INFLUENCE DTAGEAMS FOR THE DETERMINATION OF MAXIMUM MOMENTS IN TRUSSES AND BEAMS BY MALVEED A. HOWE, C.E. PROFESSOR OF CIVIL ENGINEERING, ROSE POLYTECHNIC INSTITUTE MEMBER AMERICAN SOCIETY OF CIVIL ENGINEERS FIRST EDITION FIRST THOUSAND NEW YORK JOHN WILEY & SONS, INC. LONDON: CHAPMAN & HALL, LIMITED 1914 Copyright, 19U BY MALVERD A. HOWE THE SCIENTIFIC PRESS ROBERT DRUM MONO AND COMPANY BROOKLYN, N. Y. PREFACE THE use of influence lines for the determination of maximum moments and criterions for wheel loads is pre- sented in many modern text books. The object of these few pages is to bring attention to the fact that for loads on all ordinary trusses, the influence diagrams for bending moments are drawn by following a single simple rule, and that the diagrams so con- structed require no computations for their direct applica- tion. In addition to this the influence diagrams for loads on continuous trusses, cantilever trusses and arches are shown to be based upon the one general diagram for simple trusses. While the diagrams, as a rule, are constructed for moments yet they can be as easily drawn for stresses or even areas of truss members. The use of the influence diagrams in the determination of criterions for the positions of wheel loads which produce maximums is explained and shown to be very simple. M. A. H. DECEMBER, 1913. iii TABLE OF CONTENTS CHAPTER I SIMPLE TRUSSES PAGE Rule for Drawing Influence Diagrams 2 Diagonal of Pratt Truss 3 Load Position for Maximum Moment 5 Neutral Point 7 Vertical of Pratt Truss 8 Web Member of Warren Truss 9 Web Members in Simple Trusses having Parallel Chords 9 Bottom Chord of Curved Chord Simple Truss 12 Top Chord of Curved Chord Simple Truss 13 Bottom Chord of Curved Chord Warren Truss 14 Trusses with Sub-divided Panels 15 Lower Segment of Diagonal in Sub-strut Truss 15 Upper Segment of Diagonal in Sub-strut Truss 17 Vertical of Sub-strut Truss 18 Upper Segment of Diagonal in Sub-hanger Truss 19 Lower Segment of Diagonal in Sub-hanger Truss 20 Vertical of Sub-hanger Truss 21 Bottom Chord of Sub-strut Truss 21 Top Chord of Sub-strut Truss 23 Top Chord of Sub-hanger Truss 24 Bottom Chord of Sub-hanger Truss 25 The Length of the Cut Stringer 26 Diagonal of Simple Truss having Center of Moments between the Supports 27 Vert ical of Simple Truss having Center of Moments between the Supports . 28 v vi TABLE OF CONTENTS CHAPTER II DOUBLE INTERSECTION TRUSSES PAGE Double Intersection Trusses 30 Top Chord of Whipple Truss 30 Vertical of Whipple Truss 31 Diagonal of Whipple Truss 32 Top Chord of Sub-divided Double Triangular Truss 32 Bottom Chord of Sub-divided Double Triangular Truss 33 Lower Segment of Diagonal in Sub-divided Double Triangular Truss ... 35 Chords of the K Truss 35 Diagonal Web Members of the K Truss 36 Upper Segment of Vertical in the K Truss 38 Lower Segment of Vertical in the K Truss 39 CHAPTER III CONTINUOUS TRUSSES The Moment Influence Diagrams 41 General Moment Formula 41 Cantilever Truss 43 Top Chord of Anchor Span 43 Top Chord of Cantilever Span 44 Diagonal of Anchor Span 46 Diagonal of Cantilever Span 47 Partially Continuous Truss 48 Top Chord of a Partially Continuous Truss 49 Diagonal of a Partially Continuous Truss 52 Continuous Truss of Two Equal Spans 53 CHAPTER IV ARCHES Arches 55 Diagonals of a Three-hinged Arch with Open Web 55 Chord of a Two-hinged Arch with Open Web 57 Diagonal of a Two-hinged Arch with Open Web 58 TABLE OF CONTENTS vii Moment Influence Diagram for Two-hinged Arch with Solid Web 59 Vertical Shear for Two-hinged Arch with Solid Web 59 CHAPTER V BEAMS OF CONSTANT CROSS-SECTION The Influence Diagram 60 Restrained Beams 60 Simple Beam on Two Supports 61 Beam Fixed at One End and Supported at the Other End 62 Beam Fixed at Both Ends 63 DEFINITION An influence diagram is one which shows the effect of a unit load moving across a structure upon any function of the structure for any position of the load. viii INFLUENCE DIAGRAMS FOR THE DETERMINATION OF MAXIMUM MOMENTS CHAPTER I SIMPLE TRUSSES* An Influence Line for the bending moments at the center of moments for any member of a truss on two sup- ports for vertical loads can be constructed by a method which is perfectly general in its application. The truss may be of any shape, and the loads may be applied to either the upper- or lower-chord joints. A section is passed through the truss, in the usual manner, cutting the member whose influence line is to be drawn. This section must evidently also cut a stringer in the panel of the truss containing the cut loaded chord. Let: d=the horizontal projection of the length of the stringer which is cut by the section. This stringer will be called the cut stringer. ^}q c b = the horizontal distance from the left support of the truss to the left end of the cut stringer. * See General Method for Drawing Influence Lines for Stress in Simple Trusses," by Malverd A. Howe. Engineering News, June 12. 1913. 2 INFLUENCE DIAGEAMS s = the horizontal distance from the left support of the truss to the center of moments, s will be considered positive when measured to the right. h = the vertical distance, at the left end of the cut stringer, between the two chord members which are cut by the section. h f = the vertical distance between the same chords at the right end of the cut stringer. The General Rule for drawing the influence line is as follows: % (a) Through any point A (refer to Fig. 1; the other figures are lettered correspondingly) in a vertical line pass- ing through the center of moments^ draw a horizontal line cutting the vertical line through -the left support of russ at C and the vertical line through the right support (6) From C, at any convenient scale, lay off vertically downward the distance CD=AC =s, and connect B and D by a straight line. (c) Through A and D draw a straight line and prolong it until it cuts the vertical line drawn through the left end of the cut stringer at E. (d) Draw a vertical line through the right end of the cut stringer, intersecting the line ACB at F, and connect E and F by a right line. (e) The polygon DEFBD contains the influence line sought; the ordinates between the line DEFB and the line DB are the respective moments for unit loads on the truss vertically above them, i.e., the lines ef=Zi, rc=z 2 , gk=zz, are proportional to the stresses produced by loads of one pound at Wi, Wz and Ws, respectively. The ordinates are measured at the scale used in laying off the distance CD=s. The application of the above rule to several forms of trusses will now be considered. SIMPLE TRUSSES 3 Diagonal of Pratt Truss with Inclined Top Chord. Fig. 1 shows this truss and the influence diagram for the member UL'. The shaded area is the influence diagram. To prove the construction correct, let the angles made with AB by the lines AE, EF and DB be respectively 0i, 6 2 and 03, and let the distance from the left support of the truss to the unit load be, in general, represented by '* Criterion, =(Wi + HY) FIG. 1. a, While the load W\ is to the left of the cut stringer, or to the left of the vertical through E, the moment is: 4, d) or tan0i, . (2) = TTi ( - pe +fp) = W l ( +ef) . (3) 4 INFLUENCE DIAGRAMS Since the resultant moment is positive, the moment UL'(y) is negative and therefore UL' is in compression. If the load, W 2 (or W 2 f ) is on the cut stringer, or between the verticals through E and F, the moment is, for W 2 , . . (4) -a 2 f-^~,. . (5) d < or M 8 = -W 2 (l-a 2 ) tan03 + TF 2 (&+d-a 2 )tan02; (6) and hence M s = W 2 (-qr+qc)=W 2 (+rc) ..... (7) Since the resultant moment is positive the stress in UL' is compressive. For the load TF 2 ' M s =W 2 '(-rc), ...... (8) and UL' is in tension. When the load is between the cut stringer and the right support of the truss, the moment is: M,= -W^^s= -Wz(l-a3) S = -TF 3 (Z-a 3 )tan03, or M s = W3(-gk), . ..... (9) and the member UL' is in tension. SIMPLE TRUSSES 5 If the diagonal inclines in the opposite direction, as in a Howe truss, the construction of the influence diagram remains unchanged but the character of the stress is reversed. Load-position for Maximum Moment. For uniform loads, the influence diagram DEGFB at once indicates the portions of the span which are loaded to produce like moments and hence maximum stresses. The criterion for giving the position of wheel loads producing the maximum moment is readily found from the influence diagram. For cases which usually occur in practice, the portion of the span on the left of the cut stringer may be considered as unloaded. For convenience W* is assumed to represent all of the loads between the verticals through E and G concentrated at their center of gravity, WY all of the loads between the verticals through G and F, and Ws all of the loads between the verticals through F and B. Let 22, 2 2 ' and 23, be the ordinates of the influence diagram directly below the wheel loads W2, W% and TF 3 . The moment is M s = W 2 (z2)-W 2 '(z2')-W 3 (z 3 ). . . . (10) If the loads move toward the left a distance dx, and no additional load comes on the span from the right and no load moves off the left end of the cut stringer, the moment becomes : Ml = ^2(22 - dx tan 3 + dx tan 2 ) -TF 2 '(z 2 ' + dx tan 3 - dx tan 2 ) -W 3 (z 3 + dx tan 3 ). (11) The difference between these moments is, M s ' -M a = dM s = W 2 ( - tan 3 +tan 2 ) dx -TF 2 '(tan 3 -tan e 2 )dx-W 3 (tan 3 )5o;. (12) 6 INFLUENCE DIAGRAMS Dividing through by 5x and placing - = 0, (13) (14) where W = W 2 + W 2 ' + W 3 = the total load on the span. Therefore, from equation (14), the desired criterion is W . (15) where s+b d -- or d The value of d' is found graphically from the influence diagram by drawing a line through D parallel to CB until it cuts the vertical line through F, then d! is the distance indicated in the figure (Fig. 1 and those which follow). This value may also be found graphically without drawing the influence diagram by simply drawing in the truss dia- gram the line Mm parallel to the bottom cut chord LU and prolonging it until it cuts a diagonal line drawn from the intersection of the cut top chord and a vertical through the left end of the cut stringer to the intersection of the cut bottom chord and the vertical through the right end of the cut stringer. This point is indicated by the letter m in the figure. The horizontal distance of this point m from SIMPLE TRUSSES 7 the vertical through the right end of the cut stringer is the value of d'. This is easily shown as follows: Fig. 1. mn : h :: d' : d, or d f = j-mn = -rMN] . . (17) but MNisiih :s+b, or MN =-^h; . . . (18) therefore ...'.. (19) which is the value given in (16). Neutral Point. The position of a load which produces no stress in the web member UL' is indicated by the point G in the influence diagram, Fig. 1, since a load in the truss immediately above G produces no stress in the member as shown by a zero ordinate in the influence diagram. The point w'shown in the truss diagram is directly above G. This point is located by the intersection of the line M N f and the diagonal drawn as explained in determining d'. Let x' be the horizontal distance from m f to the vertical through the right end of the cut stringer, and H-}-x f the horizontal distance of m' from the vertical through the right support of the truss. Then in Fig. 1 m'n' :h::x' : d, or m'ri' = ^x' = MN(H+x')\; (20) Ci i but MN :s::h : s+6, or MN = h; .... (21) 8 hence and INFLUENCE DIAGRAMS mn = KA-V, .... (22) . . (23) this becomes (H+x r ) tan 03 =z' tan 8 2 . (24) The ordinate above the point G in the influence diagram satisfies this equality, and therefore G and ra' are in the same vertical line. ^. Tan 63 =4- a < FIG. 2. Vertical of Pratt Truss with Inclined Top Chord. The truss diagram and the influence diagram are shown in Fig. 2. SIMPLE TRUSSES 9 All of the demonstrations given for Fig. 1 apply to Fig. 2. For all loads upon the left of the vertical through G the vertical is in tension and for those upon the right of G it is in compression. Web Member of Warren Truss with Inclined Top Chord. The web member U'L, Fig. 3 has been selected. A comparison with Fig. 1 shows that the influence diagram / Criterion. - = (PF 2 + HY)~ = sd FIG. 3. is constructed in exactly the manner followed in Fig. 1, and that all of the demonstrations of Fig. 1 apply equally well to Fig. 3. The locations of points m and m' are clearly shown in the figure where the verticals h and h r and the diagonal UL' are not members of the truss. Web Members in Simple Trusses Having Parallel Chords. The case shown in Fig. 4, presents an apparent exception to the general rule for drawing the influence line DEFB. The center of moments for any web member of a truss having parallel chords is at infinity. Taking N, the left 10 INFLUENCE DIAGRAMS support of the truss, as a reference point s = oc. For a load on the left of the vertical through E y the moment is M,= _}Ti +T7i(oo+ai). (25) Dividing both members of equation (25) by +00. (26) But, Wi j-^ + Wi is the expression f or vertical shear on the right of TFi, and, consequently, practically at the load Wi. If CD is laid off equal to s^-oo =1, the ordinates of the influence diagram represent vertical shears instead of moments. DE drawn through A, which is an infinite SIMPLE TRUSSES 11 distance from N, will be sensibly parallel to ACB. It appears that the vertical shear influence diagram is con- structed in precisely the manner outlined for the moment influence diagram when CD is made equal to unity. For a load on the left of the vertical through E, the vertical shear between E and F is Z-ai)-TTi(l);. (27) or -S=+Wi(l-a 1 )tui0 5 -Wi(i)=Wi(+pe-pf)-, therefore +S = Wi(+ef). . . . (28) In a similar manner it is shown that the ordinates directly below W^ Wz and T7s, represent the vertical shears produced by unit loads at these points. A negative ordinate in the influence diagram indicates that the resultant shear acts upward; therefore the vertical component of the stress in the web member, for which the influence diagram is drawn, acts downward. In Fig. 4, the member U'L' is in compression for loads on the left of G. The vertical UL has the same influence diagram but it is in tension for the same loads. The criterion for maximum shear is determined in the manner given above for maximum moment. Without writing all of the equations and referring to equation (13} and Fig. 4, ~Cf -tan 05+tan 4 )-TCY(-tan 4 +tan 5 ) dx -W 3 tan 02=0 (29) 12 or which becomes INFLUENCE DIAGRAMS tan 05 = I d' 2 ') tan 4 , . (30) (6L) The points m and m' and the value of d' are found in the manner explained for Figs, 1, 2 and 3. Bottom Chord of Curved-chord Simple Truss. The influence diagram for this case is constructed according to the general directions as shown in Fig. 5. The points E and A coincide. ~ ., . W Wi Wi Criterion. = =-: . ISO Tan 0i=l. Tan 8t=*-^. Tan0j=4- a I FIG. 5. The positive sign indicates that the resultant moment is positive and that the moment of the stress in LU is negative. This shows that LU is in tension. The criterion for the maximum moment when wheel loads are on the bridge is found as follows: (see equation 13). 8M S 6x Wi( -tan 0i +tan 3 ) +W 2 tan 3 +TF 3 tan 3 =0 (32) = (TFi+T7 2 +TF 3 ) tan 3 -TFi tan 0i=0; (33) SIMPLE TRUSSES this becomes 13 (34) An equivalent criterion is l-s _ _ ; __ " ' o __ i /n K\ : s AB -AT W W\ AC' This criterion holds good for any influence diagram which is triangular in shape. Top Chord of Curved-chord Simple Truss. The construc- tion of the influence diagram follows the same general rule uu Tan 01=1. Tan 2 =^. Tan fc =^-. FIG. 6. as in the previous case. The points A and F, Fig. 6 coincide, while E is on the right line connecting D and A since the line EF coincides with this line. For wheel loads the criterion is W 1+ W 2 W 3 W = ^ = ~' ' (36) where W = the total load on the span. 14 INFLUENCE DIAGRAMS Bottom Chord of Curved-chord Warren Truss. If both sets of web members are inclined as shown in Fig. 7, the influence diagram remains unchanged in its construction, but the line EF does not coincide with the line DA as in the previous case. Criterion. ~ = ' + W* - . I a sd Tan ft=l. Tan FIG. 7. - Tan* -. The criterion for the maximum moment produced by wheel loads is found as follows : 8M, 6x -tan 0i+tan 3 )+TF 2 (-tan 2 +tan tan 03=0. (37) (Wi+W 2 +W 3 ) tan 03-TFi tan -W 2 tan 02=0, (38) Substituting the values of the tangents, equation (38) becomes =0. . . (39) SIMPLE TRUSSES 15 Hence If s b = ^d, then - %-*+&* ....... (41) Trusses with Sub-divided Panels. The usual type of truss with sub-divided panels has one set of web members vertical. The trusses here considered will have one set of web mem- bers vertical and the chords not parallel. The only dif- ficulty in constructing the moment influence diagram is the determination of the proper length of the cut stringer. This once determined the construction of the diagram follows the general rule. Lower Segment of Diagonal in Sub-strut Truss. Let VL", Fig. 8, be the lower segment considered. This is a portion of the diagonal VL" of the main truss and also forms a chord member of the auxiliary truss LVL". Considering the member VL" as a part of the diagonal of the main truss, the influence diagram DD'FBD is constructed in the usual manner. The auxiliary truss LVL" may be considered as a simple truss supported at L and L" . The center of moments for VL" may be taken anywhere in LL" or LL" produced and in order that the lever arm of VL" shall be the same as that for VL" the center of moments will be taken at the inter- section of UU" and LL" produced, the center of moments for VL". The influence diagram, constructed according to the general rule, is D'EB f . Combining the two diagrams just constructed, the final influence diagram for VL" is DEFBD. This diagram is the same as if constructed 16 INFLUENCE DIAGRAMS with the horizontal projection of VL" as the length of the cut stringer. The values of x' and d' can now be found in the usual manner from the truss diagram. The criterion for the position of wheel loads producing maximum moment is found in the manner previously ex- plained. Referring to equation (13) and Fig. 8. dx or - = W 2 ( tan 63 -ftan 2 ) TF 2 '(tan 6$ tan d>>] tan -W 3 tan 03=0, . (42) ') tan 2 . . (43) Substituting the values of the tangents .. (44) SIMPLE TRUSSES 17 and which is the required criterion. In case the chords are parallel s becomes oo and the criterion is W W 2 +W 2 ' l'~ d ' . (46) Upper Segment of Diagonal in Sub-strut Truss. Refer- ring to Fig. 9, it is clear that the diagonal UV has the A ---ife^S " ^SsSrs FIG. 9. stress of UL f , the diagonal of the main truss, as the auxiliary truss LVL" simply serves the purpose of a trussed stringer extending from L to L" . The influence diagram is constructed according to the general rule taking the length of the cut stringer as the horizontal projection 18 INFLUENCE DIAGRAMS of LL". The graphical determination of d f and x' is evident from the truss diagram. The criterion for wheel loads producing maximum moment i ')j. . . (47) sd If the chords are parallel the criterion becomes (48) Vertical of Sub-strut Truss. As indicated in Fig. 10, the vertical of a sub-strut truss is a member of the main truss. FIG. 10. The auxiliary truss acts simply as a trussed stringer. The construction of the influence diagram follows the general rule using the horizontal projection of LL" as the length SIMPLE TRUSSES 19 of the cut stringer. The graphical determinations of d' and x' are clearly shown in Fig. 10. The criterion for wheel loads producing maximum moment is ) . . "(49) Ll Upper Segment of Diagonal in Sub-hanger Truss. The member UV in Fig. 11, is a part of the diagonal UL" of h-h // Tan 0i=l. Tan 2 = . Tan 63 = . 'E * d l FIG. 11. the main truss and also a chord member of the auxiliary truss UVU". The influence diagram for UL" js DEF'BD and that for the chord UV is D'FB'. The combination of these two diagrams gives the influence diagram DEFBD. This diagram is constructed according to the general rule when the horizontal projection of UV is taken as the length of the cut stringer. The graphical determination of x' and d f requires no explanation as the constructions are evident in Fig. 11. 20 INFLUENCE DIAGRAMS The criterion for wheel loads producing maximum moment is W W 2 +W 2 ' I ~ d' (50) Lower Segment of Diagonal in Sub-hanger Truss. As in the case of the upper segment in the sub-strut truss this member is a part of the diagonal of the main truss and has K-h . Tan0i= . * FIG. 12. its stress only. Taking the horizontal projection of UL" as the length of the cut stringer the influence diagram DEFBD, Fig. 12, is constructed according to the general rule. The distances d f and x' are found in the usual manner. The criterion for wheel loads producing maximum moment is W I d' (51) SIMPLE TRUSSES 21 Vertical of Sub-hanger Truss. In Fig. 13, the cut stringer for the vertical UL is the horizontal projection of LL', since no part of the load on UL" is supported at L. The construction of the influence diagram follows the general rule. In determining x' and d f from the truss diagram, h 1 and the diagonal UL' form no part of the truss proper. $--:;; The criterion for wheel loads producing maximum moment is d' (52) Bottom Chord of Sub-strut Truss. The bottom chord in Fig. 14 is a part of the chord of the main truss and also a part of the chord of the auxiliary truss L 3 FL 5 . Con- sidering only the chord L 3 L 3 of the main truss the influence diagram is DABD. Now considering the chord L^L 5 of the auxiliary truss L 3 VL 5 and taking the center of 22 INFLUENCE DIAGRAMS moments at V, the influence diagram, constructed according to the general rule, is C'EB'F'C'. The distance C'D' (7 3 L 3 ) is laid off equal to (AK) ,yj in order that the moments may be equivalent to taking Us as a center of moments. The diagram is inverted and when combined with the FIG. 14. previous diagram the two form the influence diagram for FLo, as shown by the shaded area DAEFBD. Referring to Fig. 14, D'C':EK::AF:KF or but therefore AF KF=d' AF =AK - SIMPLE TRUSSES 23 If the line DA is prolonged until it cuts the vertical through K the ordinate cut-off above K equals AK and hence it cuts the vertical at E, and EK in the small diagram equals EK of the large diagram. This shows, if the horizontal projection of VL 5 is taken as the length of the cut stringer and an influence diagram constructed according to the general rule, that this is the true influence diagram for VL 5 . The criterion for -wheel loads producing maximum moment is found as follows: -=TFi(-tan 0i+tan 3 )-pW 2 (-tan 0i+tan 3 ) +TT 2 '(tan 0?+tan 03-) +TF 3 tan 3 =0; (53) then W tan 3 = (Wi +W 2 ) tan 0i - W 2 ' tan 0' 2 . . (54) Substituting the values of the tangents and dividing through by s -- KY-^-. (55) If b-s=d, then W TF.+TFz-HY. I (56) Top Chord of Sub-strut Truss. In Fig. 15 the top chord UU' has no double duty to perform as it is simply a chord member of the main truss. The influence diagram is drawn according to the general rule using the horizontal projection of the chord UU' as the length of the cut stringer. 24 INFLUENCE DIAGRAMS The criterion for wheel loads producing maximum moment is W Wi+W 2 (57) U. . ^ ^ ___-J Criterion, -y = ( TFi + Wz). Tan 0i = 1. Tan 03 =4- FIG. 15. Top Chord of Sub-hanger Truss. Referring to the expla- nations given for Fig. 14, the horizontal projection of UV, Fig. 16, is taken as the length of the cut stringer and the influence diagram constructed according to the general rule. The criterion for wheel loads producing maximum moment is W_Wi W 2 (s-b) I ~ s + " sd ' . . . (58) If s -b=2d, then W Wi+2W 2 (59) SIMPLE TRUSSES 25 Criterion. If = W> + W 2 (s-b) ^ t S SCI Tan 0i=l. Tan 2 FIG. 16. . d Tan 63 Bottom Chord of Sub-hanger Truss. This case follows the general rule for simple trusses. 4 The influence diagram is shown in Fig. 17. -d- > Tan 0i = 1. Tan =-. Tan FIG. 17. 26 INFLUENCE DIAGBAMS The criterion for wheel loads producing maximum moment is The Length of the Cut Stringer for trusses with sub- struts or sub-hangers of the type considered above, can be readily determined by the following rules. (a) For diagonal members: whenever the section cuts the segment which is a part of the main diagonal and also a part of the auxiliary truss, the length of the cut stringer is the horizontal projection of the segment cut. (6) For diagonal members: whenever the section cuts the segment which is a part of the diagonal of the main truss and forms no part of the auxiliary truss, the length of the cut stringer is the horizontal projection of the diagonal of the main truss. (c) For chord members: whenever the section cuts the chord which forms a part of the main truss chord and also a part of the auxiliary truss, the length of the cut stringer for this chord is the horizontal projection of the diagonal segment cut by the section. Otherwise the length of the cut stringer is the horizontal projection of the chord, of the main truss, which is cut. (d) For vertical members of the main truss: whenever the section cuts any portion of the auxiliary truss the length of the cut stringer is the horizontal projection of the diagonal of the main truss. In case the section does not cut the auxiliary truss the length of the cut stringer is the horizontal projection of *the segment of the diagonal adjacent to the vertical. SIMPLE TRUSSES 27 SPECIAL CASE Diagonal of Simple Truss, Having Center of Moments be- tween the Supports. In Fig. 18 make a section cutting two chords and the diagonal UL'. The intersection of the two chord members is between the supports, a condi- tion which has not obtained in any of the previous examples. The influence diagram for moments is constructed according FIG. 18. to the general rule and is shown by the shaded area in Fig. 18. This diagram shows that each and every load placed upon the span produces the same kind of stress in the diagonal UL 1 '. The criterion for the position of wheel loads which produce the maximum moment is found as follows: - oX> 0i+tan 3 )+TF 2 (-tan 2 +tan 3 ) + TF 3 tan 3 =0, . . . .' (61) 28 or INFLUENCE DIAGKAMS W tan 6 3 = Wi tan B l +W 2 tan 2 . . . (62) Substituting the values of the tangents W W 1 s-b (63) where W is the total load on the span. FIG. 19. Vertical of Simple Truss Having Center of Moments between the Supports. The vertical member UX, Fig. 19, has its center of moments between the supports. The moment influence diagram is constructed according to the general rule and is the shaded area in Fig. 19. This diagram SIMPLE TRUSSES 29 is similar to that in Fig. 18 and therefore the criterion locating wheel loads which produce the maximum moment is CHAPTER II DOUBLE INTERSECTION TRUSSES Double Intersection Trusses are made up of two or more simple trusses and the influence diagram for any member is found by first drawing the influence diagram for each simple truss and then connecting these diagrams so as to form one diagram. This is shown in the fol- lowing examples. Top Chord of Whipple Truss. The Whipple truss shown in Fig. 20 is made up of two simple trusses as indicated by the full and dotted lines of the truss diagram. The chord UzU is a part of the top chord of both trusses. Considering it as a top chord member of the truss shown by the full lines, the center of moments is at Z/4 and the length of the cut stringer is L 4 L 6 . The influence diagram DAFBD is constructed according to the general rule. When the member UzU* is considered as a part of the top chord of the dotted truss, the center of moments is at Ls, the length of the cut stringer is LsZ/7 and the influence diagram is DA'F'B'D' '. This diagram is constructed upon the line DB of the first diagram by making D'C" equal s'. Now a load at L 2 does not affect the members of the dotted truss and hence the ordinate ef in the influence diagram for the truss shown by full lines is the correct moment for a unit load at L 2 . A load at L 3 by similar reasoning has for its true ordinate e'f in the influence diagram for the dotted truss. For a 30 DOUBLE INTERSECTION TRUSSES 31 load between L 2 and L 3 the influence line is a straight line connecting e and e'. In a like manner the other panel points are considered and the final influence diagram, shown by the shaded area, obtained. The loads at LI and LH are assumed to be equally divided between the two trusses and hence the influence line passes midway between the two influence diagrams for these points. Span= I Ui (J-2 Us 5 U4 !U 5 !U 6 !U "1 kj L 2 ; L 3 ; L ? L 5 | L | L 7 C ' i m i V FIG. 20. The position of wheel loads producing maximum moment is best found by trial. A criterion can be deduced but it is too complicated for practical use. Vertical of Whipple Truss. The vertical C/ 4 L 4 is a member of the truss shown by the full lines, Fig. 21. The influence diagram for vertical shear in the panel cut is DEFBD. For loads at panel points L 2 , Z/ 4) LQ, etc., the ordinates of this influence diagram are correct for unit loads. The loads at L 3 , L 5 , Z/?, etc., are supported entirely by the truss shown by dotted lines and hence do not con- tribute any shear in the panel being considered, of the 32 INFLUENCE DIAGRAMS truss shown by the full lines. Therefore, the ordinates in the influence diagram directly below these points are zero. Connecting these points with the ends of the ordinates which are correct, as shown in Fig. 21, the influence diagram shown by the shaded area is obtained. The loads LI and I/ 11 are assumed to be equally divided between the two trusses. FIG. 21. Diagonal of Whipple Truss. The diagonal C/2^4, Fig.- 22 is a member of the truss shown by full lines and its influence diagram is constructed in a manner similar to that explained for the vertical U^L^. The influence diagram is shown by the shaded areas. The position of wheel loads producing the maximum vertical shear is found by trial. Top Chord of Sub-divided Double Triangular Truss. The two principal trusses are shown in Fig. 23 by full and dotted lines. The top chord member UeUs forms a part of each truss. The influence diagram for U&Us as a member of the truss shown by full lines is DABD and that for the truss shown by dotted lines is D'A'B'. Connecting these DOUBLE INTERSECTION TRUSSES 33 two diagrams in the manner explained for the Whipple truss the influence diagram shown by the shaded area is obtained. FIG. 22. BB FIG. 23. The position of wheel loads producing the maximum moment is found by trial. Bottom Chord of Sub-divided Double Triangular Truss. The bottom chord member L 4 I/6, Fig. 24 is a part of the 34 INFLUENCE DIAGRAMS two simple trusses and also a part of the auxiliary frame LiMzLs. Neglecting the auxiliary truss LMzL, the influence diagram is drawn in the manner outlined for a top chord member. This diagram is DED"B"Fri, etc., and for a load of unity anywhere between Z/4 and L& the moment is equal to the ordinate immediately below the load between the lines D"B" and DB, and the stress in L 4 Z/6 equals this moment divided by the depth of the truss. The stress in LL as a part of the auxiliary truss LiMzLs can FIG. 24. be found by drawing the influence diagram for L^Ls and dividing the ordinate directly below the load by the length L 5 M S . If the scale of this diagram is properly taken the ordinates may be added directly to those of the large diagram. Following the general rule for constructing influence diagrams, with D"C", Fig. 24, equal to s" =LL 5 multi- plied by LiUi+LsMz, the diagram D"A"B" is obtained. The ordinates are increased in the same ratio as the ratio of the lever arms of LLs. The shaded area is the influence diagram for L^Ls. This is simply the proper combination DOUBLE INTERSECTION TRUSSES 35 of three diagrams for simple trusses. For the truss shown C" falls in the line CB. This method of constructing the influence diagram requires no preliminary calculations. As in the previous case the position of wheel loads producing the maximum moment is best found by trial. Lower Segment of Diagonal in Sub-divided Double Tri- angular Truss. Selecting the member L^M-s, Fig. 25, it is at once seen that it forms a part of the diagonal of the truss E D"E" FIG. 25. shown by full lines and also a part of the auxiliary truss L^MzLe. The influence diagram for vertical shear, neg- lecting the effect of the auxiliary truss, is DeD"B"mnopBD. The shear diagram for L 4 M 3 as a member of the auxiliary truss is D"F"B"D". This is constructed according to the general rule using the line D"B" as the line DB in the usual diagram. The influence diagram for L^Ms now becomes the shaded area in Fig. 25. The position of wheel loads producing the maximum vertical shear is found by trial. Chords of the "K" Truss. The truss shown in Fig. 26 is called the " K " truss since the web members form 36 INFLUENCE DIAGRAMS a series of the letter K. Making a section cutting U 2 U3, UzMz, MoLs and L 2 L 3 , it at once appears that the upper and lower chord members in a panel have equal horizontal components since the centers of moments are in the same vertical line. The construction of the influence diagram for/ moments FIG. 26. follows the general rule. The criterion for wheel loads which provide the maximum moment is W (65) where W is the total load on the span and Wi the total load between the left support of the truss and the center of moments of the chords considered. Diagonal Web Members of the " K " Truss. The usual method of sections cannot be employed as any section made through the truss cuts at least four members. From the arrangement of the diagonals it is clear that the pair panel must have horizontal components equal in DOUBLE INTERSECTION TRUSSES 37 magnitude but opposite in character. The magnitude of this component is equal to the difference between the horizontal components of the stresses in the chords in the panel containing the diagonals being considered and the adjacent panel opposite the point of intersection of the diagonals. In Fig. 27 the horizontal components of the diagonals U^M 2 and M 2 L% is the difference between the horizontal components of the chords in this panel and the panel immediately upon the right. JLJL The influence diagram for the horizontal component of .U$U- and L^L^ can be constructed according to the general rule by laying off CD equal to s+h 2 . On DB as a base construct a similar diagram for the chords U 2 Uz and L 2 L 3 , the difference between these diagrams as shown by the shaded areas in Fig. 27 being the influence diagram for the horizontal components of the diagonals U 3 M 2 and M 2 Lz. The stress in any diagonal equals its horizontal component multiplied by the secant of the angle it with the horzontal. 38 INFLUENCE DIAGRAMS The criterion for the position of wheel loads which produce maximum stresses is found as follows : -Tf 7 2 (-tan 3 '+tan 2 '- tan 2 ) + TF 3 tan0 3 '=0, (66) or F(tan0 3 0=(^2+TF 2 ')tan02. . . . (67) Substituting the values of the tangents, '):r-^7r-. (68) In case the chords are parallel, then hi=h<2 and ss'=d' and the criterion becomes the same as developed for Fig. 4. Upper Segment of Vertical in the " K " Truss. In Fig. 28 a section cutting the upper segment of the vertical C7 3 Z>3 FIG. 28. also cuts the members t/ 3 C/4, MzLz, and I/2l/ 3 . Taking the center of moments at the intersection of the chord members UsU^ and L^L^ the moment influence diagram DOUBLE INTERSECTION TRUSSES 39 is constructed according to the general rule. In this case the stress in U^M^ is not equal to the ordinates of the diagram divided by the lever arm s+b+d, since a portion of each ordinate corresponds to the moment of the stress in the diagonal M^L^. The simplest way to utilize the influence diagram is to first determine the stress in U?Mz for a unit load at L 3 . (This can be very quickly done graphically.) Then below F in the influence diagram lay FIG. 29. off to any convenient scale the distance Fg equal to this stress. Draw a straight line through B and g and extend it until the point D is located. Complete the diagram according to the general rule. The ordinates in this diagram correspond to the actual stresses in the piece UsM 3 produced by a unit load moving across the span. Lower Segment of Vertical in the " K " Truss. As in the case of the upper segment it is impossible to make a section which does not cut at least four members of the truss. Practically the same method is employed as outlined for 40 INFLUENCE DIAGRAMS the upper segment. Graphically determine the stress in MzL 3 produced by a unit load at L and lay off this stress below F as Fg. Locate D by drawing a straight line through B and g and then complete the influence diagram according to the igeneral rule. This gives the diagram DEFBD. A unit load at L 3 produces a stress in M 3 L 3 due to its position in the principal truss and also a local stress con- sidering M 3 L 3 as a piece which simply transfers a portion of the load at L 3 to the main truss. While the stress in M 3 L 3 produced by a unit load at Z/ 3 is readily computed yet it is usually more satisfactory to find this stress graph- ically by means of an ordinary stress diagram. After this stress is found, lay off mn equal to it and connect / and n and F and n. The shaded area shown is the stress influence diagram for the member M 3 L 3 . CHAPTER III CONTINUOUS TRUSSES The Moment Influence Diagrams for various forms of simple trusses have been explained in detail and found to follow a single general rule in their construction. After the dis- tance of the center of moments from the left support of the span is determined the diagram is constructed without further calculations arid the ordinates are the moments for unit loads, on the truss, directly above. For trusses which are continuous, or partially con- tinuous, the moment influence diagrams are founded upon the influence diagram for simple trusses. The diagram for simple trusses will be called the base diagram for con- venience. General Moment Formula. Considering any span of a continuous girder, let M s = the moment about a center of moments distant s from the left end of the span. M L = the bending moment at the left end of the span. M R =the bending moment at the right end of the span. S L =the vertical shear at the left end of the span. W = any concentrated vertical load in the span. a = the distance of W from the left end of the span. Ri =the left reaction produced by W considering the span as a simple beam on two supports. I = the length of the span. 41 42 INFLUENCE DIAGRAMS w, = the moment about a center of moments distant s from the left support of the truss, considering the truss as a simple girder on two supports, s = the distance of the center of moments from the left end of the span. Positive when measured to the right. Then s>a s = M L +S L s-W(s-a), .... (a) but M K -M, L therefore /l/f l/f s>a M,=M L +- Ji s+R 1 s-W(s-a). . (c) s>a Now, RiS W(sa) = m g , hence w This expression shows that the influence diagram for M s may be considered as being composed of two separate diagrams combined algebraically. One of the diagrams is that for the common moment m a or the base diagram. It will be shown in the problems which follow that the influence diagram for M s can be easily constructed upon the base diagram and with a small amount of calculation. CONTINUOUS TRUSSES 43 Cantilever *Truss. Truss bridges which are called cantilever bridges are composed of anchor spans, cantilever spans and suspended spans. The simplest form of this combination is shown in Fig. 30. Top Chord of Anchor Span. Assume any chord as UzU^, Fig. 30, with its center of moments at L 2 . For loads in this span the influence diagram is the base diagram rep- resenting the moments m s in equation (d), since the moments M L and M R are zero. FIG. 30. The criterion for wheel loads in this span which produce the maximum moment is (69) For loads in the cantilever span M L =Q, M R and from equation (d), M,= - (70) 44 INFLUENCE DIAGRAMS This shows that the lines DB and CB of the base dia- gram are extended until they cut the vertical through the right end of the cantilever. The influence diagram is shown by the shaded area BKL. For loads in the suspended span, M L = 0, M R = - W 3 1 2^1 2 , m s = Q, and hence M s = -Wz(h-^Y"f (^1) This is the equation for the straight line KM when tan 65 = j~ T~ = ~T~- The shaded area in Fig. 30 is the LI 63 63 complete moment influence diagram for the chord UzUz. The criterion for wheel loads which produce the max- imum negative moment is _ 1 2 "" h ' Top Chord of Cantilever Span. The chord member, Ui U 2 , Fig. 31 will be considered. For loads in this span M L W2d2, M R =Q and hence, from equation (d), (72) The base diagram is first constructed for the term m, as shown by the heavy lined triangle in Fig. 31. Prolong the line DA to K; then from the triangle DKB, b'c' : KB::a 2 : Z 2 , and CONTINUOUS TRUSSES 2 T ^ r> b'c' = -KB = 45 But this expression for b'c f is the same as the coefficient of W2 in the expression for M s and hence the triangle DKB is the influence diagram for TF 2 a 2 -^i . Since the 1 2 ordinates of the base diagram are positive and the ordinates of this diagram are negative, the combination of the two Tan 0i =1. Tan 3 =4. Tan 0- a =-^- S . FIG. 31. diagrams produces the shaded area AKB as the influence diagram for loads in the cantilever span. For loads in the suspended span the influence diagram is the shaded area KBM. The criterion for wheel loads which produce the max- imum moment is W 2 ' (73) 46 INFLUENCE DIAGRAMS Diagonal of Anchor Span. As shown in Fig. 32, the diagonal 11^2, has its center of moments at the inter- section of the two cut chords UWs and L 2 Z/ 3 . For loads in this span the moment influence diagram is the base diagram as shown by heavy lined shaded figure in Fig. 32. Tan 6 3 =4- Tan ft =T' 4- h & h FIG. 32. The criterion for wheel loads in this span which produce the maximum moment is (74) For loads in the cantilever span, M L =Q, M R = m, =0, and =W2a2T. . . . (75) Since p is the tangent of the angle #3 the influence diagram ^i for M, is the shaded area BKL. CONTINUOUS TEUSSES 47 For loads in the suspended span the influence diagram is the shaded area KLM. The criterion for wheel loads which produce the max- imum positive moment is (76) FIG. 33. Diagonal of Cantilever Span. Referring to Fig. 33, for loads in the cantilever span M L = W 2 a 2 , M B = 0, and +m. (77) The influence diagram for m s is the base diagram shown by the heavy lines in Fig. 33. From the triangle DEL, n'n : EL ::a 2 : 1 2 . 48 INFLUENCE DIAGRAMS and , ^jyr CL2, 7 N h~ S nn Y~BL = 7- (s 1 2 ) = az 7 . 12 1^2 12 Therefore the triangle DEL is the influence diagram for TPoCte 7 , where W2 1- Combining this 'with the base 12 diagram, the shaded figure EFBL is obtained as the influence diagram for loads in this span. For loads in the suspended span the influence diagram is the shaded area ELM. The criterion for wheel loads producing the maximum moment is found as follows: Without stating the intermediate equations or (78) W 2 ' W 3 s-h d s-b-d^s-b-d Z 8 " ' ' ' ( } Its-I 2 =h, W 2 _W 2 '+W 3 d sbd (80) CONTINUOUS AND PARTIALLY CONTINUOUS TRUSSES OF TWO EQUAL SPANS Partially Continuous Truss. In Fig. 34 is shown a common form of the revolving draw-bridge. The two end spans are equal and are separated by an unbraced CONTINUOUS TRUSSES 49 short span over the turntable. The moments M% and 1/3 over the turntable are assumed equal and, for con- venience, their values are determined as if the truss was a beam of constant cross-section. In any case the values of M 2 and M 3 can be found by a more rigid method, if desired, without increasing the labor of constructing the influence diagrams. FIG. 34. Top Chord of a Partially Continuous Truss. Referring to Fig. 34, the top chord U^Ui has its center of moments at La. For loads in this span, L = and but where 50 INFLUENCE DIAGRAMS Therefore, from equation (d), _ l fci-fei 3 , Dividing through by s, T = ~ Wl 4+Qn H (82) The influence diagram for - L is represented by the s base diagram drawn with CD = l. This is shown in Fig. 34 by the heavy line triangle. ki ki 3 The value of , A is now computed for the positions 4 ~r~\)Tl of Wi= unity, corresponding to the panel points of the truss. In the truss shown ki has the values 0, 1/5, 2/5, 3/5, 7 73 4/5, and 1. The corresponding values of Aia are laid off as ordinates above the line DB directly below the panel points and the ends of the ordinates connected by straight lines forming the polygon Dk'd'm'gB. The influence M diagram for * is the shaded area shown in the figure, s To obtain the value of M t for any particular load the cor- responding ordinate in the influence diagram is multi- plied by s. For any other center of moments as 1/2 it is only necessary to draw one straight line as DA' to obtain the influence diagram. In case the center of moments lies in a field covering about one-fifth the span Zi, adjacent to the turntable, the line DA will cut the polygon Dd'B and thereby indicate CONTINUOUS TRUSSES 51 the fields of loading in this span which produce moments of opposite character. The position of wheel loads producing the maximum moment is best found by trial. For loads in the third span 4+6n and . . . . (83) 4+6n Remembering that Ii=l 3 =l and dividing through by s, s _ _ 8 '' (84) The influence diagram corresponding to this expression is shown by the shaded area BKM, Fig. 34. The position of wheel loads producing the maximum moment is found by trial. This position when found will remain constant for all centers of moments in the first span. VALUES OF NCMBER OF PANELS IN SPAN. 4 5 G 7 8 9 Ll .0586 .048 .0406 .0350 .0308 .0274 Lo .0938 .084 . 0740 . 0656 .0586 .0527 L* .0820 .096 .0937 .0874 .0806 .0740 L 4 .072 .0925 .0962 .0938 .0891 U .0637 .0875 .0952 .0960 Le .0568 .0820 .0925 L 7 .0513 .0767 L, .0466 52 INFLUENCE DIAGRAMS The above table contains the values of for unit loads placed at the panel points of trusses having from four to nine panels. These values multiplied by give the ordinates represented by k-k 3 + Qn If the ordinates 4+6n in any particular case are laid off in an inverse order the resulting polygon will represent the expression 4+6/1 TIG. 35. Diagonal of a Partially Continuous Truss. The diagonal LzUz in Fig. 35 has its center of moments at the inter- section of UoUs and L 2 L 3 . Since the point of intersection lies upon the left of the left end of the span, s is negative. For loads in this span, if, and m (85) CONTINUOUS TRUSSES 53 tn The influence diagram for - * is indicated by the heavy lines in Fig. 35. This is the base diagram with CD = unity. The ordinates represented by the expression . fi are laid off above the line DB. The influence diagram for is the shaded area DEGFB. For any other diagonal r*~a'->i w FIG. 36. The expression for H is -- This indicates that H varies directly as the distances of the load from the supports and hence the influence diagram 56 INFLUENCE DIAGRAMS for H will be a triangle between the support and the center hinge. Placing a unit load at the center hinge, Lay off this distance above DB directly below the center hinge and thus locate the point K. Draw KD and KB', then the shaded area is the influence diagram M for * corresponding to the diagonal U 2 L S . Fig. 366 y is a similar diagram for the diagonal UsL^. For chord members the same method is employed. The triangle DKB is constant for all influence diagrams. The position of wheel loads producing the maximum moment is best found by trial as some of the criterions become too complex for easy application. To illustrate the shape of one criterion, consider the influence diagram in Fig. 366 in deriving a criterion for the diagonal / 3 L 4 . s f 05+tan 2 ) + TF 2 '(+tan 5 -tan 2 ) + W 3 ' tan 05 + WY tan 4 = 0, . (88) or (TPi'+T^'+flYHan 5 -(ttY + TF 2 ') tan 2 +WY tan 4 =0. . . (89) Substituting the values of the tangents (HY + HY + TfY)| - (WS+W, 1 )*-^ - W 4 '(~ -i). (90) AKCHES 57 Chord of Two-hinged Arch with Open Web. As in the case of the three-hinged arch the moments M L and M R are zero, and y (91) FIG. 376. WEB MEMBERS FIG. 37. The influence diagram for --is the base diagram drawn o with CD equal -. This is indicated by heavy lines in Fig. 37a. The values of H for a unit load at each panel point of the loaded chord are computed and laid off above DB and the upper ends connected by straight lines. The 58 INFLUENCE DIAGRAMS algebraic difference of these two diagrams is the shaded area shown in Fig. 37a and is the influence diagram for -* corresponding to the member LiZ/2. FIG. 386. VERTICAL SHEAR FlG. 38. Diagonal or a Two-hinged Arch with Open Web. The method followed in drawing the influence diagram for the diagonal is the same as explained for a chord member. The influence diagram for UiL 2 is shown by the shaded area in Fig. 376. ARCHES 59 The position of wheel loads producing the maximum moment is found by trial. Moment Influence Diagram for Two-hinged Arch with Solid Web. For the solid arch the influence diagram is quite simple as it is made up of the combination of the base diagram and the diagram for H, corresponding to unit loads, which is a smooth curve. The moment influence diagram is shown by the shaded area in Fig. 38a. For any other center of moments than the one indicated it is necessary to change only the lines CB and DA. Vertical Shear for Two-hinged Arch with Solid Web. The influence diagram for vertical shear is the same as explained for simple trusses on two supports by making the length of the cut stringer zero so that E and F of the base diagram lie in the same vertical line. This diagram is shown in Fig. 386. CHAPTER V BEAMS OF CONSTANT CROSS-SECTION The Influence Diagrams for beams are but little different from those which have been explained. In all cases the final diagram can be constructed upon the base diagram. Restrained Beams. Referring to Fig. 39, the moment at any section X is given by equation (d) or l s M s = M R ~- s ? r +w*. (d) i< yw x FIG. 39. If the vertical shear at the section X is represented by S SJ s>a M R M L s> ? s>a But Ri 2W=S =the vertical shear for a simple beam resting upon two supports; therefore o s = 4-0. 60 BEAMS OF CONSTANT CROSS-SECTION 61 Simple Beam on Two Supports. For a simple beam rest- ing upon two supports M L and M R are zero, hence M s = m s and S s = S. FIG. 40. The influence diagrams for ??i s and $ are base diagrams constructed according to the general rule. Figs. 40a and 406 show moment influence diagrams. In one diagram CB is drawn horizontal according to rule, while in the other DB is horizontal. Evidently the ordinates are the same in both figures. 62 INFLUENCE DIAGRAMS The influence diagrams for S are shown in Figs. 40c and 40d. The criterion for wheel loads which produce the maxi- mum moment is (93) Beam Fixed at One End and Supported at the Other End. kk 3 In this case referring to Fig. 41, M L =Q, M R = -Wl , and ^ = _T7~ 3 +^ (94) The influence diagram for --- is the base diagram drawn s with CD - unity as shown by the heavy lines in Fig. 41a- kk 3 The influence diagram for -^-~ is the figure DGBD and M the influence diagram for : - is the shaded area. For s any other center of moments the only line changed is DA. For vertical shear The influence diagram for S is shown by the heavy lines kk 3 in Fig. 416. The curve representing ~ is drawn above DB and then the shaded area is the influence diagram for &. BEAMS OF CONSTANT CROSS-SECTION 63 Beam Fixed at Both Ends. In this case, and or FIG. 41 b. VERTICAL SHEAR. FIG. 41. M, = -Wl(k-2k 2 +k*) l -j----Wl(k 2 -k*)j- Mt s . (96) 64 INFLUENCE DIAGEAMS The base diagram indicated by the heavy lines in Fig. 42a, is the influence diagram for . The ordinates to s ,- . , J 1 > |w |w/ s [ R FIG. 42 6. VERTICAL SHEAR FIG. 42. the curve D(75 are obtained by giving k corresponding values in the expression M 8 Then the shaded area is the influence diagram for -. s BEAMS OF CONSTANT CROSS-SECTION For vertical shear 65 (97) The shaded area in Fig. 426 is the influence diagram for S s . VALUES OF k*-W AND k-2k*+k* k 2 -A;' k k*-k* 1.00 .05 .002375 .95 .50 . 125000 .50 .10 .009000 .90 .55 . 136125 .45 .15 .019125 .85 .60 . 144000 .40 .20 .032000 .80 .65 . 147875 .35 .25 .046875 .75 .70 . 147000 .30 .30 .063000 .70 .75 . 140625 .25 .35 .079625 .65 .80 . 128000 .20 .40 .096000 .60 .85 . 108375 .15 .45 .111375 .55 .90 .081000 .10 .50 . 125000 .50 .95 .045125 .05 1.00 k-2k 2 +k* k k-2k*+k* k k-3k*+2k* = (k-2k 2 +k*)-(k*-k*) THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON*fTHE DATE DUE. THE PENALTY WILL INCREASE TO SO C ENTS ON THE FOURTH DAY AND TO $l.OOiC% THE SEVENTH DAY OVERDUE. AUG ^LIBRARY USE THE UNIVERSITY OF CALIFORNIA LIBRARY