Fundamentals of Structural Analysis
Fundamentals of Structural Analysis
5th Edition
ISBN: 9780073398006
Author: Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher: McGraw-Hill Education
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Chapter 8, Problem 1P
To determine

Find the horizontal and vertical displacement of joint B.

Expert Solution & Answer
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Answer to Problem 1P

The horizontal deflection at joint B is 0.70in._.

The vertical deflection at joint B is 0.28in._.

Explanation of Solution

Given information:

Area of all bars is 4in.2 and value of E is 24,000kips/in.2.

Procedure to find the deflection of truss by virtual work method is shown below.

  • For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.
  • For virtual system: Remove all given real loads, apply a unit load at the joint where deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces (Fv) in all the member of the truss.
  • Finally use the desired deflection equation.

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 (Fx=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 (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
  • For summation of moment about a point is equal to zero (Matapoint=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).

Condition for zero force members:

  1. 1. If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.
  2. 2. If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the bar forces FP produced by the P-system as follows:

Let Dx and Dy be the horizontal and vertical reactions at the hinged support D.

Let Ey be the vertical reaction at the roller support E.

Find the reactions at the supports using equilibrium equations:

Summation of moments about E is equal to 0.

ME=0Dy(20)100(30)=0Dy=150kips

Summation of forces along y-direction is equal to 0.

+Fy=0DyEy=0Ey=150kips

Summation of forces along x-direction is equal to 0.

+Fx=0Dx+100=0Dx=100kips

Find the member forces using method of joints:

The force in the member AB, AE, and DE are zero as it satisfies zero force member condition.

Apply equilibrium equation to the joint C:

+Fx=0100FBCcos36.87°=0FBC=125kips

+Fy=0FCDFBCsin36.87°=0FCD(125)sin36.87°=0FCD=75kips

Apply equilibrium equation to the joint D:

+Fx=0100+FBDcos36.87°=0FBD=125kips

Apply equilibrium equation to the joint E:

+Fy=0150+FBE=0FBE=150kips

Sketch the bar forces produced by the P-system as shown in Figure 1.

Fundamentals of Structural Analysis, Chapter 8, Problem 1P , additional homework tip  1

Consider a dummy load of 1 kip directed vertically at joint B with the bar forces FQ1.

The vertical reaction at E is Ey=1kip.

The force in the member AB, AE, DE, BD, BC, and CD are zero as it satisfies zero force member condition.

Apply equilibrium equation to the joint E:

+Fy=01+FBE=0FBE=1kips

Sketch the bar forces FQ1 and reactions produced by the vertical dummy load as shown in Figure 2.

Fundamentals of Structural Analysis, Chapter 8, Problem 1P , additional homework tip  2

Consider a dummy load of 1 kip directed horizontally at joint B with the bar forces FQ2.

Sketch the bar forces FQ2 and reactions produced by the horizontal dummy load as shown in Figure 3.

Fundamentals of Structural Analysis, Chapter 8, Problem 1P , additional homework tip  3

Summation of moments about E is equal to 0.

ME=0Dy(20)1(15)=0Dy=0.75kips

Summation of forces along y-direction is equal to 0.

+Fy=0DyEy=0Ey=0.75kips

Summation of forces along x-direction is equal to 0.

+Fx=0Dx+1=0Dx=1kips

The force in the member AB, AE, DE, BC, and CD are zero as it satisfies zero force member condition.

Apply equilibrium equation to the joint D:

+Fx=01+FBDcos36.87°=0FBC=1.25kips

Apply equilibrium equation to the joint E:

+Fy=00.75+FBE=0FBE=0.75kips

Find the product of FPFQ1L and FPFQ2L for each member as shown in Table 1.

Bar

L

(ft)

FP(kips)FQ1(kips)FQ2(kips)FPFQ1L(kips2ft)FPFQ2L(kips2ft)
AB2500000
BC251250000
CD30750000
DE2000000
AE2000000
BE1515010.752,2501,687.5
BD2512501.2503,906.25
    2,2505,593.75

Find the vertical deflection at joint B (δBy):

(1kip)(δBy)=FPFQ1LAEδBy=2,250kips2ft(4in.2)(24,000kips/in.2)=2,250kips2ft(12in.1ft)(4in.2)(24,000kips/in.2)=0.28in.=0.28in.

Therefore, the vertical deflection at joint B is 0.28in._.

Find the horizontal deflection at joint B (δBx):

(1kip)(δBx)=FPFQ2LAEδBx=5,593.75kips2ft(4in.2)(24,000kips/in.2)=5,593.75kips2ft(12in.1ft)(4in.2)(24,000kips/in.2)=0.70in.

Therefore, the horizontal deflection at joint B is 0.70in._.

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