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
Question
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Chapter 3, Problem 28P
To determine

Sketch the free body diagram of each segment above support A and calculate the internal forces on each free body.

Expert Solution & Answer
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Answer to Problem 28P

The horizontal reaction at A is Ax=6kips()_.

The vertical reaction at A is Ay=21kips()_.

The moment at A is MA=113.75kipft_.

The vertical force at B for AB is By=19kips()_.

The internal moment at B for AB is MB=74.25kipft_.

The vertical force at joint B is By=21kips()_.

The internal moment at joint B is MB=83.75kipft_.

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions 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 counter clockwise moment as negative and the clockwise moment as positive.

Calculation:

Let Ax and Ay be the horizontal and vertical reaction at the support A and MA is the moment at support A.

Sketch the free body diagram of the structure as shown in Figure 1.

Fundamentals of Structural Analysis, Chapter 3, Problem 28P , additional homework tip  1

Refer to Figure 1.

Use equilibrium equations.

Find the horizontal reaction at A:

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

+Fx=0Ax+6=0Ax=6kips()

Hence, the horizontal reaction at A is Ax=6kips()_.

Find the vertical reaction at A:

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

+Fy=0Ay2×6.58=0Ay=21kips()

Hence, the vertical reaction at A is Ay=21kips()_.

Find the moment at A:

Summation of moment about A is equal to 0.

MA=08×6+2×6.5×(66.52)+6×5MA=0MA=113.75kipft

Hence, the moment at A is MA=113.75kipft_.

Sketch the free body diagram of the beam BC as shown in Figure 2.

Fundamentals of Structural Analysis, Chapter 3, Problem 28P , additional homework tip  2

Refer to Figure 2.

Use equilibrium equations.

Find the vertical force at B:

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

+Fy=0By2×5.58=0By=19kips()

Hence, the vertical force at B is By=19kips()_.

Find the internal moment at B:

Summation of moments about B is equal to 0.

MB=08×5.5+2×5.5×5.52MB=0MB=74.25kipft

Hence, the internal moment at B is MB=74.25kipft_.

Sketch the free body diagram of the joint B as shown in Figure 3.

Fundamentals of Structural Analysis, Chapter 3, Problem 28P , additional homework tip  3

Refer to Figure 3.

Use equilibrium equations.

Find the vertical force at joint B:

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

+Fy=0By192×1=0By=21kips()

Hence, the vertical force at joint B is By=21kips()_.

Find the internal moment at joint B:

Summation of moments about B’ is equal to 0.

MB=074.25+19×12MB=0MB=83.75kipft

Hence, the internal moment at joint B is MB=83.75kipft_.

Sketch the free body diagram of the column AB as shown in Figure 4.

Fundamentals of Structural Analysis, Chapter 3, Problem 28P , additional homework tip  4

Refer to Figure 1.

Use equilibrium equations.

Find the vertical force at A:

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

+Fy=0Ay=21kips()

Hence, the vertical force at A is Ay=21kips()_.

Find the moment at A:

Summation of moment about A is equal to 0.

MA=083.75+6×5MA=0MA=113.75kipft

Therefore, the moment at A is MA=113.75kipft_.

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