560 STRUCTURAL MECHANICS. CHAP. XVI. 



f l-P-sec 0. With a maximum stress in 4-5 the stress in 4-7 will be = ( 66/14 + 7/i4)-P' 

 sec = ^fP-sec 0. This is the maximum stress, for the stress in 4-7 when there is a 

 maximum shear in the panel is = 10 X 11/2 X T \P-sec = f|P-sec0. In a similar 

 manner it will be found that maximum stresses in members 8-9 and 8-1 1 occur with a maximum 

 shear in 8-9. On the right side it will be seen that minimum stresses in the diagonals occur for a 

 minimum shear in the odd-numbered panels from the right. 



(c) Results. The dead and live loads were assumed as applied on the upper chord. The 

 upper chords are in compression, while the lower chords are in tension the same as for a through 

 truss. The live and dead load stresses are given separately on the left side of the lower truss. 



PROBLEM 6. MAXIMUM AND MINIMUM STRESSES IN A THROUGH BALTIMORE TRUSS BY ALGEBRAIC 



RESOLUTION. 



(a) Problem. Given a through Baltimore truss, span 320' o", panel length 20' o", depth 

 40' o", dead load 800 Ib. per lineal foot per truss, live load i ,800 Ib. per lineal foot per truss. 

 Calculate the maximum and minimum stresses due to dead and live loads by algebraic resolution. 

 Scale of truss, i" = 40' o". 



(6) Methods. Construct three truss diagrams as shown. 



Dead Load Stresses. The shear in each of the hangers is W, while the stress in each of the 

 diagonal auxiliary members is %W-secO. The stress in the upper part of the end-post is 

 (+ 6^ + 5) W^-sec = + 7W'sec0, where + 6JW-sec0 is the stress due to the shear and 

 + \ W- sec is the stress due to the half load carried toward the center by the auxiliary diagonal 

 member. The stress in the main diagonal in the third panel is $%W-sec 0, where 5%W is the 



shear in the panel; while the stress in the diagonal in the fourth panel is ( 4! |) TV- sec = 



sW-sec 0, where 4^W-sec is the stress due to the shear in the panel and ^PF-sec is the 

 stress carried toward the center of the truss by the auxiliary member. The chord coefficients 

 are calculated as in Problem 5. 



Live Load Stresses. The maximum shear in the third panel occurs with 13 loads to the 

 right of the panel and with no loads to the left of the panel. The shear in the panel is then equal 

 to the left reaction, equals 13 X K J 3 + i) X -P/i6 = H-P- The stress in the main diagonal 

 in the third panel is then equal to. ^P-sec 0. The stress in the main diagonal in the fourth 

 panel is ( *\P -f- ^P) sec = ffP sec 0, = a maximum, the maximum shear in the panel 

 being 12 X 5(12 + i) X P/l6 = ff-P. In like manner the maximum stresses are found in 

 5th and 6th panels when there is a maximum shear in the 5th panel, and in the 7th and 8th panels 

 when there is a maximum shear in the 7th panel. Minimum stresses in the 3d and 4th panels 

 from the right abutment occur when there is a minimum shear in the 3d panel; and in the 5th 

 and 6th panels when there is a minimum shear in the 5th panel. 



(c) Results. The double panels next to the center require counters. It should be noticed 

 that in calculating the stresses in these counters the diagonal auxiliary ties will have the dead 

 load stress of + 5.66 tons as a minimum. 



PROBLEM 7. MAXIMUM AND MINIMUM STRESSES IN A CAMEL-BACK TRUSS BY ALGE- 

 BRAIC MOMENTS. 



(a) Problem. Given a Camel-back truss, span 100' o", panel length 20' o", depth at hip 

 20' o", depth at center 25' o", dead load 300 Ib. per lineal foot per truss, live load 800 Ib. per 

 lineal foot per truss. Calculate the maximum and minimum stresses due to dead and live loads 

 by algebraic moments. Scale of truss, i" = 20' o". 



(b) Methods. Calculate the arms of the forces as shown and check the values by scaling 

 from the drawing. 



Dead Load Stresses. To calculate the stress in the end-post L Ui, take center of moments 

 at Li, and pass a section cutting L Ui, U\L\ and L\L^ and cutting away the truss to the right. 

 Then assume stress LoUi as an external force acting from the outside toward the cut section, 

 and stress L Ui X 14.14 Ri X 20 = o. Now Jf?i = 6 tons and stress L Ui = + 8.48 tons. 

 To calculate the stresses in L Li and L\Li take the center of moments at U\, and pass a section 

 cutting members UiU%, U\Li and LiL 2 , and cutting away the truss to the right. Then assume 

 the stress in LI L% as an external force acting from the outside toward the cut section, and LiLzX 20 



RI X 20 = o. Now RI = 6 tons and the stress in L Li = LiL 2 = 6 tons. To calculate 

 the stress in U\ U 2 take the center of moments at Li, and pass a section cutting members Ui Uz, 

 UzLz and LzL z ', and cutting away the truss to the right. Then assume the stress in L\Ui as an 

 external force acting from the outside toward the cut section, and Ui Uz X 24.25 RI X 40 + W 

 X 20 = o. Now Ri = 6, W = 3 tons, and the stress in Ui U 2 = + 7.42 tons. To calculate 

 the stress in UiLz take the center of moments at A, and pass a section cutting members UiUz, 

 UiLz, and LiZ-2, and cutting away the truss to the right. Then assume the stress in U\L Z as an 



