Explanation Given information: Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below. For summation of forces along x -direction is equal to zero ( ∑ F x = 0 ) , consider the forces acting towards right side as positive ( → + ) and the forces acting towards left side as negative ( ← − ) . For summation of forces along y -direction is equal to zero ( ∑ F y = 0 ) , consider the upward force as positive ( ↑ + ) and the downward force as negative ( ↓ − ) . For summation of moment about a point is equal to zero ( ∑ M at a point = 0 ) , consider the clockwise moment as negative and the counter clockwise moment as positive. Method of joints and section. The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates Tension (T). Calculation: Show the free body diagram of the truss as shown in Figure 1. Refer Figure 1. Consider the angle made by the member CE with the horizontal is denoted by θ . Calculate the value of the angle θ as follows: tan θ = 3 4 θ = tan − 1 ( 3 4 ) θ = 36.869 ° Consider the horizontal and vertical reactions at A are A x and A y . Consider the vertical reaction at C is C y . Take the sum of the forces in the horizontal direction as zero. ∑ F x = 0 A x + 60 + 60 + 60 + 30 = 0 A x = − 210 kN Take the sum of the forces in the vertical direction as zero. ∑ F y = 0 A y + C y = 20 + 40 + 20 A y + C y = 80 kN (1) Take the sum of the moment about point A as zero. ∑ M A = 0 [ − ( 60 × 3 ) − ( 60 × 6 ) − ( 60 × 9 ) − ( 30 × 12 ) − ( 40 × 4 ) − ( 20 × 8 ) + ( C y × 8 ) ] = 0 C y = 1 , 760 8 C y = 220 kN Substitute 220 kN for C y in Equation (1). A y + 220 kN = 80 kN A y = 80 kN − 220 kN A y = − 140 kN Consider a section a-a passing through the members DF , EF , and CH . Show the portion of the truss above the section a-a as shown in Figure 2. Refer Figure 2. Take the sum of the moment about F as zero. ∑ M F = 0 − ( 30 × 6 ) − ( 60 × 3 ) − ( 40 × 4 ) − ( 20 × 8 ) − F C H × 8 = 0 F C H = − 680 8 F C H = 85 kN ( C ) Take the sum of the forces in the horizontal direction as zero. ∑ F x = 0 30 + 60 + 60 + ( cos 36.869 ° ) F E F = 0 F E F = − 150 × 5 4 F E F = − 187.5 kN ( C ) Take the sum of the forces in the vertical direction as zero. ∑ F y = 0 − 20 − 40 − 20 − F D F + 85 + ( sin 36.869 ° ) F C H = 0 − 20 − 40 − 20 − F D F + 85 + ( 3 5 ) × 187.5 = 0 − 20 − 40 − 20 − F D F + 85 + ( 3 5 ) × 187.5 = 0 F D F = 117.5 kN ( T ) JOINT A . Show the joint as shown in Figure 3. Refer Figure 3. For Equilibrium of forces, ∑ F x = 0 F A B = 210 kN ( T ) ∑ F y = 0 F A D = 140 kN ( T ) JOINT C . Show the joint as shown in Figure 4. Refer Figure 4. For Equilibrium of forces, ∑ F y = 0 220 + F C H + F C E sin ( 36.869 ° ) = 0 220 + ( − 85 ) + F C E sin ( 36.869 ° ) = 0 F C E = − 135 0.6 F C E = 225 kN ( C ) ∑ F x = 0 F C B = − ( − 225 ) cos ( 36.869 ° ) F C B = 225 cos ( 36.869 ° ) F C B = 180 kN ( T ) JOINT B . Show the joint as shown in Figure 5. Refer Figure 5. For Equilibrium of forces, ∑ F x = 0 − F B A − F B D cos ( 36.869 ° ) + F B C = 0 − 210 − F B D cos ( 36.869 ° ) + 180 = 0 − F B D = − 30 cos ( 36.869 ° ) F B D = 37.5 kN ( C ) ∑ F y = 0 F B E + F B D sin ( 36

Structural Analysis, Si Edition (mindtap Course List)

6th Edition

ISBN: 9781337630948

Author: KASSIMALI, Aslam

Publisher: Cengage Learning

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Plane and Space Trusses

In civil engineering, a truss is the most important structure that forms the most fundamental member of any building structure and commercial complexes. Trusses are composed of individual members or links that are interconnected by suitable joints, such …

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Chapter 4, Problem 50P

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Find the forces in the members of the truss.

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Structural Analysis, Si Edition (mindtap Course List)

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Find the forces in the members of the truss.

Solution Summary: The author explains the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium.

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Find the forces in the members of the truss.

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