Answered: Use the VIRTUAL WORK method to answer…

Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)

5th Edition

ISBN:9781305084766

Author:Saeed Moaveni

Publisher:Saeed Moaveni

Chapter8: Time And Time-related Variables In Engineering

Section8.5: Engineering Variables Involving Length And Time

Problem 2BYG

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Question

Use the VIRTUAL WORK method to answer the following question(s). Show clear and full solution.

In the structure shown, find the horizontal movement of H knowing that the horizontal members of the structure are 2-L-35 x 35 x 3 mm angle bars, vertical members are 2-L-50 x 30 x 3 mm angle bars and diagonal members are 2-L-35 x 35 x 3 mm angle bars.

D
H
20 kN
4 m
C
40 kN
G
4 m
В
F
40 kN
4 m
A
E
3 m

Expert Solution

Step 1 Given

Step 2 Calculation of reactions with external loads

$\sum_{} M_{A} = 0 4 \times 40 + 8 \times 40 + 12 \times 20 - 3 \times R_{E} = 0 R_{E} = 240 k N \sum_{} F_{y} = 0 R_{A} + R_{E} = 0 R_{A} = - R_{E} R_{A} = - 240 k N \sum_{} F_{x} = 0 H_{A} + 40 + 40 + 20 = 0 H_{A} = - 100 k N$

Step 3 Calculation of member forces with external loads

Joint D

$\sum_{} F_{x} = 0 20 + F_{D H} = 0 F_{D H} = - 20 k N F_{D C} = 0$

Joint H

$\sum_{} F_{x} = 0 F_{H C} \cos θ + F_{H D} = 0 F_{H C} = \frac{- F_{H D}}{\cos θ} F_{H C} = \frac{20}{\frac{3}{5}} = 33.33 k N F_{H C} = 33.33 k N \sum_{} F_{y} = 0 F_{H C} \sin θ + F_{H G} = 0 F_{H G} = - F_{H C} \sin θ F_{H G} = - 33.33 \times \frac{4}{5} F_{H G} = - 26.66 k N F_{G F} = - 26.66 k N F_{C G} = 0 A t j o i n t E, \sum_{} F_{y} = 0 F_{F E} = - 240 k N F_{A E} = 0 k N$

Step 4 Calculation of member forces with external loads

Joint A

$\sum_{} F_{x} = 0 F_{A F} \cos θ + H_{A} + F_{A E} = 0 F_{A F} = \frac{- H_{A}}{\cos θ} F_{A F} = \frac{100}{\frac{3}{5}} = 166.66 k N F_{A F} = 166.66 k N \sum_{} F_{y} = 0 R_{A} + F_{A B} + F_{A F} \sin θ = 0 F_{A B} = - [R_{A} + F_{A F} \sin θ] F_{A B} = - [- 240 + 166.66 \times \frac{4}{5}] F_{A B} = 106.672 k N J o i n t B F_{B C} = F_{A B} F_{B C} = 106.672 k N F_{B F} = - 40 k N J o i n t C$

$J o i n t F \sum_{} F_{x} = 0 F_{A F} \cos θ + F_{C F} \cos θ + F_{B F} = 0 F_{C F} = \frac{- 166.66 \times \frac{3}{5} + 40}{\frac{3}{5}} F_{C F} = - 100 k N$

Step 5 Calculation of reactions with unit load on joint H

$\sum_{} M_{A} = 0 3 \times R_{E} + 12 \times 1 = 0 R_{E} = \frac{- 12}{3} R_{E} = - 4 k N R_{A} = 4 k N H_{A} = 1 k N$

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