# 13.13 through 13.25 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.13–P13.25 using the method of consistent deformations. FIG. P13.21

### 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 13, Problem 21P
Textbook Problem
76 views

## 13.13 through 13.25 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.13–P13.25 using the method of consistent deformations.FIG. P13.21

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

### Explanation of Solution

Given information:

The structure is given in the Figure.

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:

Find the degree of indeterminacy of the structure:

Degree of determinacy of the frame is equal to the number of unknown reactions minus the number of equilibrium equations.

The frame is supported by 4 support reactions and the number of equilibrium equations is 3.

Therefore, the degree of indeterminacy of the frame is i=1.

Select the horizontal reaction Cx at the support C as redundant.

Sketch the support reactions for the frame and the frame with unit value for the unknown redundant Cx as shown in Figure 1.

Find the reactions at the supports using equilibrium equations:

Summation of moments of all forces about A is equal to 0.

MA=0Cy(20)+3(20)(202)3(5)(52)+15(5)=020Cy=60037.5+75Cy=31.875k

Cy31.88k

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

+Fy=0Ay3(25)+Cy=0Ay3(25)+31.875=0Ay=43.13k

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

+Fx=0Ax15=0Ax=15k

For unit value of the unknown redundant Cx:

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

+Fx=01Cx=0Cx=1k

Summation of moments of all forces about A is equal to 0.

MA=0Cy(20)1(15)=0Cy=1520Cy=34k

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

+Fy=0AyCy=0Ay34=0Ay=34k

Let the deflection at point C due to external loading is ΔCO and the flexibility coefficient representing the deflection at C due to unit value of redundant Cx is fCC.

The moment of inertia is 2I for span CDE.

Calculate the value of ΔCO as follows:

ΔCO=1EI[12020(31.88x1.5x2)(34x)dx+05(15x)(x)dx+515(15x+15(x5))(x)dx]=1EI[12020(23.91x2+1.125x3)dx+05(15x2)dx+515(75x)dx]=1EI[[23.91x36+1.125x48]020+[15x33]05+[75x22]515]=1EI(23.91(20)36+1

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