PROBLEM 14A. MAXIMUM AND MINIMUM STRESSES IN A QUADRANGULAR WARREN TRUSS BY ALGEBRAIC RESOLUTION. (a) Problem. Given a quadrangular Warren truss, span 135′ o", panel length 15′ 0′′, depth 20' 0", dead load 400 lb. per lineal foot per truss, live load 700 lb. per lineal foot per truss. Calculate the maximum and minimum stresses due to dead and live loads by algebraic resolution. Scale of truss, I' 20' 0". = PROBLEM 15. MAXIMUM AND MINIMUM STRESSES IN A WHIPPLE TRUSS BY ALgebraic RESOLUTION. (a) Problem.-Given a Whipple truss, span 260′ o", panel length 20' 0", depth 40′ 0′′, dead load 1,200 lb. per lineal foot per truss, live load 2,000 lb. per lineal foot per truss. Calculate the maximum and minimum stresses due to dead and live loads by algebraic resolution. Scale of truss, I' 30' 0". = (b) Methods. The dead load stresses and the maximum live load chord stresses can be calculated by beginning at the center and calculating the shears, and then calculating the chord stresses as in Problem 14. The maximum and minimum live load web stresses are statically indeterminate as were the web stresses in Problem 14. The usual solution of this problem is to divide the truss into two trusses of single intersection. The dead and the live load chord stresses and the maximum and minimum web stresses are then calculated as for independent trusses. The loads at the foot of the hip verticals are assumed as equally divided between the two systems. The final chord stresses are the sums of the chord stresses in the separate trusses. The stresses in the web members, except the hip vertical, are as given in the separate trusses. In solving the problem the partial truss diagrams should be drawn. The trusses will be unsymmetrical, one being the same as the other turned end for end. With the joints all loaded the dead load chord and web coefficients, and the live load chord coefficients are calculated. In calculating the maximum live load web coefficients the loads are moved off to the right, and the maximum stresses in the webs on the left of the center will occur when the longer segment is loaded, and the minimum stresses in the webs on the right will occur when the shorter segment is loaded. Then with all joints in the truss loaded move the loads off to the left, calculating the maximum web coefficients on the right of the center and the minimum web coefficients on the left of the center. In calculating the stresses from the shears it will be seen that functions of two angles are used. The relation between the two angles is tan 0′ = 2 tan 0. Web coefficients in terms of @ are enclosed in a ring. The calculation of the chord coefficients may be illustrated by calculating the coefficient of the end panel of the upper chord = − [− 156/26 – 70/26 — 2(60/26)] W⚫tan 0 346/26W⚫tan 0. = (c) Results. The chord stresses calculated as above do not agree with those calculated by beginning at the center of the truss as in Problem 14. The student should calculate the dead load chord and web stresses and the live load chord stresses as in Problem 14. Whipple trusses were usually built with an odd number of panels. The Whipple truss was formerly quite generally used for long span highway and railway bridges, but is now rarely built, being replaced by the Petit truss. PROBLEM 15A. MAXIMUM AND MINIMUM STRESSES IN A WHIPPLE TRUSS BY ALGEBRAIC (a) Problem.-Given a Whipple truss, span 260′ o", panel length 20′ 0′′, depth 40′ 0′′, dead load 1,200 lb. per lineal foot per truss, live load 2,000 lb. per lineal foot per truss. Calculate the maximum and minimum stresses due to dead and live loads. Calculate the dead load chord and web stresses and the live load chord stresses as in Problem 14. Scale of truss, I" = 30′ 0′′. R1 +275.8 +2570 $26.0 18.38 -39.0 2020 W tan o 두리+ Algebraic Resolution. +132.00 +124.00 -7 -7 -7 R1 W W W tan O M -120.00-120.00 96.00 -96.00 $56.00-56.00 60.00 W=8 Tons, W sec 0-11.31 Tons, Wtan 0-8 Tons. Dead Load Coefficients. 60.0 Dead Load Stresses in Tons. 20'20' -195.0-195.0-182.0-182.0-312.0-312.0-390.0-390.0-120.0 -120.0 -96.0 -96.0-56.0 -56.0 -60.0 -60.0 Maximum Stresses in Tons. R2 -18.1 +28.6 -51.8 -12.0 5.66 +7917 +84.83 -3 0 0 34.00-39 Rz PROBLEM 16. 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' 0", depth 40' o", dead load 800 lb. per lineal foot per truss, live load 1,800 lb. 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′ 0′′. (b) 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 sec 0. The stress in the upper part of the end-post is (+6} + {) W⋅sec = +7W.sec 0, where +6 W sec is the stress due to the shear and W sec 0 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 - 5W sec 0, where 5W is the shear in the panel; while the stress in the diagonal in the fourth panel is ( 41 ) W sec 0 = - 5W-sec 0, where 4W sec is the stress due to the shear in the panel and W sec is the stress carried toward the center of the truss by the auxiliary member. The chord coefficients are calculated as in Problem 13. = 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 × }(13 + 1) × P/16 P. The stress in the main diagonal in the third panel is then equal to Ps sec 0. The stress in the main diagonal in the fourth panel is (PP) sec 0 -Psec 0, = a maximum, the maximum shear in the panel being 12 × (12 + 1) × P/16 = }}P. In like manner the maximum stresses are found in the 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 16A. MAXIMUM AND MINIMUM STRESSES IN A THROUGH BALTIMORE TRUSS BY ALGEBRAIC RESOLUTION. (a) Problem.-Given a through Baltimore truss, span 320′ 0′′, panel length 20' 0", depth 45′ o", dead load 800 lb. per lineal foot per truss, live load 1,800 lb. per lineal foot per truss. All the auxiliary ties are to be in compression as in the 1st and 2d panels in Problem 16 and as in Problem 6. Calculate the maximum and minimum stresses due to dead and live loads by algebraic resolution. Scale of truss, I" = 40' 0". PROBLEM 17. MAXIMUM AND MINIMUM STRESSES IN A CAMEL-BACK TRUSS BY ALGEBRAIC (a) Problem.-Given a Camel-back truss, span 100' o", panel length 20′ o", depth at hip 20' 0", depth at center 25′ o", dead load 300 lb. per lineal foot per truss, live load 800 lb. per lineal foot per truss. Calculate the maximum and minimum stresses due to dead and live loads by algebraic moments. Scale of truss, I" (b) Methods.-Calculate the arms from the drawing. 20' 0". of the forces as shown and check the values by scaling = = == Dead Load Stresses.-To calculate the stress in the end-post LoU1, take center of moments at Li, and pass a section cutting LoU1, UL1 and L1L2, and cutting away the truss to the right. Then assume stress LoU1 as an external force acting from the outside toward the cut section, and stress LoU X 14.14 R1 X 20 0. Now R1 = 6 tons and stress LoU1 +8.48 tons. To calculate the stresses in LoL and LL2 take the center of moments at U1, and pass a section cutting members U1U2, UL2 and LiL2, and cutting away the truss to the right. Then assume the stress in LiL2 as an external force acting from the outside toward the cut section, and LIL X 20 R1 X 20 = 0. Now Ri 6 tons and the stress in LoLi LiL26 tons. To calculate the stress in U1U2 take the center of moments at L2, and pass a section cutting members U1U2, U2L2 and L2L2', and cutting away the truss to the right. Then assume the stress in Lil' as an external force acting from the outside toward the cut section, and U1U2 X 24.25 - R1 X 40 + W X 20 = 0. Now Ri 6, W = 3 tons, and the stress in U1U, +7.42 tons. To calculate the stress in UL2 take the center of moments at A, and pass a section cutting members UU UL2, and LL2, and cutting away the truss to the right. Then assume the stress in UL, as an external force acting from the outside toward the cut section, and UL X 70.7 + R1 X 60 - W X 80 = 0. Now R1 = 6 tons and W = 3 tons, and U1La X 70.7 120 ft.-tons, and stress 1.70 tons. The other dead load stresses are calculated as shown. Live Load Stresses.-The live load chord stresses are equal to the dead load chord stresses multiplied by 8/3. The maximum stress in UL2 will occur with loads at L2, L', and Li', while the maximum stress in counter UL1 will occur with a load at L, only. The maximum tension in UL2 will occur with all the live loads on the bridge, while the maximum compression will occur when there is a maximum stress in the counter UL2', loads at L' and L'. The details of the solution are shown in the problem. U1L2 = = = (c) Results.-The stress in the counter U2L' and the chords U2U2' and L2L2' may be calculated by the method of coefficients, and will be the same as for a truss with parallel chords having a depth of 25' 0". The maximum stress in U2L2' will occur with loads L2 and Li' on the bridge, when the left reaction equals 2 X 3P/5 6/5P. The stress in U2L2' = — 6/5P.sec 0 = 6.15 tons. = PROBLEM 17A. MAXIMUM AND MINIMUM STRESSES IN A CAMEL-BACK TRUSS BY ALGEBRAIC MOMENTS. (a) Problem.-Given a Camel-back truss, span 120′ o", panel length 20′ o", depth at hip 25' 0", depth at U1⁄2 30′ 0′′, depth at U. 30′ 0′′, dead load 300 lb. per lineal foot per truss, live load 800 lb. per lineal foot per truss. Calculate the maximum and minimum stresses due to dead and live loads. PROBLEM 18. MAXIMUM AND MINIMUM STRESSES IN A THROUGH WARREN TRUSS BY GRAPHIC MOMENTS. (a) Problem.-Given a through Warren truss, span 140′ 0′′, panel length 20' 0", depth 20′ 0′′, dead load 800 lb. per lineal foot per truss, live load 1,200 lb. per lineal foot per truss. Calculate the maximum and minimum stresses by graphic moments. Scale of truss, I" = 20' 0". Scale of loads, I' = 50,000 lb. (b) Methods. Chord Stresses.-Calculate the center ordinate of the parabola = = w• •L2/8d = 98,000 lb., and lay it off at 5 to the prescribed scale. Now lay off the vertical line 1-5 at the left and right abutments. Make 1-2 = 2-3 3-4=2 (4-5). Draw the inclined lines 1-5, 2-5, 3-5, 4-5, 5-5. The intersections of these lines with verticals let drop from the lower chord points are points in the stress parabola for the upper chord stresses. The stresses in the lower chords are the arithmetical means of the stresses in the upper chords on each side. By changing the scale the live load stresses may be scaled directly from the diagram. = Web Stresses. At the distance of a panel to the left of the left abutment lay off the vertical line 1-8 equal to one-half the total live load on the truss, to the prescribed scale, equal 1,200 X 70 84,000 lb. Now divide the line 1-8 into as many equal parts as there are panels in the truss, and mark the points of division 2, 3, 4, etc. Connect these points of division with the panel point 7, the first panel point to the left of the right abutment. Drop verticals from the panel points of the lower chord of the truss to the line 1-8, and the intersections of like numbered lines will give points on the curve of maximum live load shears. To construct the dead load shear diagram, lay off 3W, downward to the prescribed scale under the left abutment, and reduce the shear under each load to the right by W, until the dead load shear is - 3W at the right abutment. The dead load shear diagram is then constructed as shown. Maximum and Minimum Web Stresses.-The maximum shear in any panel is then the ordinate to the right of the panel point on the left end of the panel, and the stresses in the web members are calculated by drawing lines parallel to the corresponding member as shown. Negative stresses are measured downwards from the live load shear curve, and positive stresses are measured upwards from the live load shear curve. (c) Results. This method is an excellent one for illustrating the effect of the different systems of loads, but consumes too much time to be of practical use. It should be noted that the maximum ordinate to the chord parabola is not a chord stress in a Warren truss with an odd number of panels. PROBLEM 18A. MAXIMUM AND MINIMUM STRESSES IN A THROUGH WARREN TRUSS BY GRAPHIC MOMENTS. (a) Problem.-Given a through Warren truss, span 160′ o", panel length 20′ o", depth 24′ o", dead load 900 lb. per lineal foot per truss, live load 1,200 lb. per lineal foot per truss. Calculate the maximum and minimum stresses due to dead and live loads by graphic moments. Scale of truss, 1" = 25′ 0′′. Scale of loads, I' = 50,000 lb. |
