# Determine the member end forces of the frames shown in Figs. P10.6–P10.8 by utilizing structural symmetry. FIG. P10.6, P10.21

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

Chapter
Section

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 10, Problem 21P
Textbook Problem
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## Determine the member end forces of the frames shown in Figs. P10.6–P10.8 by utilizing structural symmetry.FIG. P10.6, P10.21

To determine

Find the member end forces of the frame using structural symmetry.

### Explanation of Solution

Given information:

The structure is given in the Figure.

The young’s modulus E and area A is constant.

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.

Calculation:

Refer the given structure.

The structure is symmetric with respect to the vertical s axis passing through the hinge E.

Sketch the half loading for the given structure as shown in Figure 1.

Sketch the substructure with symmetric boundary conditions as shown in Figure 5.

Find the reactions and member end forces of substructure using equilibrium equations and to the left of s axis.

The member end forces to the right of s axis are obtained by the reflection.

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

+Fy=0(30)(8)+Ay=0Ay=240kN

Summation of moments about E is equal to 0.

ME=0Ax×12Ay×5+30×8×4=0Ax×12240×5+30×8×4=012Ax=240

Ax=20kN

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

+Fx=020Ex=0Ex=20kN

Summation of forces along x-direction is equal to 0

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