STRESSES IN BRIDGE TRUSSES. 503 



(b) Methods. Now II - H' - 1 ,000 Ib. V - V, and by taking moments about B, 

 V 30 X 2,000/15 4,000 Ib. V'. 



Algebraic Moments. In passing sections care should be used to avoid cutting the end-posts 

 for the reason that these members are subject to bending stresses in addition to the direct stresses. 



^Irulate the stress in member 3~F take the center of moments at joint (i) and pass a section 

 rutting members 4-6, 3-4 and 3~F, and cutting the portal away to the left of the section. Then 

 assume stress 3~F as an external force acting from the outside toward the cut section, and 3~F 

 X 10 X 0.447 + II X 30' o. The stress in 3~F = 6,710 Ib. The remaining stresses are 

 cMii-ul.itt-il as shown. 



Graphic Resolution. Lay off a-A = A-b = H = 1,000 Ib., and A-Y V 4,000 Ib. 

 Then beginning at point B in the portal the force polygon for equilibrium is a-A-Y-l'-a, in 

 which I'-o is the stress in the auxiliary member i-a, and Y-l' is the stress in the post i-Y when 

 the auxiliary member is acting. The true stress in i-F is equal to the algebraic sum of the vertical 

 components of the stress I'-a and Y-l', and equals V = 4,000 Ib. Next complete the force 

 triangle at the intersection of the auxiliary members. Stress x'-o is known and the force triangle 

 is a-l'-2'-a, the forces acting as shown. The stress diagram is carried through in the order shown, 

 checking up at the point A. The correct stresses are shown by the full lines in the stress diagram. 

 The true stress in 3-2 will produce equilibrium for vertical stresses at joint (l) as shown. The 

 maximum shear in the posts is H = l ,000 Ib. The maximum bending moment in the posts will 

 occur at the foot of the member 3~F, joint (3), and is M = 1,000 X 20 X 12 = 240,000 in.-lb. 



(c) Results. The method of graphic resolution requires less work and is more simple than 

 the method of algebraic moments. 



Note: The portal is not pin-connected at joints (3) and the corresponding joint on the oppo- 

 site side, as might be inferred from the figure. 



PROBLEM 12. WIND LOAD STRESSES IN A TRESTLE BENT. 



(a) Problem. Given a trestle bent, height 45' o", width at the base 30' o", width at the top 

 9' o", wind loads Po, PI, Pz, PS, Pi, as shown. Calculate the stresses in the members of the 

 bent due to wind loads by algebraic moments, and check by calculating the stresses by graphic 

 resolution. Assume that the diagonal members are tension members, and that the dotted members 

 are not acting for the wind blowing as shown. Scale of truss, l" = 10' o". Scale of loads, 

 i" = 2,000 Ib. 



(b) Methods. Algebraic Moments. To calculate the stresses in the diagonal members take 

 centers of moments about the point A, the point of intersection of the inclined posts. Then to 

 calculate the stress in 3-4, pass a section cutting members 3-.X", 3-4 and 4~F; assume that the 

 stress in 3-4 is an external force acting from the outside toward the cut section, and 3-4 X 15.9' 

 + 3,000 X 19.3' + 3,000 X 11.3' = o. The stress 3-4 = 5,800 Ib. Stresses in 4-5, 5-6, 

 6-7, 7-8 and 8-Z are calculated in a similar manner. To obtain reaction RI take moments about 

 RI, and RI X 30' 2,000 X 15' 2,000 X 30* 3,000 X 45' 3,ooo X 53' = o. Then RI 

 .= 12,800 Ib. = R 2 . 



To calculate the stress in 4~F, take center of moments at joint Pi, and pass a section cutting 

 members $-X, 4-5 and 4~F, and assume the stress in 4~F as an external force acting from the 

 outside toward the cut section. Then 4~F X 15.6' 3,000 X 15' 3,000 X 23' = o. Then 

 4-Y = + 7,300 Ib. 



Graphic Resolution. The load PO is assumed as transferred to the bent by means of the 

 auxiliary members. The loads P , PI, Pj, P 3 , P 4 are laid off as shown, and with the load PO the 

 stress triangle F-.Y-2 is drawn. The remainder of the solution is easily followed. 



(c) Results. The stress in the auxiliary member 2- F acts as a load at the top of post 4- F. 

 Load Po is the wind load on the train and is transferred to the rails by the car. For the reason 

 that the wind may blow from the opposite direction, both sets of stresses must be considered in 

 combination with the dead and live load stresses in designing the columns. 



