The diagram presents a truss structure with members connected at joints labeled A, B, C, and D. The truss is subject to external forces, including a 12 kN force acting to the left at joint D and a 9 kN force acting to the right at joint C.Key dimensions and elements of the truss are as follows:- The horizontal distance between points A and B is 1.5 m.- The vertical height between points A and D, and points B and C, is 2 m.- All members of the truss have a square cross-section with side dimensions of 100 mm × 100 mm.- The modulus of elasticity (E) is given as 200 GPa.You are tasked with determining:(a) The force in member AC.(b) The forces in all members of the truss.(c) The horizontal displacement at point C.This analysis involves using the force method, a structural analysis technique to calculate internal forces and displacements.

A truss with hinged support at A and roller support at B is given. Horizontal 9 kN and 12 kN loads are applied at C and D respectively. E = 200 GPa Cross-sectional area, A = 104 mm2 To determine (a)The force in member AC (b)Forces in all members (c)Horizontal displacement at point CDegree of static indeterminacy, Ds DS= m +r-2j=6+3-2×4=1 Thus, the truss is indeterminate of degree 1. Assume force in member AC as redundant Using Castigliano's theorem At joint C FCA=RFCD = 9 -Rcos a=9-0.6RFCB=-Rsin a=-0.8R At joint D FDB=-FDC-12cos a =-9+0.6R-120.6 =-35+RFDA=-FDBsin a = 28-0.8R At joint B FBA=-FDBcos a=21-0.6R Member Length AE Fi ∂Fi∂R Fi∂Fi∂RLi CA 2.5 2000000 R 1 2.5R CD 1.5 2000000 9-0.6R -0.6 -8.1+0.54R CB 2 2000000 -0.8R -0.8 1.28R DB 2.5 2000000 -35+R 1 -87.5+2.5R DA 2 2000000 28-0.8R -0.8 -44.8+1.28R BA 1.5 2000000 21-0.6R -0.6 -18.9+0.54R ∆CA=∑Fi∂Fi∂RLiAE0=8.64R -159.3R =18.44 kN Thus, the force in member AC is 18.44 kNThe member forces due to applied external loads are Member Fi∂Fi∂RLi For R = 18.44 kN Member forces CA 2.5R 46.1 CD -8.1+0.54R 1.8576 CB 1.28R 23.603 DB -87.5+2.5R -41.4 DA -44.8+1.28R -21.20 BA -18.9+0.54R -8.94At joint C FCA=RFCD = 1-Rcos a=1-0.6RFCB=-Rsin a=-0.8R At joint D FDB=-FDC-12cos a =-1+0.6R0.6 =-1.67+RFDA=-FDBsin a = 1.33-0.8R At joint B FBA=-FDBcos a=1-0.6R Member Length Fi ∂Fi∂R Fi∂Fi∂RLi For R = 1.153 kN Member forces CA 2.5 R 1 2.5R 2.88 CD 1.5 1-0.6R -0.6 -0.9+0.54R -0.277 CB 2 -0.8R -0.8 1.28R 1.476 DB 2.5 -1.67+R 1 -4.175+2.5R -1.293 DA 2 1.33-0.8R -0.8 -2.128+1.28R -0.6523 BA 1.5 1-0.6R -0.6 -0.9+0.54R -0.277 ∆CA=∑ Fi∂Fi∂RLi = -5.603+4.86RR = 1.153 kNMember Length AE Member forces due to external loading, Fi Member forces due to unit load, fi FifiLiAE CA 2.5 2000000 46.1 2.88 0.00016596 CD 1.5 2000000 1.8576 -0.277 -0.00000038591 CB 2 2000000 23.603 1.476 0.000034838028 DB 2.5 2000000 -41.4 -1.293 0.00006691275 DA 2 2000000 -21.20 -0.6523 0.00001382876 BA 1.5 2000000 -8.94 -0.277 0.000001857285 ∆C = ∑FifiLiAE=4.7168×10-5 m=0.0472 mm Thus, the horizontal deflection at C is 0.0472 mm

Engineering

Civil Engineering

Consider the truss below. Using the force method, determine (a) the force in member AC, (b) forces in all members, (c) horizontal displacement at point C. E = 200 GPa. Consider that the cross-section area of the members is 100 mm x 100 mm square. 12 kN D C 9 kN 2 m B

Consider the truss below. Using the force method, determine (a) the force in member AC, (b) forces in all members, (c) horizontal displacement at point C. E = 200 GPa. Consider that the cross-section area of the members is 100 mm x 100 mm square. 12 kN D C 9 kN 2 m B

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Load on Structures

Civil engineers primarily deal with beams in their day-to-day applications. A beam is a structural member that resists shear loadings. There are various types of beams such as cantilever beam, simply supported beam, and over the hanging beam. All these beams have a different end and loading conditions.

Question

The diagram presents a truss structure with members connected at joints labeled A, B, C, and D. The truss is subject to external forces, including a 12 kN force acting to the left at joint D and a 9 kN force acting to the right at joint C.

Key dimensions and elements of the truss are as follows:

- The horizontal distance between points A and B is 1.5 m.
- The vertical height between points A and D, and points B and C, is 2 m.
- All members of the truss have a square cross-section with side dimensions of 100 mm × 100 mm.
- The modulus of elasticity (E) is given as 200 GPa.

You are tasked with determining:
(a) The force in member AC.
(b) The forces in all members of the truss.
(c) The horizontal displacement at point C.

This analysis involves using the force method, a structural analysis technique to calculate internal forces and displacements.

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Step 4 Forces in all members

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Step 5 Member forces due to unit applied load at C, unit load method

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