SOLUTIONS OF PROBLEMS.

PROBLEM 5. DEAD LOAD STRESSES IN A BALTIMORE TRUSS BY GRAPHIC RESOLUTION. (a) Problem.-Given a Baltimore truss, span 280′ 0′′, panel length 20′ 0′′, depth 40′ 0′′, Calculate the dead load stresses by graphic resoludead load 0.5 tons per lineal foot per truss. 40' 0". Scale of loads, 1" = 40 tons.

tion. Scale of truss, I'

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(b) Methods.-The loads beginning with the first load on the left are laid off from the top Calculate the stresses at the left reaction by constructing downwards. Calculate R1 and R2. triangle 1-Y-X as shown. Then calculate the stress in 1-2 by constructing polygon Y-1-2-Y. Draw 3-2, which is the stress in member 3-2. Then calculate the stress in 3-4 and 4-Y by constructing polygon Y-2-3-4-Y. Calculate the stresses in the remaining members in order, finally checking up at R2.

(c) Results. It will be see that the Baltimore truss is a Pratt truss with subdivided panels The stresses in the first and second panels of the lower chord are larger than the stresses in the third and fourth panels of the lower chord. The stress in 6-7 is equal to the inclined component of the shear in the panel, plus the stress due to the half load that is carried toward the center of the bridge by 5-7. The Baltimore truss is used for long spans in which short panels can be used with an economical depth.

PROBLEM 5A.

DEAD LOAD STRESSES IN A BALTIMORE TRUSS BY GRAPHIC RESOLUTION. (a) Problem.-Given a Baltimore truss, span 320' 0", panel length 20′ o", depth 50' 0", Calculate the dead load stresses by graphic resolution. dead load 0.5 tons per lineal foot per truss. = 50 tons.

Scale of truss, I′′ = 50' 0". Scale of loads, I'

PROBLEM 6.

DEAD LOAD STRESSES IN A PETIT TRUSS BY GRAPHIC RESOLUTION.

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(a) Problem.-Given a Petit truss, span 350' o", panel length 25′ 0′′, depth at hip 50′ 0′′ Calculate the dead load stresses depth at center 58′ o", dead load 0.9 tons per lineal foot per truss. 50' 0". Scale of loads, 1" = 45 tons. by graphic resolution. Scale of truss, I' (b) Methods.-The loads beginning with the first load on the left are laid off from the top downwards. Calculate R1 and R2. Calculate the stresses in the members at the left reaction by constructing force triangle 1-Y-X. Then calculate the stress in 1-2 by constructing polygon Y-1-2-Y. Draw 3-2, which is the stress in member 3-2. Then pass to joint W, where there appears to be an ambiguity, stress 4-5 being unknown. To remove the ambiguity proceed as follows: At W, on the left side of the stress diagram assume that W, is the stress in 5-6 (the member 5-6 is simply a hanger and the stress is as assumed). Calculate the stress in 4-5 by comFirst complete pleting the triangle of stresses in the auxiliary members. The stresses are now all known at W2 except 3-4 and 5-Y, but the stress in 4-5 is between the two unknown stresses. the force polygon 2-3-4-5'-Y-Y-2. Then by changing the order the true polygon 2-3-4-5-Y-Y-2 The stress diagram is may be drawn. This solution is sometimes called the method of sliding in a member. The apparent ambiguity at joint W, may be removed in the same manner. carried through as shown and finally checked up at R2. It will be seen that there is no apparent ambiguity on the right side of the truss.

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(c) Results.-It will be seen that the Petit truss is an inclined Pratt or Camel-back truss with subdivided panels. The auxiliary members are commonly tension members in all except the end primary panels as in the Baltimore truss in Problem 5. It will be seen that the stresses The loads in this type of Petit truss are in the first four panels of the lower chord are the same. carried directly to the abutments. The Petit truss is quite generally used for long span highway and railway bridges.

PROBLEM 6A.

DEAD LOAD STRESSES IN A PETIT TRUSS BY GRAPHIC RESOLUTION. (a) Problem.-Given a Petit truss with the same span, panel length, depths, and dead load as in Problem 6; the auxiliary members being arranged as in the Baltimore truss in Problem 5.

W4

W3

W2

Wi

R2

Web

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= Coefficient x Wx sec 0.

PROBLEM 7. Dead Load STRESSES IN A QUADRANGULAR WARREN TRUSS BY GRAPHIC RESOLUTION.

(a) Problem.-Given a quadrangular Warren truss, span 120' o", panel length 15' 0", depth 18' o", dead load 400 lb. per lineal foot per truss. Calculate the dead load stresses by graphic resolution. Scale of truss and loads as shown.

(b) Methods. The loads beginning with the first load on the left are laid off from the bottom upwards. Calculate R1 and R2. The stresses at R1 may be calculated by constructing force polygon 1-X-Y. However, on passing to the next joint in either the top or bottom chord the solution is indeterminate. To solve the problem calculate the stress in the top chord 9-X by taking moments about the center joint in the bottom chord, the stress in 9-11 being zero. Lay off 9-X in the stress diagram and complete the diagram to the left and the right of the center as shown. It will be seen that the stresses in certain members occur twice in the diagram. The truss diagram can be divided into two systems as in Problem 14, and the stresses can be calculated for each system, the chord stresses in the two systems being added together for the final stresses. (c) Results. The quadrangular Warren truss is a double intersection truss in which the stresses are statically determinate for dead loads but are statically indeterminate for live load web stresses, as will be shown in Problem 14. This truss, built with riveted connections, is extensively used by the American Bridge Company for highway bridges for spans from 80 to 152 feet.

PROBLEM 7A. Dead Load Stresses in a QUADRANGULAR WARREN TRUSS BY GRAPHIC RESOLUTION.

(a) Problem.-Given a quadrangular Warren truss, span 150′ o", panel length 15' 0", depth 18' 0", dead load 500 lb. per lineal foot per truss. Calculate the dead and live load stresses by graphic resolution. Scale of truss, I' 20' 0". Scale of loads, I'

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10,000 lb.

PROBLEM 8. Dead Load Stresses in a Warren Truss by Algebraic Resolution (Method OF COEFFICIENTS).

(a) Problem.-Given a Warren truss, span 140′ 0′′, panel length 20′ 0′′, depth 20′ 0′′, dead load 800 lb. per lineal foot per truss. Calculate the dead load stresses by algebraic resolution.

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(b) Methods.-Beginning at the left the left reaction R1 3W. The shear in the first panel is 3W, in the second panel is 2W, in the third panel is W, and in the fourth panel is zero. Now resolving at R1 the stress in 1-Y 3W tan 0, stress I-X +3W sec 0. Cut members 1-Y, 1-2 and 2-X and the truss to the right by a plane and equate the horizontal components of the stresses in the members. The unknown stress 2-X will equal the sum of the horizontal components of the stresses in I-Y and 1-2 with sign changed, The stress in 3-Y =

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(62) W⚫tan 0 +10W-tan 0; stress in 5-Y

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-(-3-3) W·tan 0 +6W tan 0. - 8W tan 0. Stress in 4-X (82) W⚫tan 0 (+10+1) W⚫tan 0 +11W tan 0; and the stress in +12W tan 0. The coefficients of the chord stresses when multiplied by W tan 0 give the stresses, while the coefficients for the webs when multiplied by W sec 0 give the web stresses.

6-X

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(c) Results. In the method of coefficients the shears are calculated first, and the chord coefficients follow easily by summing the horizontal components. This method is the shortest and the best for the calculation of the stresses in bridge trusses with parallel chords.

PROBLEM 8A. DEAD LOAD STRESSES IN A WARREN TRUSS BY ALGEBRAIC RESOLUTION. (a) Problem.-Given a Warren truss, span 160' o", panel length 20' 0", depth 24′ 0′′, dead load 700 lb. per lineal foot per truss. Calculate the dead load stresses by algebraic resolution. Scale of truss, I" = 25' 0".

+15

+15

+12

5

3

-72

Maximum stresses.

Joint Load, P, = 12 tons.

Y

Stresses in Chords = Coeff. x P tan
Stresses

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PROBLEM 9. LIVE LOAD STRESSES IN A WARREN TRUSS BY ALGEBRAIC RESOLUTION.

(a) Problem.-Given a Warren truss, span 140′ o", panel length 20' 0", depth 20' 0", live load 1,200 lb. per lineal foot per truss. Calculate the maximum and minimum stresses due to the live load by algebraic resolution. Scale of truss, 1′′ = 20′ o".

(b) Methods.-Construct two truss diagrams as shown.

Chord Stresses.-The maximum chord stresses occur when the joints are all loaded, and the chord coefficients are found as in Problem 8. The minimum live load stresses in the chords occur when none of the joints are loaded, and are zero for each member.

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- 3P sec 0

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Web Stresses.-The maximum web stresses in any panel occur when the longer segment into which the panel divides the truss is loaded, while the shorter segment has no loads on it. The minimum live load web stresses occur when the shorter segment is loaded and the longer segment has no loads on it. The maximum stresses in members 1-X and 1-2 occur when the truss is fully loaded. The shear in the panel is 3P, or 21/7P, and the stress in I-X = 3P sec = +40.4 tons, while the stress in 1-2 40.4 tons. The minimum stresses in I-X and 1-2 are zero. The maximum stresses in 2-3 and 3-4 occur when 5 loads are on the right of the panel and there are no loads on the left of the panel. The shear in the panel will then be equal to the left reaction, = R1 (5 X 3 X P)/7 15/7P. The stress in 2-3 15/7P s ⚫sec 0 = + 28.8 tons, while the stress in 3-4 15/7P.sec 0 = 28.8 tons. The minimum stresses in 2-3 and 3-4 will occur when there is one load on the shorter segment. In the corresponding panel on the right of the truss, if the shorter segment is loaded, the left reaction

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1/7P the shear in the panel. The minimum stress in 2-3 = — 1/7P.sec 0 1.92 tons, while the minimum stress in 3-4 +1.92 tons. The stresses in remaining panels are calculated in same manner. (c) Results. It will be seen that the web members meeting on the unloaded chord (top chord) have their maximum and minimum stresses for the same loading.

PROBLEM 9A. LIVE LOAD STRESSES IN A WARREN TRUSS BY ALGEBRAIC RESOLUTION. (a) Problem.-Given a Warren truss, span 180′ o", panel length 20' 0", depth 24′ 0′′, live load 1,500 lb. per lineal foot per truss. Calculate the maximum and minimum stresses due to the live loads by algebraic resolution. Scale of truss, 1" = 25' 0".

PROBLEM 10. MAXIMUM AND MINIMUM STRESSES IN WARREN TRUSS BY ALGEBRAIC RESOLUTION.

(a) Problem.-Given a Warren truss, span 160' o", panel length 20' 0", depth 20' 0", dead load 800 lb. per lineal foot per truss, live load 1,600 lb. per lineal foot per truss. Calculate the maximum and minimum stresses in the meinbers due to dead and live loads by algebraic resolution.

(b) Methods.-Construct three truss diagrams as shown. On the first truss diagram place the dead load and the maximum live load chord coefficients, calculated as in Problems 8 and 9. The maximum live load chord coefficients are the same as the dead load chord coefficients. On the second diagram place the maximum and minimum live load web coefficients, calculated as in Problem 9. The maximum live load web coefficients are given on the left and the minimum live load coefficients are given on the right of the diagram. On the third diagram place the maximum and minimum stresses. The maximum chord stresses are equal to the sum of the dead and live load chord stresses. The minimum chord stresses are the dead load chord stresses. The maximum web stresses are the sum of the dead and maximum live load web stresses. The minimum web stresses are the algebraic sum of dead load stresses and minimum live load stresses.

(c) Results.-The web members 7-6 and 7-8 have a reversal of stress from tension to compression, or the reverse. These members must be counterbraced to take both kinds of stress. PROBLEM 10A. MAXIMUM AND MINIMUM Stresses in Warren Truss by Algebraic ResOLUTION.

(a) Problem.-Given a Warren truss, span 180' o", panel length 20' 0", depth 24′ 0′′, dead load 700 lb. per lineal foot per truss, live load 1,500 lb. per lineal foot per truss. Calculate the maximum and minimum stresses due to dead and live loads by algebraic resolution.