN STATICS. 47 together over the piers. Then the middle portions may be made longer than the others. 41. A beam supported at several points. When a beam is uniformly loaded and is supported at several equi-distant points at the same level, different points of the beam are differently strained, being in the conditions given in Arts. 39 and 40, and the pressure upon the supports is not equally distributed. Thus if the uniformly loaded Fig. 15. beam in Fig. 15, were cut in two at the C. X point, C, that post would support onehalf the lqad, and A and B each onequarter; but if the beam be continuous C would support five-eighths of W, and A and B each three-sixteenths. The same takes place in the tie beam of a roof truss. Similar results occur with any number of supports. Thus with two intermediate supports, the two middle posts each support eleven-thirtieths W, and the end ones each four-thirtieths W. The supports may be made proportionally strong, or the pressure on them may be equalized, by a slight change in their relative weights. FLEXUiRE OF REOrANGULAb BEms. 42. The general formula for prismatic beams for a beam supported at one end and loaded at the other, was 3 El [3.] Let b = breadth of a rectangular beam and h - height b h. or depth. Then I = -. (Jackson's Mechanics, par 48 ENGINEERING STATICS. 49.) Substituting this in [3] we get 6 -= 4W (7). Ebh"7 This shows that the deflection is directly as the weight, and as the cube of the length, and inversely as the breadth, and the cube of the height or depth. The ratios of deflection in other conditions of beams are the same as those of prismatic beams in general. The most convenient standard of comparison is the important case of beams supported at both ends, and WI, loaded in the middle, for which we have 6 E [8.] All the above formulas suppose the same unit of dimension to be used. It is, however, most convenient in practice to take the length in feet, and the breadth and height in inches, and to obtain the deflection in inches. The above formula [8] may then be written 6 = ha eb h'' In which L - length in feet, and e-E 432. Since 6 IT W (12 L)' WL) W [L9' 4 Ebh - 4 Eb h E — 432bh' — ebh3' Taking formula [9] for the standard, the deflections for the other cases are found by the ratios given in the case of prismatic beams in general. (See Art. 72) FIEXURE OF PRISMATIC BE.AMS OF OTHER FORMs. 43. To determine this, substitute in the general expression of the moment of flexure, = - Wi [3], the proper, 3 El value of I for the particular cross-section in question. The following are the most' useful. IENINEEIGNG STATICS. 4 For a hollow beam of a uniform rectangular cross-section, the moment of inertia, I equals b h- b'; in which 12 b' and h' are the inside breadth and height. Fig. 16. For an I shaped beam with dimension's AT, as in the Fig. 16, the value of I is the same, and the preceeding formulas apply without change, and as before I = b h - bl h13 wtlithe 12,:.. A The two are therefore equally st wfith the same amount of-material For a square beam with its diagonals vertical and horizontal, I =' c4, c being one of the sides of cross-section. This is the same as that for a square beam with its side horizontal and vertical: hence each beam has the same resistance to flexure. (N. B. It is not so for the resistance to rupture.) For a cylindrical beam of radius r, I = i Xr r'. For a square beam whose section circumscribes the section of a circular beam I- 1 (2 r)4, therefore its resistance to flexure is to that of the circular beam:: ~1 (2 r)4,: i r:: 0.59. 44. Relative deflection of similar beams, or those having an equal cross-section. Figs. 17 and 18. In Fig. 17, I =, [b (nb)' - bl (n b1)]. =, (n' bM - n' b, 4). In Fig. 18, 1-=;a bn (n' bat), Noi" n No w. n' bin'. Now n bns t r bs' - nj;. therefore bel/a Jb2. -.bl'2; therefore = 7T n' (b' - b-. )'. Then I: P:: b'-:14': (:b2-62)2; I: I':: b+bi6:p 6'-. (The letters on Figs.:17 and 18 are reversed.) 7 50 ENGINEERG STATICS. GENERAI PRINCIPLES OF RTPTURE. 45. The resistance to rupture or breaking, is examined in the same way as that to flexure or bending. Let R = resistance or strain, just before giving way, of the fibre most extended or most compressed. Call the distance from neutral axis v; and let a = area of the cross-section of the body in question. Then its resistance = R v d a, and the resistance of any other fibre at a distance'l from the neutral axis = R x vl d a. Its moment - R x 2 da. The total moment of the resistance of all thefibres equals the integral of the of the resistance of all the fibres equals the integral of the above from 0 to v on both sides of the neutral axis fv v2 Ld a.,[10.] Equating this in with the moment of the bending force, tending to produce rupture, we can determine the proper dimensons, &c. RUPrURE OF PRISETIC BEs. 46. For these the total moment of resistance to rupturebecomes = x L. [11.] Beanmsfixed at one end. 47. When loaded at the other end with W, the shearing or vertical force is everywhere W. (A force acting perpendicularly to the beam in any position may be substituted for -W.) The bending or horizontal moment at any distance x, from fixed end, length being 1, is W (l-x), and at fixed end is therefore W I. INEDERENG STATE8. 5l The equation then is W R I.= [12.] V This is the dangerous section of the beam. From this equation we can determine W, the breqking weight of any given beam; also I, which gives the size of the beam to resist any given stress WV, also, R, or the coefficient of strength, having found W by direct experiment. When the force is applied at any intermediate point of the beam, formula [12] applies, calling 1 the distance to the fixed end. 48. When the beam is uniformly loaded with w 1. Then the shearing force at a distance x from fixed end is w (! - X), and at the fixed end = w 1. The bending moment at x is w (- x)' (1 — X) w (- x)2, and at the' fixed end (the "dangerous section") is I w 12: R I * [13.] V That is, the -stress is one-half as much as before. The weight of the beam itself acts as a uniform load. BEAMS SUPPORTED AT BOTH ENDS. 49. When loaded in the middle with W. Then the shearing stress everywhere is j W.- Calling x the distance from the middle the bending moment at that point = W ( 1 - x). Calling xl the distance from the nearest end of the beam this becomes. W x1. It is therefore greatest when x -0; i. e, at the middle of the beam where it is R w 4W l_ R *[14]. Hence it will require four times the weight to break it. 62 ENGINEERING STATICS. At the end of the beam it is equal to 0. The shearing force being everywhere Wj may be represented by the perpendiculars of a rectangle whose length equals 1, and whose height -- W. The bending moment may be represented by the perpendiculars of an isosceles triangle whose base equals 1 and whose altitude equals the bending force in the middle of the beam. 50. When a beam, supported at both ends, is loaded, uniformly with w 1. Then the shearing force at any point x firom the middle = w x. It is therefore 0 in the middle and greatest at the ends of the beam, where it becomes ~ w i. It may therefore be represented by the perpendiculars Figs. 19 and 20. as in Fig. 19. The bending moment v at any point x from the middle -= - w. (- 1 + x) ( 1 - x). Calling xl distance. —- -------- from nearest end this becomes I w xl (I a Poof.-Fig. 20. Reaction of A - w 1 upwards, with leverage xl; pressure downwards of weight from A to Mi = Iw xl with leverage equal to I x1. Resulting moment around M = ~ w 1 xl - w X12 = 1 W x1 (1 - xi). It may therefore be expressed by the perpendiculars to the double ordinate, terminated by a parabola. Fig. 21. The bending moment is greatest at the middle - w l2 - W12 R I1 T Wo [15.]a t V To prove that the curve is a parabola, Fig. 22, y = ENGINEEERING STATIOS. 53 wlP- w (3 I+x) (I I -X)=I W -t W Figs. 21 and 22. l2 + X w x = -wx';or changing the letters x = I w y2, the equation of the parbola. These principles apply to trussed > bridges, which are framed beams, and show that those parts of them which have to resist a vertical stress, as do the posts and braces, should increase in size from the middle of the bridge to its ends in the uniform ratio shown in Fig. 19, and that those parts which have to resist the horizontal bending, as the top and bottom beams or chords, should decrease in size-from the middle to the ends in the ratio shown by the perpendiculars in Fig. 21. 51. The lines of principal stress in beams uniformly loaded,/are curved lines, such that the tangents to them at any points indicate the directions at these points of the lines of stress. Thay all intersect each other at right angles, and make angles of 450 with the neutral axis, on which line or plane, therefore, forces are acting in those directions. The lines convex upwards are lines of thrust, those convex downwards are lines of tension. The stress along each of these lines is greatest where it is horizontal, and at the end of each the stress is 0. To show this experimentally, draw a number of small circles' on the side of a beam before it is loaded. Then when it is loaded the circles will become ellipses. Those near the top of the beam will have their long axes vertical, and their short ones horizontal; those near the bottom of thie beam, vince ersa; and along the neutral axis these axes 54 3NGINEEIaNG aTATIOS. will make angles of 450 with the horizon. At interme. diate points their figures will be intermediate in form. 52. A beam loaded with Wat distances 1 and 12 from the ends, 4 being the smaller. The shearing force between W and the nearest end is-l x W, on the other side of W, it is x W. The corresponding bending moments at any point distant x from the centre of the beam, are i x) W, and ( 1 + ) W xmeasuredtothe left of the middle is considered positive; to the right negative. The: greatest moment is where W is applied, where it is. 4,4 x W [16.] In words this bending moment = W multiplied by the product of the two parts of the beam, divided by its length. Calling xt the distance from the middle at which W is applied, the shearing force for points between W and nearest and fartherest ends of beam are respectively: and I-L x W.. The corresponding bending moments at distance xl from the middle, for points between W and nearest end are (xil W, and on the other side of W, (1 - () 1 + x) 1.. )t Rx is positive or negave according BENG*EBING. 8TTAU. 55 as it is to the left or right of the middle. The greatest bending moment is ( I-x) (1 + x) W-RI [17 I,- [17. * I v In the first half of the above formulas, the point of application of W, is measured from the ends of the beam, and in the latter half from the middle. In the former Z4 and a are.used; in the latter these distances equal IZ$1, and - l + Il. 53.:With two forcex span1 w 2V ~B C~ rise cot. A. If a weight, W, be uniformly distributed, as is that of the rafters themselves, its effect is equal to that of 2 W applied at B, and in that case t - W x s legth rise, and h - j W x span. Wnow means the entire uniformly rise 88 NOGINN3GEERIG STAfBn. distributed load. Otherwise; the horizontal thrust of one beam will be the same as in Fig. 31, that is, the horizontal thrust of A B its wt.x A. We might suppose BC since the horizontal thrust of A B is the same as that of A B, that the horizontal thrusts of the two beams would equal twice the above expression; but such is not the case, for either horizontal force merely corresponds to the reaction of the wall in Fig. 31, consequently, the thrust of both rafters = 4 weight of both x -span - X W x rise span rlse The inclination which produces the least horizontal presure is where the angle A or the pitch = 350 15', or where the height: i span 1: a/2, or where the height is a little more than one-third span.; 1! 1.414. Pressure of Water on Lock Gates. 90. A canal lock gate is a frame of two struts. It is anaFig. 35. lyzed thus, Fig. 35. Let the whole presF 4 —t s sure on one gate, of the water acting perpendicularly to it, = P. This is resisted by 2 P, acting at B and 2 at A. DecomA D- -Ad pose I P, acting at B, into a pressure h parallel to A D and another t acting along B A. Then P;h::EF: BFor::B D':BA. Hence h= i BA BD P P P X i P = P Then again lB ED B.A sin. A 2 sin. A. wehave,i P:t::EF:BE::BD:D A. Hence,t ENGXINERG STATICS 89 pD4 A = P. cot. A — P DB P. cot. A = 2 tan. A91. Braced struts. The horizontal thrust of a pair of rafters may be resisted by walls, or they may be relieved of it by a tie-beam connecting their feet, as in Fig. 36. Fig. 36, When the tie beam is long its middle is supported by a rod or stick, called a " ing post." The only strain on it is one-half the weight of the tie-beam. More precisely five-eighths of weight of ltie-beam when this is level. (See notes on flexion.) Influence of change of temperature on iron tie-beams, or tie-rods, of roofs, &c. Let t~ — temperature when the frame was constructed, and 6P0 - temperature at a subsequent period. Then to - t~== fall of temperature. The shortening (of iron)- THT= A x length x (t~ - tol). Now an elongation of,0008 corresponds to a strain of 20,000 pounds per square inch. Then the additional strain caused by cooling will be, 7TT x (to - t) x.-~ =- 175 (t~- t). Thus a change of 1000 would cause an extra strain of 17,500 pounds per square inch. Sometimes the tie-beam is replaced by a "collar beam," as indicated by the dotted lines in Fig. 36. In a roof of this kind, the strain on the "collar-beam," is greater than that on a tie-beam in proportion to its elevation. If midway up it will strain twice as much, &c. This must be so to satisfy the condition of moments. 12 -90 ENGInEMING STATICS. Fig. 37. Sometimes the tie beam is replaced by two oblique ties as in Fig. 37. MoE/1\,F ment of weight- 2 W x I A N. MoA ment of tie A E- its strain x C D. N —— "~ *Hence for equilibrium, - W x 2 A N strain on tie x CD. Hence strain on this tie must equal ~ IV A N i W x A or I that if there be two ties. GD GD Fig. 38. In Fig. 38, the tie beam is formed B \ *' by two pieces, inclined upward and 3D \meeting in the middle, and supported B by a vertical tie rod. Then this last c O 27~ F}AX rod is subjected to a stress from the s,D\ weight of the rafters, and the load on Fi /'F them. Then the weight W on the vertex B tension of B D- W CD BD If C D = B D, the tension = W. If CD = 2BD,the stress = 2 W, &c. Where C D = 0, the tension on the rod produced by W- 0. The stress on each part of the tie rod AD or A' D-'2 W B D A uniformly distriblike that of rafters and roof, produces this uted load, like that of rafters and roof, produces this stress. To determine the stress on the parts of this frame. Let G D represent the stress on AD in Fig. 38. Decompose it into G F and D F. The horizontal component is balanced by thatbelonging to A' D. The vertical component is D F. The vertical stress: horizontal stress:: D F: G F or:: D C: A C. But the horizontal stress ENGnEERING: STATICS. 91 AAl BC AwAC -4- W x XBD W B Substituting this into the above proportion, the vertical stress =- W A xC CD =_2 WOD. The other tie' D produces B D AC D an equal stress on B D. The total vertical stress on B D = W CD Stress on A D: 2 vertical stress at D:: A D BD D C, or stress on A D: W D C::1 D: DC. Hence B D stress on A D = 2 W B-D Stress on A B = horizontal AB A C AB AB stress at A x W x X AB = 1 W AC BD A C B.D. A uniformly distributed load produces one-half this stress. Second method. Let W = weight at B, and t = tension of A D. Draw B E perpendicular to A D produced. Equating moments, t x B E- - W x A C.. t = W AC 14 C AC B E x BD sin. BDE = B D sin. A D C AC AD = W W. A C. B D - W W Tension of B D =C D AD CD CD -2t D2 x W x x D x WD. AADAB D BD Third nmethod. Construct a diagram of the forces by drawing from some point o, parallels to the bars, intersected by a vertical line. The length B D and B1 D' will represent the supporting forces at A and A', and their sum equals the total load at B, which equals W. But for equilibrium the load must equal B BL. Then the stress 92 ENGINEERING STATICS. on the tie B D, must be represented by D D1. Numerically one-half this stress' I:: C D: B D. Whence this stress on B D = D. Stress on A D::: B D AD: B D. Hence stress onAD = W -D. Stress B D on A 2B W AB. Horizontal stress:stress of B D C 1D A C_::.A.ac: 2 C D. -Hence h =W BD=: -2 A C BD 92. Trussed Struts. Roofs of large span are usually subdivided into other triangular frames variously comFig, 39. bined. Fig. 39. The struts E F and B El F1 are perpendicular to the rafters V i V at their middle points. Hence A E = A E E A' EB. Let w= weight of one rafter and its load. The horizontal tension h of E E' = x 2w sean _ W The parts A E and A1 E1 besides rise 2 tan. A this have extra tension, t' from the effect of the struts F E and F' E' which also cause a tension, t", of the inclined ties B E and B E'. The vertical pressure at F = I W. Its component, R, along FE = cos. A. Components (t' = t") of this along A E and B E (they being equal in length) w co = w h. This. 2 sin..A 4 tan. T being added to the -tension h of A E and A' F' makes their total horizontal strain h' -I h + 2 h-T h. Finally ENGINEERING STATICS. 93 calling W 2 w, the tension of E E' = / W sp.an. Tenrise sion of A E and A' E' = W sp.an Tension of B E and B E'= 2 tension of E' - 1 W SPan.These strains rise could also be obtained by decomposing the strains by graphic construction. Roofs are also Fig. 40. sometimes sustained as shown in Fig. 40. The strain on B is 2 W,. on C and C'Z W, and on D and D' W. The strain on C E is ~1 Wz, and on B F W. K When the tie-beam is raised (as in PA E F' As Fig. 41), which is usual in iron roofs to give a lighter look, Fig. 41. the strain on it is increased in the I X proportion of former height to new height. The other stresses on it are A e found similarly to Fig. 38. Draw t3H IJs- a diagram by parallels to the bars. ci-/F V/,4/,Then I F I'- 2F ~ W, and the lines F G and F' G in the diagram, f~r zI j L are proportional (on the same scale) -^' e PR to the stress B E and B E' on the frame. A modification of the preceding is to add the ties shown by the dotted lines, drawn from E and E'. Various arrangements for wide iron roofs for railroad depots, are given in Morin's " Resistance des Materiaux," P1. VI. 94 FENGITEERING STATICS. A Pair of Ties. 93. This isjust the converse of Art. 88. Fig. 42. The stresses are just the same in amount but cause tension instead of compression. Fig. 42. Horizontal strain produced by W= 4 W Span, and the strain on depression each tie = I W length depression Fig. 43 shows a beam A A' supported by a strut, B C, Fig. 43. below, and by tie rods A B and A' B k e. which can be screwed to any tension. With a load W in the middle, the strain on each tie rod, to be in equilibrium with the load, must be equal to t W length of rod A B depression = B C' For a passing load, a tension sufficient to resist the action of the load in the centre would exercise the same force after the load had passed to bend the beam in a contrary direction. To make the actual flexure of the beam a minumum let each produce an effect equal to one half this, and the rod should have a tension = ~ t/ x length of rod. The beam will then require to have only depression one-half the strength to resist flexure which would otherwise be necessary. A strut and a tie. 94. Such a combination in its simplest form is called a "bracket." This is a frame consisting of two pieces, a ENGoNERING STATICS. 95 strut and a tie, supporting a load; the two points of support of the strut and the tie being on the same side of the load. In all the forms the upper piece is a tie and the lower a strut. We are to calculate the strain on each. The beams may slope in a contrary direction, or in the same direction. The investigation applies to all. Fig. 44. Let A D, Fig. 44, represent the units _-M in weight at a scale of one inch for any convenient number of pounds. On A D as a diagonal construct the paralleloF gramAFDE. Then willAErepre. sent the thrust on the strut A C, and AF will represent the strain on the tie A B. We then have A F: A D:: A B:B. Hence, A F= sin. A B W cosA. 4 C G A D sw A And.AE: A D::.A C sin. B A C- sin. BA CA s C s sin. AinB:BC. Hence, AE= = in. AB = BC sin. BA C cos.. B H a.In words this becomes. The strain on sin.B A C' either piece equals W x cos. of the angle which the other piece makes with the horizon, divided by the sine of the angle which they make with each other. When one of the beams becomes horizontal the formulas may be simplified. When the tie is horizontal, B Al C = A C G, and A B H = 0, and the above formulas becomes AF = W W x cot. BA = W sin.B A C A B cos. 0 A C A and A E= W COS0 - wA C This also will B C' sin. B A C BC 96;i~i ENGIN~e31PWEERING STATIOB. follow directly from the triangle A B C, whose sides will then be.parallel to the directions of forces exerted. When the strut is horizontal, A B H = B A C, and A C G 0. Then the above. formulas become; A F W s. B 0 sin. B A C t.. A'~ ". cos. BA'C _=, W' a=W,and A E = W.s BA= sin. B AC0 BC' sin. B A C A aC In this form the strain is greater on the tie than on:B C' the strut. The expressions for A F and A E being interchanged: The preceding form is therefore preferable, because any defect in the wood lessens its power of extension more than compression. 95. Modification of brackets. A beam fixed at one end Fig. 45. and supported between its ends. Fig. 45. A...._.. sE The weight acts in the line D G. The E; AyceThL strut acts in the line C A, meeting D G produced in E. The force at B must act in the line B E, since three forces in equilibrium must meet in a common --.~ __ _ i point. Since the weight acts parallel to B C the three forces in equilibrium are parallel to the three sides of the triangle B C E, and proportional to them. B C will represent the units in the weight, C E the units of pressure on the strut, and iB Ethe force at B in the direction B E. Pressure on the grt at C in the direction A C =. The force B E at B in the directions B E W Required;, the korCWB ENGNEERING STATICS. 97 izontal and vertical components of these strains. The force C E may be resolved into C F and F E and the force B E into B F and F E; hence we see that the horizontal thrust at C and B are equal but in opposite directions. The vertical forces acting at C and B are as C Fand B F; o F acting downwards- and B F upwards. Their difference, C B, represents the weight. There are many forms of brackets, depending on the directions of the strut and Fig. 46. tie, and their angle with each other; but the above formulas apply to all, by making the proper modifications for the dif0 _ _ a e mferent conditions. E/Ij1, When the beam is horizontal as in Fig. 46, the strains are as the sides of the triangle B C E. The pressure at W - W B but BC WB B C' BE AEF CDa Then the pressure at B= W CD__x BA B A Trigonometrically we have B cosec. B A C. Then the pressure at B =W Xa cosec. B A C- W CA CA sin. BAO' To find the horizontal pressure at C, resolve the pressure at B into BC and C A and we get, horizontal pressure = pressure at 13 98 ENGINEERING STATICS. BCAWCD BA X CA CD AB CW xA BC BA- = C' Otherwise. The weight W, acting at D, is equivalent to CD C D acting at A. Let B C equal this. Then the other strains are proportional to BA and C A, and we have; CD BA strain onBA.- W Strain on A C ==- W CA XB C CD C A CD C A B C B W When the tie is horizontal, the same formulas apply, interchanging the letters B and C. BD CA Stress on t A W x X andhorizontalstress at BD =WB C Fig. 47. Another case of a strut and a tie is shown in Fig. 47. The triangle A C E has its sides parallel to the direction of the stresses, and hence we find the piece A C is compressed with a force equal to cos. C B D sin. (CAD-C-BD)' cos. CA D The tension of the rope B CY= - o D) sin.(CA D —CB D.) The horizontal strain tending to make the piece A C slide wcos'CB D cos. CAD sin.(C a D - C B D)' If -the weight was not merely suspended at C, but supported by a cord pass ENGINEERING STATICS. 99 ing through one or more pulleys (as in the case of a crane), we should have to consider instead of W, the resultant of W and of the tension of the cord. Two struts and a strlaining beam. 96. With loads W and W' on C and C' the stresses Fig. 48. are analagous to Fig. 34. Consider the upward reaction at A and A' which are each equal to V. Then.A.//..... St the triangle A B C gives stress on AC AB A Ca= W -an horizontal stress A' B' All the stresses are the same as if the struts, A C and A' C', were produced upward to meet, and the whole load (W+ W') placed at that point, as in Fig. 34. Suppose the load on A A to be uniform, and supported. by rods B C and B' C', A A' not to be continuous but to be divided at B and B'. This corresponds to the weight on C C' in Fig. 48. If the beam A A' be continuous and. level, then I — W1 is on B C and B' C, and 4-~ on A and A'. This is safest to take, being greatest, though it is not generally done. In either case calling IT' the load on B Cor B' C', then the horizontal thrust on C C and the pull on A A' = Wj A B A 4' span span A GC W BCrise rise The 100 ENGNBE~RtING STATICS. thrust on C A and C'' = W' If a similar load B C were on top, the stresses would be the same. When a bridge of this form is reversed the stresses remain the same except that the former stresses of compression have become extension, and vice versa. This arrangement may be extended to any number of panels. It is preferable for materials like wood and wrought iron, because the shortest pieces are exposed to compression. Multiple Frames. 97. Suppose a uniform load W on a beam A A', Fig. Fig. 49, 49. Each post or vertical D E E tie supports 4 W = WI. The struts E B and E B'...A B ~ B: A' resist a stress _ 2 W' BE I E' 2as found from Fig. 32. They produce a horizontal C E strain at E and onBB' = B WV'B. The weight on CE Dor D' = Wt + -2 W = 3 W' ( W'being transferred from E.) Consequently D A and D' A' have a stress3W'4 D They produce a horizontal strain on D D' -:BD' and A A'- W' B The horizontal strain B B' = the sum of the two horizontal strains - (a W' + a W') A span W= I W span_ W spcan rise rise rise ENo.GNm'ERlInG. 1sTAT08. 18 So proceed for any number of panels. It will be found that the strains on the posts and on the struts, increase in a direct ratio to the distance from the centre. Their strength and size should therefore. be increased in the same ratio. The strains on the top and bottom beams increase from the ends to the middle, but; not in a direct ratio. The increase is most rapid as you proceed from each end and becomes less rapid on approaching the ceatre or middle. It is analogous to that of a solid beam, in which latter case the relative increase is indicated graphically by a parabolic curve (see Fig. 21.) The usual formula for the horizontal stress on a- frame, span caused by a uniform load ( W ) supposes the weight to be uniformly applied at the ends of the struts, as well as uniformly over the road way. This iS the case in frames which have an even number of panels; but is not so with those of an uneven number. For example with three panels, the horizontal stress -- W span; for rlse five panels -1 W pan, for seven panels.P 81 rise 81 rise' and generally forn panels (n being uneven) = -') span - rise 10o2 GINGRlING STATICS. In the preceding forms the ties were vertical and the Fig. 50, struts inclined. In Fig. Z50, both the ties and struts are inclined. The L,,k \/ V \stresses, however, follow similar laws. With a uniform load W, such as its own weight, the vertical strain increases uniformly from the middle, where it equals zero, towards the end where it equals i W. At x' from end, or x" from middle, it equals W ~'p'. The strain on any diagonal whose middle is x"'span from the middle of the bridge xi/ length of diagonal span depth of panel The horizontal strain at the centre =. W span. At any point x" from middle or x' from end, it W x' (span - x") 2 span x depth' diminishing from the centre to the ends in ratio before shown by a parabolic curve. When the loads are applied along top or bottom, or along both. The distance x' and x", in the preceding formulas, are measured to the tops of the diagonals when the load is attached to the bottom of the beam, to the lower ends of it if it be on top, and to their middle if the load be equally on the top and bottom, as its own weight. Bridges of this form are called "Triangular Girders," or " Half Lattice," or "Warren's," or " Neville's." If the ENGINEERING STATICS. 103 number of the oblique pieces be doubled, then each sustains half the above strains. This is a lattice bridge. Any number of such pieces may be combined. 98. Best angle for the diagonals in a triangular girder, when all make the same angle with the horizon. Their total length is less in proportion to the smallness of their angles with the horizon. The strain on them is greater in the same ratio, and this increases their necessary crosssection. Then, there will be some one angle which will require the least amount of material and, consequently weight and cost. Let the angle B A C = a, in Fig. 51, be this angle. Call Fig. 51. the weight B C = 1. Then the length A B =- B C. cosec. a = cosec. a. The i/1CJ A strain, and consequently the cross-section, is also as cosec. a. Combining these, we have, the material in one varies as cosec.2 a. The reach, or horizontal pressure, of each = cot. a. Then their number (calling length of girder 1= ) is Then the total material is as cot. a cosec.2a x I 1 cos.a This cot. a sin.2 a sin. a sin. a cos. a is to be a minumum, or sin. a x cos. a, a maximum. It is so when a = 450~. Other practical considerations render it desirable to make the angle greater. It is usually between 450 and 600. 104 NERXING. STATICS. Best angle when the diagonals are alterinzteltyi:: and oblique. In Fig. 52. Let B A C = a. Calling the depth of the Fig. 52. bridge, or the length of the vertical beam, B C= 1, the length of the oblique beam, A A B = cosec. a. The stress on the oblique beam (calling that on the vertical = 1), is also as cosec. a. Their weight, &c., is, consequently, as cosec.2 a. The number of each kind of braces (calling the length; of bridge 1) = cot. a' Then the weight of all the vertical braces are as cot. a and those of the oblique braces as cot. a cosec. a x Io. Hence the weight of all'are as t (1 + cosec.' a] This is to be a minimum. It is so when a = 350 16'. We have been assuming that the weight, length, crosssection, material, &c., of each unit of length of the vertical brace was capable of conveying the same strength as a unit of the oblique brace. If not, let the strength of the former be to that of the latter:: n: 1. Then in the preceding expression the weights, &c., will not be as 1 cosec." a, but as n cosec.2 a. Consequently the expression to be minimum is (n + cosec.'a). This is a minimum when cot. a = /(1 + n.) When n = 1, as in our first hypothesis, cot.: a = V 2 = 1.414, and hence, a = 350 16' as before. When n = ~, as might be for ENGINEERING STATICS 1Q05 wood aqi4wrought iron where the oblique braces are compressed, then a = 50~ 46'. When n = 2, as might be for cast iron, a = 60~. Compound frames. 100. In this combination, shown in Fig. 53, the stress Fig. 53. on each pair of struts may be decomB B/ posed, as in Fig. 33, or they may be found by one of the methods of Art. 89. If the frame be inverted the stresses A g are the same, but the parts before compressed are now extended, and vice versa. In the frame, shown in Fig. 54, composed of four distinct trusses, the thrust is Fig. 54. found as before. The B B strain on the tie beam A A', is the sum of that due to each set of struts. The A Al strain on the straining beam, B B', increases from the ends to the middle, the outer portions receiving the stress of the outer pair of struts, the next portion the additional stress of the next pair, and so on. Latrobe's Bridge is on this plan. In such a frame as shown in Fig. 55 (sometimes used Fig. 55, for a turning bridge) the platform -A A', being supported by oblique iron rods, the strain or pull on each rod = load at lower end x length + by the height of its upper end above the lower. The sA tress, or thrust on the lower beam, A A', 14 106 ENGINEERMING STAIaCS. increases from the ends to the middle. Between the two outside tie rods it- equals the load at A x distance from centre + height of tops of outer rods. So for each remaining rod. Polygonal Frames. 101. Such frames, even when of only four sides, are A Fig. 56. F liable to change their angles from the ~,. / action of a slight force. Let the force W act in the direction E B, Fig. 56, and the 0 | / corresponding reaction act in A F. Let M- the frame be stiffened by the brace C D across one of its angles. Required the strain on the brace. The force W tends to rack the frame, that is, tends to turn it around A. Its moment therefore - W x A M. Let P -- force with which the brace resists, and let A N be the perpendicular to the brace. Then the moment of brace - P x A N. Hence, for equilibrium, P x AN= lW x AMorP= WA'M From this we A NV see that the nearer the brace is to A, the more it is strained. Every frame having more than three sides, should be braced so as to form triangles. In a frame like a gate, where the brace is a strut, it should be placed in that diagonal of the rectangle which the weight tends to shorten. If it is a tie it should be the converse. ENGINEERING STATICS. 107 Fig. 57. The form in Fig. 57 is called a "curbed," or "Mansard," or "hipped" roof. Practically to find the best angles for such a roof. Prepare four sticks, proportional to the intended length of the rafters, and fastened by pins; attach their extremities to a vertical board, so that when inverted they may hang freely. The position which they then assume gives their proper form in a roof. If the rafters are to be loaded unequally, weightsproportional to their respective loads, should be suspended from the centres of gravity of the pieces. In such a frame, in which the lower pieces made an angle of 30 with the vertical, and the upper made an angle of from 450 to 630 with the vertical, the horizontal thrust of the former was from.197 to.227 of weight, or about 5, the weight being uniformly distributed along the length of the two upper pieces. Geometricaly. Where a polygonal frame is in equilibrium the horizontal thrust is the same at each angle. Let frame be loaded at each angle. The strains on each beam, in the direction of its length, are proportional to lines drawn from any point, 0, parallel to those directions, and limited by a vertical line, as shown in Fig. 57. The portions of the vertical line cut off by these diverging lines, are proportional to the weights suspended from the joints of the beams parallel to those lines. 0 O' is proportional to the horizontal stress. 108 ENGINEWENG - STATIS.Trigonometricaly. The strain on any beam in the direction of its length = horizontal strain x secant of its angle with the horizon. The weight suspended from any angle = horizontal strain x difference of thetangents of the angles which the two beams, meeting at that point, make with the horizon. NoTE.-The general principles of inflexible frames are these: 1st. Strength and economy are the two objects sought. 2d. Give little excess of strength to the parts which themselves add to the load to be supported. 3d. Endeavor so to arrange the beams that the strain shall be transmitted as nearly as possible through their axes. 4th. Avoid as much as possible transverse strains. 5th. Load obtuse angles lightly. 6th. Divide polygonal frames into triangles by means of braces. 7th. Distinguish carefully between struts and ties and construct them accordingly. 8th. Give all the parts of the frame equal relative strength, since the strength of the frame can never exceed the strength of the weakest part, and partial strength produces general weakness. CURVED FRA&eS. Convex upwards, or Linear arches, 102. Conceive a polygonal frame to have the number of its sides indefinitely increased and their lengths indefinitely diminished. It then becomes a linear arch. A diagram of forces may by constructed for it, as for a polygonal frame. The various lines drawn from some pointO, as in Fig. 57, will then represent the thrusts at those parts of the arch to whose tangents they are respectively parallel. If i be the angle of one of these tangents the thrust there equals the horizontal thrust at the top, multiplied by sec. i. If the arch sustains a vertical load, uniformly distributed along its length, its form when in equilibrium would ENGINEERING STATICS. 109 be a" catenary." If the arch sustains a vertical load distributed horizontally, its curve when in equilibrium would be a parabola. If the arch sustains a uniform normal pressure, like that of a fluid, its form would be circular. If the arch sustains a pressure not uniform, but symmetrical on opposite sides, its form would be elliptical and the ratio of the axes should be the square root of the two pressures. Thus, for a tunnel arch very deep under ground we should use an ellipse in which the horizontal semi-axis, divided by the vertical semi-axis = the square root of the horizontal pressure of earth, divided by the square of the vertical pressure of earth. An arch sustaining a pressure every where proportional to the depth of the point in question below a horizontal plain (such is the pressure of standing water), should have a curve such that the radius of curvature at the point is inversely proportional to the depth. This curve somewhat resembles a cycloid. It has been called a " hydrostatic arch." An approximate form for an hydrostatic arch is a semi-ellipse of the same height, and having its greatest and least radii of curvature, in the ratio of the greatest and least depth of load. Let xl = depth of load at the crown of such an arch. Let xll = depth of load at the spring. Then approximately its half span = (X11 - 1)!'X. An arch sustaining a pressure of earth at any depth should have a form named the " Geostatic arch." _. Its equation is very complicated. (For a full investigation of these last two curves, see Rankin's Applied Mechanics, pp. 190-208.) 110 ENGNEENG SrATsMS.:Experiments on wooden arches formed of bent planks, so that they could approximately be considered linear arches, gave the following results, the arches being semicircular: When the weight is uniformly distributed along the circumference, the horizontal thrust = about onesixth weight. This rule also gives their thrust by their own weight. When the weight is uniformly distributed horizontally, the thrust = about two-ninths weight. When the weight is all at the summit of the arch the horizontal thrust = about one-third weight. When the weight rests vertically above a point one-fourth the diameter from the spring, the horizontal thrust is about twosevenths weight. If the arch be depressed or elevated the thrust varies as the ratio,one-half span: rise. Frames of straight pieees are preferable to curves for a given amount of material. Structures framed of straight pieces resist flexure twice as well as those of curved pieces. Cast iron, timber and similar arches, are usually arches only in form. Such an arch may be regarded as composed of two parts resting against each other at the crown. Let G, Fig. 58, be the centre of gravity of the semiFig. 58. arch. Its weight acts through the E A verticle drawn through G. From Ho -E_ _1~ the middle point of the crown, H, D ~ d;a L draw a horizontal lue meeting the j\ verticalthrough G in F. From' F draw a line F I to the spring at eB its middle point, and draw I M parallel to F. Set of F G = weight of semi-arch, aind complete the parallelogram F L G K. Then F K will re ENGnOtEEMUG STATIC& ll present the horizontal thrust, and F L its pressure on the abutment. By similar triangles we haveF K- = I F G FM Hence, horizontal thrust = horizontal distance from the middle of the spring to the vertical through the centre of gravity of the semi-arch, x. weight of semi-arch - vertical distance from the centre of the spring to the centre of the crown. Approximately, calling centre of gravity mid-way between spring and crown, the horizontal thrust = j- weight of whole arch x span This errs on the safe side. The rise oblique thrust at the abutment = v/(horizontal thrust' + weight 2 arch2.) 103. Concave upwards or Catenaries. Let x = depth of curve, y = i span, t = horizontal tension at the lowest point, T = tension at point of suspension, a = angle which the tangent at the point of suspension, makes with horizon, and I = length of curve. Let p = the length of a similar chain whose weight = tension at the lowest point. Given; and y to find a~. A long investigation gives log. tan. (45~ + a)= Y. The method of finding the see. a x ver. sin. a.2.3x value of a from this formula; is by trying different values for it and continually approximating until one is found which satisfies the equation. Care must be taken to subtract ten from; the log. tan. of the numerator, in order to make it of the same radius as the denominator, that is, to 112 ENGINEERING STATICS. make the equation homogenous. We also find from the same investigation the formulas: $ 1'I - -x 2=x sin. a sec.a-1 aver. sin. a ver. sm a The weight and consequently the whole length of any portion of a uniform chain will be proportional to the tangent of the inclination of the catenary at the upper end of this portion. This is the characteristic property of the catenary. The strain exerted tangentially, at any point, is proportional to the secant of the inclination at that point. Practically, the catenary does not differ sensibly from the parabola as applied to suspension bridges, since in them the platform of the bridge and the load upon it are uniformly distributed along a horizontal line, and not along the chain. The curve thus approximates nearer a parabola than a catenary, and would be an exact parabola if its own weight as well as that of the platform were distributed horizontally. The Catenary as a Parabola. Fig. 59. Let the span A B of Fig. 59, equal 2 y, the deflection = x, t horizontal tension at D, and w = weight of catenary per unit of span. Required; horizontal \VE strain at D. Moment of horizontal strain -t x. Moment of chain itself - wy x 2 -- w Y xy; 2 2 Hence, for equilibrium t x- x y, t = X wt 2 2x ENGINEERING STATICS. 113 of chain x. Span -_ wt. of chain x span Thisinvesrise rise tigation assumes the weight to be uniformly distributed horizontally, in which case the curve will be a parabola, having for its equation y'= 2 t Hence in such a parawx bola, 2 t parameter. w 2d. To find the tension at any other point. It is the resultant of the horizontal tension and of the weight between the required point and vertex. This resultant is the hypothenuse of a right angled triangle, and is numerically equal to the square root of the sum of the squares of the two forces. Let yl - ordinate of any point as D1; then the weight of the portion D D1 = w yl, making t one side of a right angled triangle, and the weight of DD1 as the other side, then tl or hypothenuse = v [t2 + (wyl)2]. The tension at the point of suspension, represented by T;- / t' + WY2( = / 2 + wyV' +1. 3d. To find the angle which the tangent to this curve makes with the horizon, i. e., C A E. When the curve is a parabola we have, tan.CA E 2 0D 2 x -2 dflecion An approximate formula for the length 4 x' 8 sq. of depth of curve is, 2 y +, or span + of depth 3y 3span. 114 ST AIN; T 4.th Depression caused by elongation, produced by stretching, as by heat. Depression in middle equals elongation x A x span ver. sin. 5th. When produced by a weight in the centre; depression=n__ weight in centre x ver. sin. 2 (wt. of chain + load) + ~ wt. in centre. Fig. 60. 6th.. Chain suspended at unequal heights, considered as a parabola. Fig. 60. +ll yll- S X _I VV + V$111 ~= s x< 1/Zn Horizontal tension -- 2 x + 2 xu + 4 1/x all Length S+ yl Note on Modes. 104. The relations between the strength of frames and small models of them, No model can have all of its parts enlarged in the same ratio and be as strong, comparatively, as before. There are three different cases. 1. Case of extension. Suppose the structure ten times the size of the model. The cross-section, and strength of a piece in the structure is increased one hundred times that in, the model, but its load has been increased one thousand times. Suppose the model could have supported TGNERDIG -STATWS 115 twice the original load upon it,;the full sized structure can bear only one-fifth the increased load. A good example of this is a suspension bridge. It will, say, with 1,000' span, bear a heavy-load, but increase its span eight or ten times and all its parts proportionally, and it will fall by its own weight. Large spiders spin threads much larger in proportion to their thickness than small ones. Suppose one nine times as thick as the other, his weight would be (9') tims that of tie smaller. His thread would be twenty-seven times as thick. 2. Caase of compression as a prop or support. A model being increased n Utimes as before, its strength is increased n' times, whilst its load is increased n' times. Nature affords an example of this kind also. Were every part of a man enlarged proportionally until his body equalled that of the elephant, his legs would break under his own weight. Hence the great proportional size of the elephant's legs. 3. Case of cross-section. For this we have the general formula W = S I-n, in which b, 1, and h, are the dimeisions of the beam and S the coefficient found by experiment. Required, what.-will be the effect of increasing all the dimensions n times, W S__ =S nt l This shows that the strength has been increasedi n' times, but the load..has increased n' times,. and the beam is therefore only - as strong as before. Therefore for cross strain in enlaging a model n times, the breadth miust 116 ENGINERZING STATCS.& again be increased n times, that is, n2 times as great as originally, if the depth be increased n times only. The depth must be made n t n times as great as originally, if the breadth be increased n times.- Although increasing the dimensions of the beam n times, makes it n2 times stronger to resist breaking, yet it becomes only n times stronger to resist bending, since we have, d W - (nt) 1 W P c n b (nh)s n c b-h! that is, of the former deflection, the load being supposed to remain the same. Hence the deflection of beams from their own weight, or from a uniform load proportionally increased (that is n, times) will be n or as the square of the ratio of increase. To preserve the same stiffness against deflection of beams, we see by the formula that the depth must increase in the same ratio as the length, if the breadth remains constant, or if the depth remain the same, the breadth must be increased as the cube of the length. For example; a beam whose length has been increased ten times, must, in order to retain the same stiffness as before, be ten times as deep, or one thousand times as b~road. Calculation of the strength of structures from that of their models. Let p = weight that the model will bear at its centre, w = weight of model, r = ratio of the scale of the model to that of the structure, P = load the structure must be able to bear at the centre. Then it is evident that r3 w = weight of structure. Resistance of struc ENGINEEING STATICS. 117 ture: resistance model:: cross-section of structure: cross-section of model, or:: r' 1. A weight uniformly distributed = same weight at the centre. Hence p + 2 w = resistance of model, and since the resistance of the structure - r' times that of model r W -' W we have P - r (p + w) - r'w - r2pr w r w =r p- r' (r -1)w r [-(r-i)w ] Conversey. —P given and p required. From preceding formula we have, p = + (r - 1) w BLOCK WOREK OR STRUCTURES WITH WIDE JOINTS. Singlepieces, or cemented mases. 105. General prinples of stability. A solid resists overturning by its own weight. This action may be considered as concentrated at, and acting in a line passing through, its centre of gravity. The "moment of stability " of a body - its weight, x horizontal distance from this vertical line to the edge about which the body would turn if overthrown. In Fig. 61, the moment of the wall A B CD - weight Fig. 61. X A H. In all cases where the contrary B 77 fC is not mentioned we consider a portion of wN.. wall one foot long. The moment of any force, as P, tendingto overthrow it - P x 11D length of the line drawn from the front edge of the wall, perpendicular to the direction of the force or -- P x A F. There are two ways of determining whether a wall will stand or fall under a given pressure. 1. Graphically. Construct a parallelogram of forces at the point where the direction of the force meets the vertical through the centre of gravity. Draw its diagonal. If this passes within the front edge the wall will stand; if not, it will fall 2. Nwmerically. Calculate the moments of the wall, and of the force. If the former exceed the latter, the wall will stand, and vice versa. If the force acts horizon NGINEERING sATIs. 119 tally, its leverage is the height of the point of application above the edge. If force acts obliquely, Fig. 62, the leverage is obFig. 62. fained thus: Numerically. It will be in equilibrium when P x A F= W x A H. I // Required to find A F. Let a = angle which, / the direction of the force makes with the.-v ehorizon. Then A F=- A V x cos. a(C D - C Z)scos. a= C Dcos.a- CZ cos. a. CZ=VZx tan.a —- VZx Hence, A F=. D x cos.a - VZx i. a x Cos. a Cos. a cos. a = C D x cos. a - V Zx sin.a. That is, the perpendicular from the point of overturning to the direction of the force = height of the point of application of the force x cos. a - the distance from the point of overturning to the vertical through point of application of the force, x sin. a. The "moment of stability " of a body, = W x H A. The "dynamicalstability" of a body= W x height, through which it is necessary to raise G in order to overturn it. In Fig. 60, W x M N. It is the "work' required to overturn it. The oeicintofstab ty" x H Fig60 For double stability HA 2. When = = = 24, HR HR then 2* is the coefficient of stability. When R is at A, then HA -1 and we have simple stability. When R is HR 120 ENGINERING STATICS. at 1,; then A HA = mo; that is, no possible force a HR 0 can overturn the wall. If the resultant passes through the centre of the figure and of the base, the pressure at all parts would be uniform, and equal the whole pressure, divided by the area of the base. If it does not, the pressure increases on the side to which the resultant is nearest. It is found desirable in structures of masonry, resting on earth, that the maximum pressure should nowhere exceed double the mean pressure. To effect this for a rectangular base, make HA -3, which brings R at H9 R one-third the thickness from the front edge; for a circular base make HA 4; for a hollow square, 1-; for a circular ring, 2. These last two expressions apply to tall factory chimneys. Masses resisting paralld forces as wind, like towers, chimneys, &c. 106. Any force, acting upon a convex surface, is twothirds what it would be on a plane surface. Regarding the wind as a number of parallel forces, we consider the resultant at the centre of gravity. We should examine whether the lower courses of bricks might not be crushed by the data given. The crushing force for common bricks is from 900 to 1,900 pounds per square inch, or 60 to 120 tons per square foot. The force of waves acts similarly lo that of winds; but if the body be immersed in water, as in the case of the ENGINEERING STATICS. 121 lower stones of a light house, the weight of the object will be diminished by that of an equal bulk of water. Dams, Wacls, &c. 107. It is cheaper to make the wall stable by adding -to its thickness than to its height, for the reason that its breadth enters twice as a factor in the formula, 1st in weight, 2din moment of stability. Hence a wall twice as thick is four times as strong. To find a general expression for a rectangular wall which shall just stand while resisting a fluid. Let w = weight per cubic foot of fluid, wl = weight per cubic foot of wall, h = height of wall, t = thickness, 1 = length = 1. The pressure of the fluid = Ah x 1 x w, and the moh h. w hs ment of the fluid h x - x w x =. The weight'2 3 6 of wall- h t x 1 x wl, and the moment of the wall= h t x 1 x wl x2= wth. For equilibrium; w6h t2 2 6 wIh x.2. Hence, t h 3 ( )- h /( ) In order to secure double stability w1 h - 2 -2, 2 2 whence t h v (2 w-) and this for water becomes t.816 h 6.45 16 i22 ENGINEERING STATICS. Masses resisting earth, as retaining wa&. 108. To find an expression for the thickness of a wall, resisting earth. In the expression for double stability against water, substitute for w the weight of earth x tan.' 2 the angle of the natural slope of the earth with the vertical (as in Part I). Let a = natural slope of the earth with the vertical. Then we have; th (- w xtan. )_.8 x tan.Ia x h ~/(i) W1 WI Walls with sloping faces or batters. A wall with vertical back and sloping face, will resist overturning with nearly twice the moment of a rectangular one of the same material. To find the bottom thickness of a wall with a triangular cross-section, which shall have the same stability as a given rectangular wall. Let h- = height of wall. x = bottom thickness, t = thickness of rectangular wall,' = weight per cubic foot. Moment of rectangular wall -h t x 1 x w' t I h t' w'. Moment of triangular wall= -x x 1 x tow = ~ h x2 w1. Hence, x = 1.22 t, and the mean thickness of such a triangular wall =.61 t, thus saving about.4 of the material. If the slope of the wall was on the inside, the wall would have only one-half its.former stability. In practice a triangular wall is never built, for its top must have some thickness, and not less than two feet. The section of the wall then becomes a trapezoid. ENGINEEPRIO STATlCS. To find the thickness of battering wals with the same stability as rectangular walls of the same height; (See:R. & R. R., page 184,) subtract from the thickness of latter four.tenths of the entire projection of the batter. The remainder is the mean thickness of the required wall. AApproximately. At one-ninth of the height of the rectangular wall from the bottom, draw through its front a line having the desired batter. It will be the desired wall near enough for the usual batter. Fig. 63. The final form, for economy of material, is shown in Fig. 63, the rear of the wall also sloping. The face is sometimes curved concave outwardly, as shown by the dotted line. Walls supported by Counterforts or Buttresses. 109. These should be on the front of the wall so as to throw the centre of gravity farther from the edge of rotation, but for other reasons, are generally placed on the back of the wall. In calculating the moments of these counterforts, we may conceive them to be extended sidewise so as to fill the space between them and diminish proportionally in weight, and then reason about them as about a continuous wall. WalZs of Uncemented Blocks. 110. Let several stones be laid without cement, one upon another, and let a given pressure, P, act upon the' uppermost: stone at some point a. The weight of the stone acts vertically. At the point a construct a paral 124 ENGlMEMBING STATICS. lelogram- of these two forces whose diagonal will meet the second istone in some point b. _ On this. diagonal, and the vertical at b, construct another parallelogram whose diagonal will meet the third stone in, c; and so, on for all the points, until a diagonal is found whose direction does not differ sensibly from the vertical, or direction of gravity. By these means we find the respective pressures at a b c, &c. The curved line joining these points is called the "line of resistance." This line must intersectthe common surface of each two contiguous portions of the structure within its mass. Equtwion of the Line of Resistance. Fig. 64 Let a force p act at a point t C --— D P, Fig. 64, in a line P F,' —-, making an angle q. with Ethe horizontal. Let G be the centre of gravity of dL \ 1 hothe part of the buttress above A B. Compound K... r> the two forces and let their M\ v resultant pierce the joint A B in B. Required the B- s' ~- A value of RB, for any'value of P B. Let PB= x, RB — y,B = y', a;ndW =wt. of buttress. The moment of the buttress W- W? x HR -:W x (y - y'). The'rm of the forcep the perpendicular distance R F of BR from the line of action of the force., - x x cos. - y x sin. P. Hence the moment of p = p (x x cos. -- y x sin. A). Equating, ENGINEERING STATICS. 125 p(x x cos. q — y, sin. q)- W (y - y'). Hence, W y' + p x cos. 9 Y W+ p sin. p When the force acts horizontal O0, and the formula becomes Wy' +px y, + px When the buttress is rectangular. Let t = thickness, y' - t, h' height, above P, and h = total height. W = w h t, in which w = weight per cubic foot. The equation of the line of resistance becomes 2 w (h' + x) t2 + p x cos, 9 w (h' + x) t + p sin. This line is a rectangular hyperbola passing through the point E, and having a vertical asymptote whose distance from the face of the buttress =.t+ P cos. ~ The w p Fig. 65. same result for a rectangular buttress and.C D horizontal force may be obtained directly. From the similar triangles, Fig. 65, EK M KM and E H R we have Ror y-t _ p = P Hence, y= t + - - W wt x+w th' V $'When h' O we have y t +. That w t x w t h'. W w t is, the line of resistance is then parallel to the face of the wall, and coincides ith the asymptote. 126 ENGFIEERING STATICS. ARCHES. 111. An arch is a structure of wedge shaped blocks, the extreme ones of which are confined between two supports. It is usually curved in one direction and is then used to resist a pressure against its convex side. Its most common application is to cover a space as a room, or to serve as a bridge. In them it has to support its own weight in addition to what other load may come on it. An arch may fail to do this in one of three ways, viz.: 1st. Its material may be crushed. 2d. Some portion of it may turn over upon others. 3d. Some portions of it may slide upon others. It is therefore first necessary to investigate the amount and direction of the forces which are acting in any arch. Knowing this, we can determine by Part II. whether the materials will crush. Then by the preceding formulas in Part III., whether any portion will turn over, and finally, by Part IV., whether any portion will slide. The condition of an arch, neglecting the adhesion of mortar, is analogous to that of a wall of uncemented blocks, Art. 105. We have, consequently, to find the line of resistance, and then to determine its proper position, so that the arch shall be stable. To DETERMINE THE LINE OF RESISTANCE IN AN ARCH. 112. Coulomb's method. This method determines the line at its extreme limit, when the arch is just about to give way by overturning. An arch may do this in three ways. 1st. In ordinary cases it yields thus: It opens at the crown on the intrados, on each side of the crown EGIEEBnIRI~l OrA.70CS. 127 at the extrados, and at the pring at the intrados, as shown Fig. 66. in Fig. 66. The points D A ~H Z and D' are called "'points of rupture." At the mo-/6 _I-L IEI ment at which the equilib/ K 1 rium is destroyed, the four / /l Ii parts into which the arch i I divides, touch each other 8" M T XS' 8 (supposing them incompressible) only at the points A, D and D', and the ground only at B and B'. If then we conceive these points joined by straight lines, and the weights of the four parts of the arch applied at the points where the verticals passing through their centres of gravity meet these lines, then these lines represent the whole arch. We consider only a slice of the arch equal to a unit of length. Call p = weight of one of the upper pieces, q = weight of one of the lower ones, and G and G' their centres of gravity. LetD K== x, BT=x',B S =x",A L= D H= y, B E= y', and B Z- h. Since the two parts of the system are symmetrical on each side of A, we may replace one of these by a horizontal force t, acting along A Z. We shall then have to consider only two bodies in equilibrium around the point B. Then the conditions of equilibrium, so that, there shall not be rotation in A, are t y - p x. That there may not be rotation at B, we have t h - p (x' + x) + q x". But equilibrium is not sufficient; there must be an excess of stability, and we must have (' + x) + q "I greaterthan (X x h)= - x x h. 128. MrINEERRING STATICS. The-first member;of this inquality is determined by the form of the arch. The second will vary with the position of the point D. Among all possible -values of the second member, the greatest:one is that which concerns us, since it threatens most the stability of the arch. We have therefore to seek its maximum, since it is that which tends most to overturn the arch. In other words; to determine whether the arch will stand or fall, we must get the maximum moment of this thrusting force with respect to an extreme edge such as B, and also the moment of the entire semi-arch about that edge. If the latter moment be the greater, the arch will stand, and the greater its excess the greater will be its stability. The result of an analytical investigation is that-the tangent to the curve of the intrados at the point of rupture, meets the horizontal drawn through the extrados at the key, in the vertical line passing-through the centre of gravity of the part of the arch acting. Therefore to find the point of rupture, assume a point D for it, and at this point draw a tangent to the intrados meeting the horizontal line drawn through the top of the extrados. Find the centre of gravity of the part of the arch between the point D and the crown. If the vertical passing through it and the other two lines meet in one point 0 the point has been rightly taken. If not try again. The difficulty is the laborious operation of finding the centre of gravity. It may be found in two ways, viz; by trial, and by an approximate rule. By Trial. Cut the exact figure of the part under consideration from a piece of. paste-board, and find its centre of gravity by trial. NNGINREIING ~.STATIC:' 129 Fig, 67. B a Approximate: Rule.:.'Divide k.O a..D L into:. nnumer of,..equal parts. I,,. 039 Through the poimts of division,idraw i I'd I. ~ i\ verticalstothe t6p of thearch. (Cail the extreme ordinates a and z, and h\ the other vertical distances between the intrados and extrados, (i. e. the top of load) b.: c. d. e. c., &c. Let thecommon distaneebetweme theverticals = 6..:Then DK' xa +.(3n —1)+ 6 (b+2 c +.3 d + 4 e, &c.) -..'l r~ +.z+ 2 (b + c + d +, & c,)] If that part of the divisor in brackets be multiplied by ~ 6, it will give the area of the portion D T A H', H' being the highest point of the load or D TA H when H is the highest point. If -the filling above the arch-be lighter than the arch, its depth in our calculation for the centre of gravity, must: be proportionally less. - For example, if it be of'earth of I the specific-gravity of the masonry, consider it to be 4 as high as'it really is. If- the road way be supported by walls with-empty spaces between them and they occupy 4 the width- f thebridges, call the filling- as high as it is. In- semi-circular arches the point-of otpi re, oceurs from 25~ to 31~ from the springs,..In an_ arch with a. span: of 654 feet., depth of key stone 34, with a level roadway on top, the angle of rupture was 27~. In arches formed by segments of circles, in which the ratio of the rise to the span is less than.29 (which is the ratio of the versed sine to the chord of rupture in semi-circular arches) the point of rupture will be at the spring of the arch. In 17 130 EM1NEERING TbTICS. basket-handle arches the point of rupture is generally so situated that the normal to the intrados at that point makes an angle of 450 with the vertical. The above data gives the position of the point of rupture approximately. The portion of the arch below this point may be regarded as part of the pier rather than of the arch. Second manner oj giving way. Fig. 68. By rising at the key, opening at the extrados at the key, and on the intrados, at the haunches, and' at the outside'y \ base of the piers, Fig. 68. In this case we 1i \ have to find the minimum moment of ~L____fvY-L the horizontal force at E, with respect to various interior edges, such as S, and the moment of the whole semi arch, about the same interior edges. If the former exceed the latter the arch will stand, and vice versa. The-investigation in this case, resembles that of the first, except that the points of rotation are changed. This occurs very rarely, and only when the lower parts of the arch, that is its haunches, are excessively loaded, Third manner of arches giving way. This is by an arch sinking at the key, and the pieces sliding on their bases. This will be examined in Part IV. sENGINEERING 3STATICS. 131 Mery's Graphic Method. Fig. 69. 113. The pressures acting on any joint I E, Elf Fig. 69, of an arch, may be considered as having one resultant in some point ~D,A@ P. The weight of the arch is kept in XI / /,' E equilibrium by the reaction p of the /.__g\ joint, and by the horizontal thurst t, _ -_-I — which acts at the summit of the arch. If similar points P' P," &c., be obtained for all joints, the curve formed by connecting them will be the line of resistance. The line of resistance is determined as in the case of a'wall. (See. art., 110.) We start with the horizontal thrust as in Coulomb's method, compounding this with the weight of the part of the arch above the joint in question, we obtain a resultant which will meet the joint in a point which will be in the line of resistance. So for all other joints. This line shows in what points and how the arch is in danger of giving way. Thus, if the line of resistance be A I B', the arch tends to open at D, B, and E. If the line of resistance be A' I B, the arch tends to open in A, E, and C. It is desirable that the line should pass as nearly as possible through the middle of the arch, so that the pressure may be nearly equally distributed over the surface of the bed of each voussoir. -This is not absolutely necessary, but the following condition must be fulfilled: To prevent the material of an arch from being crushed, the line of resistance must always pass so far from the nearest surface of the arch, whether intrados or extrados, that the portion of the 182 aEN EJX GII STANS. stone within this distance shall be able to support safely two-thirds of the total pressure O' on the-, hole surface of the stone.- This takes place when th& line passes within one-third the thickness of the arch from the intrados or extrados. The extreme pressure on the edge which it approaches, is then' double the mean pressure on the entire bed, the other edge receiving none. We may then conceive the arch to be divided into three zones, the outer two of which shall satisT the above conditions, and the middle one of which shall contain -the line of resistance whatever variations in it may be caused by accidental loads or other pressures. In attempting to obtain this line, we meet with this difficulty.:Wei do not know at what point of the key the horizontal thrust is applied. We must therefore try several. In a heavily proportioned arch, likely to open in the first mode, take the starting point above the middle of the key, for the line of pressure will then be there. It should not be taken nearer.the top than one-third the depth, for the reasons just given. In a light arch, so loaded as to be in danger of rising up at the crown,'as in the second mode, and for a pointed arch, take the starting point at one-third the depth from the bottom of the keystone. If the lines thus obtained come too near the edge, try others until you find one that does come within the required conditions,,; or find that the arch does not contain. any such line, which indicates thata su6ch'an arch would not be safe. The angle which the line of pressure makes with the joints will determnte whether one'stone will slide on::the other.- (See tabie, ENGINEERING STATICS. 133 Part IV.) This, however, rarely happens and will be considered in Part IV. A skew arch may be calculated by Mery's method, as if it were a right arch with a section equal to that parallel to the heads. The line of resistance should, however, come farther. from the edge of the stones than in a right arch, in the ratio of 1 cos. angle of skew; or sec. angle of skew:1. To determine the bestform for the intrados of an arch, the span and rise being given. The general principle is that the intrados should be parallel to the linear arch, which would correspond to the particular loading of the arch. Its own weight, must however, also be taken into account. When an arch has to support only its own weight, as when it is only a roof, its curve should be a catenary. Then-the line of resistance on it would be a catenary, and everywhere parallel to the curve of the arch, and would therefore, pass through Fig. 70. its middle. When the arch sustains a,/'~ f load, uniformly distributed horizontally. >r —-~-~-! — A its curve should be parabolic. For since, its load is uniformly distributed, its resulB 3 tant bisects A x; therefore, the tangent bisects A x; since these three forces acting, must meet in one point, therefore, the curve is a parabola. When the load acts vertically, and is distributed in any manner. Then the curve of equilibration is obtained thus: On a board, let a chain hang down, so as to have the requisite span and rise. This would be a catenary, and would be the proper curve for an unloaded arch. But when the arch is 1g4 ENGINEERING STATICS. supporting a level roadway the weight will press more upon one part than upon another, hence, suspend weights or pieces of chains from the large chain, and terminate them as far below the curve as the roadway is to be above. This must be still farther modified when the arch and roadway are of different specific gravities. We must then so adjust the relative weights of the chain and the pieces hanging from it, that they shall be to each other in the same ratio as the stones of the arch bear to the load above them. An arch satisfying the above condition is in a state of perfect equilibrium; no one part having a tendency to yield before any other part. If in such an arch the joints are perpendicular to the intrados then the line of pressure of the voussoirs, will be perpendicular to every point. An arch of equilibration with horizontal roadway, is found by calculation to have the following curve, suppose the arch to have 100 feet span and 40 feet rise. Abscissas from centre. Corresponding ordinatel 0 6 2 6.03 4 6.14 10 6.91 15 8.12 20 9.93 25 12.49 30 15.89 35 20.07 40 26.89 45 35.13 50 46.00 For general purposes a segmental arch is to be pre. ferred. The thrust of a segmental arch is the same as the fNsGINEERING STA!TI.S 1X thrustof a semicircular arch of the same radius, when the segment is not less-than the arc between the points of rupture, and nearly so when less. For a moderate span an arc of 600, or one in -which the span equals the radius, is much used in France. Its rise is two-fifteenths span. Another favorite French form is one in which the rise is one-third the span. They call it "Surbaisse au tiers." In basket-handle arches the rise should be between one-third and one-fourth span, approaching the former for small spans and the latter for large spans. Semicircular arches are very common, but not very advisable. To DETERMINE THE EXTRADOS OF AN ARCH, THE INTRADOS BEING GIVEN. Thickness of crown, or depth of keystone. 115. Thus far the thickness of the arch has been assumed to be known. The depth at the keystone is usually at first determined empyrically. Perronet's formula is,.035 span + 1 foot - span + 1 foot. This gives too much for great arches. Dejardin's formula. For semicircular arches, r being the radius of intrados, depth at key-.1 r + 1 foot. For a segmental arch, comprising 60~, this =T- r+ 1 foot, r + 1 foot. For a segemental arch of 400, this - r + 1 foot = -- r + 1 foot. For equilateral pointed arches (r = span) thickness = rA r + 1 foot, which is the same as for a segmental arch of 600~. Ellet's formula is, thickness at crown 3= V span. Trautwine proposes to use for thickness at the crown 136 ENGINEERING STATICS. for arches of the best cut stone, V radius at crown. For second class work, 4 V rad. at crown. For rubble, or brick work, -5 V rad. at crown. Barlow's formula. He would make the depth, in a somicircular arch = i r. The material will greatly effect the decision as to thickness of keystone. Rankine's formula, for depth of keystone in feet is, Ad / rad. at crown, for a single arch; or, TV V/ rad. at crown for a series of arches. For example, the Dora Bridge, 160' radius depth at crown, by calculation 4.38; actual depth 4.9 feet. The Chester Bridge, 140' radius, depth by calculation 4.1 feet, actual depth 4 feet. Table of depth of key, actually and by calculation. Actual Depth By By DeName of Bridge. Span. Rise. depth of by Per- Ellet's jardin's key. ronet's formula. formula. formula. Neuilly,...... 127 31 1-2 5.2 5.3 4.1 7.3 Chester*..... 200 42 4. 7.9 5.3 I11. Dora......... 147 18 1-4 4.9 6.1 4.5 8.3 Victoria...... 160 70 4.6 6.5 4.7 9. Maiden-head, t 128 24 5.2 5.4 4.2 7.4 Pont de Sorelle 68 8.2 4.4 3.27 3.1 4.4 Telford's..... 10 3 1-2 1.25 1.34 1.18 1.5 Do........ 18 6 1.5 1.6 1.6 1.9 Do....... 30 12 2, 2. I. 2.5 Do,....... 0 15 2.5 2.7 2.6 3,5 * Segmental. t Elliptical Thickness of the arch at other points. 116. The pressure increases from crown to spring. The thickness should increase accordingly. An approximate ENGINEERING STATICS. 137 rule is this: the thickness of the arch at any joint should = its thickness at the key x sec. (or.by cog.) of the angle which that joint makes with the vertical. The arch truly so called, does not extend below the joint which makes an angle of 60~ with the vertical. The thickness at that point is double that at the key, since sec. 60~ = 1- =2.cos. 600 The dimensions of the parts below that point, are determined as under the next head. Rondelets graphic construction of the extrados. Fig. 71. Through the spring A, draw an indefinite BEL F vertical line. Prolong the vertical radius B C, making C -i == B C. Then with a. j radius D E describe an arc, cutting the vere I A tical first drawn in F. EFwill be the exI D trados required. This construction is much used by architects in France. It, however, increases the volume of the arch and abutment, without increasing its stability. Yet is one of the best of the mere empyrical rules. Tables have been calculated, giving the thrust and proper dimensions of arches of the most usual forms. See Moseley's Mechanics of Engineers; and Captain Woodbury's Stability of Arches. Dimensions of Piers. 117. Knowing the horizontal thrust of the arch, and the leverage, the stability can be determined by the principles of engineering statics. The horizontal thrust being supposed to be applied at the point of rupture. 18 138 ENGNEE G STATICS. The thickness thus obtained is that of mere equilibrium. It is usual to multiply the thickness thus obtained by a coefficient of from 1.25 to 1.40, which is equivalent to increasing the stability in a ratio = the square of these numbers, viz; 1.56 to 1.96. The following table gives suitable dimensions for semicircular arches. In this the spandrils are supposed to be filled up level and to have a level roadway, and over all a pavement sixteen inches thick. Thickness of abutments. Thickness Thring. At ten At tentess at key. at spring. At ten At twenty feet high. feet high. 5 1.25 1,4 2.6 2.9 10 1.40 1.7 3.5 4.2 15 1.60 2.1 4.1 4.8 20 1.80 2.5 4.8 5.8 25 1.95 3.0 5.5 6,6 30 2.10 3.5 -6.2 7.5 40 2.50 4.8 7.6 9.1 50 2.80 5.9 8.6 10.4 100 4.50 10.7 13.5 16.0 The following table gives very safe dimensions: I4GnNEEHNG BTAT[0S. 139 Telford's Highland Bridges. Span of arch. Rise. Depth of Height of Thickness of key stone. abutment. abutment. 4 1.6 1.0 2.6 1. 6 6 2.0 1.0 2.6 2. 0 8 3.-0 1.2 2.6 2. 0 10 3.6 1.3 3.0 2. 6 12 4.0 1.4 3.0 3. 0 18 6.0 1.6 3.0 4. 6 24 8.0 1.-9 40 5. 0 30 12.0 2.0 4'0 5. 6 50 15.0 2.6 6.0 6. 6 118. Brick segmental arches. Thickness of Span. Rise. Depth of key. abutment at top. 10 3. 1.2 3.0 20 6. 1.5 3. 30. 9. 1.9 4.2 40 11.25 2.25 5.6 Abutment batter one-fourth inch to one foot. For piers of bridges which have an arch on each side, and hence do not sustain any thrust, the usual thickness is two and one-half times the depth of the key stone of the arch. The comparative thrusts. of various forms of arches are in the following table: 140 ENGINEERING STATICS. Pointed Semi- Arch Arch Arch Plate lowered lowered lowered band circular. to 1-3. to 1-6. to 1-10. Thrusts.......49 1. 1.39 1.82 1.91 1.95 Thickness of 70 1. 1.18 1.35 1.39 1.42 The thickness of the piers varies as the square root of the thrust, since the thickness enters twice as a factor of their resistance. ENGINEERING STATICS. 141 PART IV. STABILITY OF FRICTION, OR RESISTANCE TO SLIDING. General Laws of Friction. 1. It is directly proportioned to the pressure. 2. It is independent of the extent of the surface. 3. It depends on the kind of surface. 4. It is independent of the velocity. Recent experiments indicate, that it increases faster than the pressure, when that is so great as to produce abrasion, and that it increases somewhat with the surface. Coeficient of Friction. 120. This is that fractional part of the pressure which resists motion, and which is expressed byf. Limiting Angle of Resistance. 121. When one body rests on another, any force however small, acting on the former, will cause it to slide, if the direction of this force makesan angle with the perpendicular to the surface of contact, greater than a certain angle, depending on the nature of the surface. If the direction of the force makes an angle with the perpendicular to the surface, less than this same angle, the body will not slide however great the force. In most practical cases the direction of the effective force will be the resultant of the externally applied force and the weight of the body. 142 ENGINEERING STATICS. The " coefcient offriction" is the tangent of the limiting angle of resistance. Let P, Fig. 72, be a force acting on Fig. 72- any particle W. Decompose P W into A V - W and WV. Then W V denotes the 9I1/J' _1 pressure on the plane, and A WTV the force tending to move particle along. Let the angle P TF V= a. Then the pressure on the plane W V - P cos. a. Resistance of friction =f x P cos. a. Pressure tending to move the particle = A W = P sin. a. When motion is just ready to ensue, these must be equal, that isf x P P sin, a cos. a = P sin.. a. Whence,f = = tan. a. The P cos. a "limiting angle of resistance" we will call q. We will then havef =tan. q. The cone of resistance is a conical surface, generated by a line passing around the perpendicular to the surface of contact, and making with it an angle equal to the limiting angle of resistance. Angle of repose. 122. If a body rests on an inclined plane, the angle which this plane makes with the horizon, when the body is just about to slide, is called the "angle of repose." It equals p, or the limiting angle of resistance whose tan. f. The angle of repose of earth is the same as the natural slope of the earth. The greater the friction, the greater is q, of which it is the tangent, and hence the more may the plane be inclined before the body will slip on it. ENGaERU I STATIOB. 14.3 Demonstration that the angle of Repose equals the limiting angle of resistance. Fig. 73. FG:FE: B C: ACor::AB x sin.a:AB x eos. Fig. 73. F G sin a sB a, or:: s'm.a a co. a' - -. a \dtX tan. a. When the body is just about to | slide, FG =f x F E, whenceFG =f A tan. a; but f is also the tangent of the limiting angle of resistance, hence, that angle equals the angle of repose, i. e., a - q. Tazble of coecfflients of friction and limiting angles of resistance. _________~f.,f.. Dry masonry and brick work..........6 to.7 310 -to 350 Masonry and brick work, cemented,...74 360 1-2 Timber on stone..........4 220Iron on stone,................ 7 to.3 350 to 160 2-3 Timber on timber,...5 to:2 2601-2 to,11ol-3 Timber on metal...6..............to.2 31o to 110 1-3. Metal on metal,................ 25 to.15 140 to' 80 1-2 Masonry on dry clay............. 270 Masonry on moist clay,........33 180 1-4 AI Dry sand and clay and mixed 38to 75 21 to370 o~ |earth................. Damp earth,.. 450 M Damp earth................... 1. I Wet clay........................31 170 1 Shingle and gravel,.. 81 to 1.11 390 to 480 1t4 ENGINEERING STATICS. Structures of uncemented Blocks. 123. If the resultant, as obtained for any joint by Art. 110, makes an angle with the perpendicular to the surface of contact, greater than the limiting angle of resistance for that surface, then the wall will give way by that stone sliding on the one below it, however small the pressure of the resultant. The line of resistance is continually approaching the vertical, hence the tendency to slide is con, tinually diminishing in descending through the structure. Line of Pressure. 124. The line of resistance is obtained (as shown in Part III) by joining the points at which the joints meet the resultants of the externally applied force. Now let the resultants be produced indefinitely they will meet ii' a series of points. The curved line passing through these points is called by Moseley the "Line of Pressure." All the resultants are tangent to it. Then to determine whether the structure will slide at any joint, draw from the point where the "line of resistance" intersects the joint, a tangent to the "line of pressure." The angle which this line makes with the joint, will determine whether the wall will slide at that joint or not, according as the angle be more or less than the "limiting angle of resist. ance." The whole theory of equilibrium of any structure, depends on these two lines. (See Woodbury on the Arch.) The " line of resistance" determines the point of application of the resultant of the pressure on each surface of contact, and the " line of pressure" determines the direction of that resultant. The arch is only one form of the wall before mentioned. ENGINEPJING STATICS. 145 Plles Supported by Friction. 125. Let B = resistance which the earth opposes to the sinking of a pile during its sinking; s - space the pile is sunken by one blow of the ram. Then neglecting the elasticity and assuming B constant during s, we have w' x h- R s - (w + w) s, whence, R w — x W + w) w+w 8 + w + wu. Neglecting w and w& as very small when compared with R, and we have approximately we' h w h - - X- -- X -. W + w 8 w1 1+W Hence, we infer that the dead weight which a pile will bear is directly as h, and inversely ass, and directly as w 41 + vi or nearly as to. For a permanent load from onefifth to one-tenth of the value of R is used For another formula see Rankine's Applied Mechanics, p. 56G. Airy's formula, modified by Rankine, is this: Let B, S, A, w, mean as before, let e =modulus of elasticity for the material of the pile. Tables of its value are given in Part I. Let a - area of the head of the pile, I i its length B - v [ awl + ( eas e a8 The factor of safety is from 3 to 10. McAlpine's formula is: extreme supporting power =P = 80(W +.228 V F- 1). In which W = weight of ram in tons, and F = fall in feet. 19 cONSraTS. 147 TABLE OF CONTENTS. PART I. ForcEs TO BE BEaIBSTD. AT. PAGi. 1. Division of the subject, - - - 3 2. Kinds of statical forces, -. - 3 3. Table of weights, - -. 4 4. Fluid pressure, - - - - - 6 5. Semi-fluids, - - - - - - 6 6. Heat, - - - - - - - 8 7. Freezing- water, - - 9 8. Aiqxjispfixc - - - - 9 D.ynamical Forces. 9. Solids in motion, - 9 10u.impact, - - - -- - 10 1. Pile driving, - - - - - 11 12.: Waves, - - - - - - - 12!-: Win, - - - - - - - 12 PART IL SnTENGTH OF COHEsION. 14. Preliminary principles,,. - 13 Resistance of Bodies to Jlitension. 15. General principles, -. - - - 15 16. Wrought iron, - - - - - - 17 17. Cast iron, - - - - - - 21 18. Other metals, - - - - 22 19. Wood, - - - - - - - 23 20. Other materials, - - - - - - 25 lAT. PAG Patic FAnrms. 21. Riveted iron plates, - - - - - 25 22. Wire rope, - - - - - 26 23. Hempen ropes,, i ~: ~'-: - 27 24. Chain cables, -'- - - 27 25. Thin hollow cylinders,.. - 27 26. Thick hollow cylinders, - - - - - 29 27. Spheres, - - - - - - A9 28. Suspension rod, - - - 0 Resistance to Slearing or Detrusion 29. General principles, - 30. Rivet work, - - -8 - 31 31. Wood, - - - - - -. 33 Resistance to Compression and Orushing. 32. General principles, - - - - 34 33. Cast iron,... - - 37 34. Wrought iron, - -;- -. 38 35. Wood, - - - - - -38 Resistance io Bending and Breaking. 36. General principles of flexure, - - - 40 Fexure of Prismatic Beams. 37. Fixed at one end, -. 43 38. Supported at both ends, - - - - - 45 39. Fixed at one end and supported at the other, - - 45 40. Fixed at both ends, - - - - - 46 41. Supported at several points, - - - - 47 42. Flexure of rectangular beams,. 47 43. Flexure of prismatic beams of other formas - 48 44. Relative deflection of similar beams, - - - 49 Resistance to Rupture. 45. ~ General principles, - - - - 50 Rupture of Prismatic Beams. 46. Moment of resistance to rupture, - S o0 47. Fixed at one end and loaded in the middle, - - 61 48. " " " " " " uniformly, - 1- - ooIm~r, IT.149 49. Supported at both ends and loaded in thie middle, - 51 50........ uniformly, - - 52'61. Lines of stress, - - - - - 53 52. Beam supported at both ends and loaded at any point, - 54 563. "'. " I 4 " " " with two weights, -55 64. Weight uniformly distributed, - - 56 65. Partially loaded beams, - - 6 56. Beam fixed at one end and supported at the oteir;:- -.56 57. " " " both ends,. 57 58, Beam extending over several supports, - -. 57 a9. Limiting length of beams, - - - 58 Rupture of Rectangular Beams with sides Hlorizontal and Vertical. 60. Fixed at one end, - - - 8 61. Supported at both ends, - - - 69 i. Loaded at any point, - 60 63. Loaded at two points, - - - - 60 64. Weight distributed over m, - - - - 60 65. Weight near one end, - - -. - 60 66. Partially loaded beams, ^ - 60 67. Fixed at one end supported at the other, - - 60 68. Fixed atboth ends, - - 61 69. Extending over several supports, - - - 61 70. Rupture of beams of other forms,- - - 61 71. Forms of iron beams, -- - 6 72. Relative deflection and strength of rectangular beams in different conditions, -. 64 73. Relative strength of cast iron beams, - 66 74. Cast iron beams with wrought tension rods, - 66 75. Wrought iron beams, - - - - 67 76. Experiments on rupture, - - - - 67 77. Solids of equal resistance, - 70 Resistance of Posts, Pillars, Columns, Struts, &c., to yielding by Fweure, 78. General principles, - - - 73 79. Stone columns, - v - - 7$ 80. Cast iron columns, 74 81. Hollow cast iron pillars, - 75 ADTZ. PAogi, 82. Wrought iron columns. - - 76 83. " " plates, - 77 84. Wooden posts,.. - - - 78 Reistanoe to Torsion. 85. Principles and Formulas, - - - - - 79 PART III. Tns STADuarrY or PosmON 0 RorBSsTAc~ TO O Tz R=uNu. 86. Introduction,. - - - - 82 87. Bear work, - - -- - 82 88. A pair of struts, - - *, - -.- 85 89, Calculation of strains, 85 90. Pressure on lock gs.tes, - - 88 91. Braced struts. - 89 92. Trussed struts, - - 92 93. A pair of ties, - - -. 94 94. A strut and a tie, -. - - 94 95. Brackets, - - - - 96 96. Modification of brackets, - 98 97. Multiple frames, - - - - - - 100 98. Angle of diagonals in triangular girders, - - 103 99. When diagonals are alternately vertical and oblique, - 104 100. Compound frames, - - - - - 105 101. Polygonal frames, - - - - - - 106 -COrved Frames. 102. Convex upwards, or linear arches, - - - 108 103. Concave upwards, on catenaries,. - 111 104. Note on models, - -. - - 114 BLocx WN. Single pieces, or. Ce d Mseaes. 105. General principles of stability, - - - - 118 106, Masses resisting parallel forces, - - - 120 107. Dams, walls, &c,, 121 108. Masses resisting earth, - - - - - 122 109. Buttresses, - - - - - - - 123 cOibutS 151 Agm'. PAGX Walls of Unceme ntod B. 110. General principles, - - - - - 123 111. Principles, - - -. 126 112. Coulomb's method, - - - - - 126 113. Mery's graphic method, - - - - - 131 114. To determine the intrades, - - - - 133 115. To determine the thickness at key, 135 116. " " 4 " " other points, - 136 117. Dimensions of piers, - 138 1I1 Brick segmental arches, - - - 139 PART IV. STABImril oF PosroON OR RESISTuANtCE To Sw no. 119. General laws of friction, - - - - - 141 120. Coefficient of friction, 1- - -41 121. Limiting angle of resistance, - - 141 122 Angle of repose, - - - - - 142 123. Structures of uncemented blocks, - - 144 124. Line of pressure, - - - - 144 125. Piles supported by friction, - - - - - 145