562 STRUCTURAL MECHANICS. CHAP. XVI. 



X 625 = o. Now Ri = 146.25 tons and W = 22.5 tons, and solving the equation gives stress 

 6-7 = 87.8 tons. 



Live Load Stresses. The maximum live load stres's in 6-7 will occur with the longer segment 

 of the truss loaded. Taking moments about point A as for the dead loads the maximum live 

 load stress 6-7 X 477.0 + RI X 575 = o. Now RI = 55/14 X 35 tons = 137.5 tons, and the 

 stress in 6-7 = 165.8 tons. 



The minimum live load stress in 6-7 will occur with the shorter segment of the truss loaded. 

 Taking moments about the point A, 6-7 X 477- + RI X 575 5? X 625 = o. Now RI = 90 

 tons, P = 35 tons, and stress in 6-7 = +29.1 tons. 



2. Stresses in Tie 4-7. Dead Load Stress. -Pass a section cutting members J-X, 4-7, 4-5 

 and 5-F, and cutting away the truss to the right. Now assume the stress in 4-7 as an external 

 force acting from the outside toward the cut section. Then for equilibrium about the point A, 

 stress 4-7 X 477-O + RI X 575 stress 4-5 X 442.0 2 W X 612.5 = o. Now the member 

 4-5 will carry one-half the load carried by 5-6, and the stress equals 1/2 X 22.5 X 1.414 = 

 + 15-9 tons. RI = 146.25 tons, and 2W = 45 tons. Then stress 4-7 = 103.6 tons. 



Live Load Stresses. The maximum live load stress in 4-7 will occur with the longer segment 

 loaded. Taking moments about A as for dead loads, stress 4-7 X 477-O + R\ X 575 stress 

 4-5 X 442.0 = o. Now stress 4-5 = + 24.8 tons, and RI = 66/14 X 35 = 165 tons. Then 

 stress 4-7 = 175.7 tons. 



The minimum live load stress in 4-7 will occur with two loads to the left of the panel. Taking 

 moments about the point A, the stress 4-7 X477.o-f-.Ri X575 2^X612.5 =o Now 

 .Ri = 62.5 tons and 2P = 70 tons. Then stress 4-7 = -+- 14-5 tons. 



The stresses in the members in the first and second panels and in the two middle panels 

 may be calculated by coefficients. Check up the dead load chord stresses by comparing with 

 the stresses obtained by graphic resolution in Problem 2. 



(c) Results. The auxiliary members carry the stresses directly toward the abutments and 

 there is no ambiguity of loading as in the case of a truss subdivided as in Problem 6. However, 

 the method of subdividing shown in Problem 6 is used in preference to that shown in this problem. 

 The Petit truss is quite generally used for long span pin-connected highway and railway bridges. 



PROBLEM 10. LIVE LOAD STRESSES IN A THROUGH PRATT TRUSS FOR COOPER'S E 60 



LOADING. 



(a) Problem. Given a Pratt truss, span 165' o", panel length 23' 6|", depth 30' o", live 

 load Cooper's E 60 loading. Calculate the position of the loads and the maximum and minimum 

 stresses due to the prescribed loading by algebraic moments. Scale of truss, i" = 25' o". 



(6) Methods. Chord Stresses. Calculate the position of the wheels for a maximum bending 

 moment at the different joints in the lower chord. The criterion for maximum bending moment 

 at any joint in a Pratt truss is, " the average load on the left of the section must be the same 

 as the average load on the entire bridge." Having determined the wheel that is at the joint for 

 a maximum moment, calculate the maximum bending moment as shown Having calculated 

 the maximum bending moments, the chord stresses are found by dividing the bending moment 

 by the depth of the truss. The moment diagram is given in Table V6, Chapter IV. 



Web Stresses. Calculate the position of the wheels for maximum shears in the different 

 panels. The criterion for maximum shear in a panel is, " the load on the panel must equal the 

 load on the bridge divided by the number of panels." The criterion for maximum bending 

 moment at LI is the same as the criterion for maximum shear in panel L$L\. Having deter- 

 mined the position of the wheels for maximum shears in the different panels, calculate the maxi- 

 nVam shears as shown. The stress in a web is equal to the shear in the panel multiplied by sec 9. 



Floorbeam Reaction. The stress in the hip vertical U\Li is equal to the maximum floorbeam 

 reaction. This is calculated as follows: Take a simple beam with a span equal to the sum of two 

 panel lengths and calculate the maximum bending moment at the point in the beam corresponding 

 to the panel point; in this case it will be the center of the span. This bending moment multiplied 

 by the sum of the panel lengths divided by the product of the panel lengths will be the maximum 

 floorbeam reaction; in this case the maximum bending moment at the center will be multiplied 

 by 2 divided by the panel length. 



(c) Results. When the maximum stresses occur in chords UzUs, UzUs and L 3 L 3 ', counter 

 U 3 'Ls is in action. It occasionally happens that there is more than one position of the loading 

 that will satisfy the criterion for maximum bending moment. In this case the moments for each 

 loading must be calculated. 



PROBLEM n. STRESSES IN THE PORTAL OF A BRIDGE BY ALGEBRAIC MOMENTS AND 



GRAPHIC RESOLUTION. 



(a) Problem. Given the portal of a bridge of the type shown, inclined height 30' o", center 

 to center width 15' o", load R = 2,000 lb., end-posts pin-connected at the base. Calculate the 

 stresses by algebraic moments and check by graphic resolution. Scales as shown. 



