Question

Problem 6.40 - Enhanced with Hints and Feedback Consider the truss shown in (Figure 1). Suppose that F1 = 45 kN and F-7 kN Part A Deter Expres Vies FCD Subi Parte Deter Expres Viev Sube Part HT Determ Expres View mm-mm- Foc

Answer 1

2 KN 4.5 kN 4.5 KN 7 KN 3 kN ZAN I

We can solve the truss by using method of sections.

First thing to do is to find the support reactions. We can use the equations of equilibrium for solving the support reactions.

Ʃ FY = 0

RA + RE = 2 + 4.5 + 4.5 + 7 + 3

RA + RE = 21

Ʃ MA = 0

4.5×5 + 4.5×10 + 7×15 + 3×20 = 20×RE

RE = 11.625 kN

RA = 9.375 kN

Ʃ FX= 0

HA = 0

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Now we can take section along CD , CF and GF

2 kN 4.5 KN 4.5 KN 7 KN 3 KN FCD 30.9638 FCF 9.375 kN 11.625 kN to FGE 21.8014

Taking moment with respect to F,

3FCD + 11.625×5 = 3×5

FCD = -14.375 kN

Taking moment with respect to G,

5FCD + 5 FCF cos(30.9638) + 9.375×10 = 4.5×5 + 2×10

5 FCF cos(30.9638) = 20.625

FCF = 4.81 kN

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Take section along CF, CD, HG and CG,

2 KN 4.5 KN 4.5 KN 7 kN 3 KN FCD » 30.9638° Fco FF A A FCF FCG4 30.96380 LTF 9.375 kN H 21.B014 11.625 kN V FHGA 68.1986 FcG FHG

Taking moment with respect to H,

3FCD + 3FCF cos(30.9638) + 5FCF sin(30.9638) + 5FCG + 9.375×5 + 4.5×5 = 2×5

3(-14.375) + 4.81×(3 cos(30.9638) + 5 sin(30.9638)) + 5FCG + 9.375×5 + 4.5×5 = 2×5

FCG = - 8.199 kN

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FCD = -14.375 kN

FCF = 4.81 kN

FCG = - 8.199 kN

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