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: The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates Tension (T). Calculation: Consider the forces in the member AB , BC , CD , DE , EF , FG , AH , HI, IJ, JK, KL, LG, BH, BI, CI, CJ, FL, FK, EK, EJ, DJ are F A B , F B C , F C D , F D E , F E F , F F G , F A H , F H I , F I J , F J K , F K L , F L G , F B H , F B I , F C I , F C J , F F L , F F K , F E K , F E J , F D J . Show the free body diagram of the truss as shown in Figure 1. Refer Figure 1. Consider the horizontal and vertical reactions at B are B x and B y . Consider the vertical reaction at F is F y . The loading on the truss is symmetrical. B y = F y = 10 + 20 + 20 + 20 + 10 2 B y = F y = 40 k Take the sum of the forces in the horizontal direction as zero. ∑ F x = 0 B x = 0 Consider the angle made by each of the inclined member with the horizontal is denoted by θ . Calculate the value of the angle θ as follows: tan θ = 25 25 θ = tan − 1 ( 1 ) θ = 45 ° Show the joint A as shown in Figure 2. Refer Figure 2. Find the forces in the member AB and AH as follows: For Equilibrium of forces, ∑ F y = 0 F A H sin ( 45 ° ) = 10 F A H = 14.14 k ( T ) ∑ F x = 0 F A H cos ( 45 ° ) + F A B = 0 F A B = − F A H cos ( 45 ° ) F A B = − 14.14 cos ( 45 ° ) F A B = 10 k ( C ) Show the joint H as shown in Figure 3. Refer Figure 3. Find the forces in the member HI and HB as follows: For Equilibrium of forces, ∑ F x = 0 F H I = F H A sin ( 45 ° ) F H I = 14.14 sin ( 45 ° ) F H I ≃ 10 k ( T ) ∑ F y = 0 F H B = − F A H cos ( 45 ° ) F H B = − 14.14 cos ( 45 ° ) F H B ≃ 10 k ( C ) Show the joint B as shown in Figure 4. Refer Figure 4. Find the forces in the member BI and BC as follows: For Equilibrium of forces, ∑ F y = 0 40 + F B H + F B I sin ( 45 ° ) = 0 F B I = ( − 40 − F B H sin ( 45 ° ) ) F B I = ( − 40 − ( − 10 ) sin ( 45 ° ) ) F B I = 42.43 k ( C ) ∑ F x = 0 − F B A + F B C + F B I cos 45 ° = 0 F B C = F B A − F B I cos 45 ° F B C = ( − 10 ) − ( − 42.43 ) cos 45 ° F B C = 20 k ( T ) Show the joint I as shown in Figure 5. Refer Figure 5. Find the forces in the member IC and IJ as follows: For Equilibrium of forces, ∑ F y = 0 − F I C − F I B cos 45 ° = 0 F I C = − F I B cos 45 ° F I C = − ( − 42

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ISBN: 9781337630931

Author: KASSIMALI, Aslam.

Publisher: Cengage,

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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 15P

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Find the forces in the members of the truss by the method of joints.

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Structural Analysis

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