   # 5.57 through 5.71 Draw the shear, bending moment, and axial force diagrams and the qualitative deflected shape for the frame shown.

#### Solutions

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Chapter 5, Problem 70P
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## 5.57 through 5.71 Draw the shear, bending moment, and axial force diagrams and the qualitative deflected shape for the frame shown. To determine

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

### Explanation of Solution

Write the condition for static instability, determinacy and indeterminacy of plane frames as follows:

3m+r<3j+ecstaticallyunstableframe        (1)

3m+r=3j+ecstaticallydeterminateframe        (2)

3m+r>3j+ecstaticallyindeterminateframe        (3)

Here, number of members is m, number of external reactions is r, the number of joints is j, and the number of elastic hinges is ec.

Find the degree of static indeterminacy (i) using the equation;

i=(3m+r)(3j+ec)        (4)

Refer to the Figure in the question;

The number of members (m) is 3.

The number of external reactions (r) is 4.

The number of joints (j) is 4.

The number of elastic hinges ec is 1.

Substitute the values in Equation (2);

3(3)+4=3(4)+113=13staticallydeterminateframe

Show the free-body diagram of the entire frame as in Figure 1.

Refer Figure 1,

Consider entire frame.

+MA=0By(30)2(30)(302)30(20)=030By900600=0By=50k

Find the vertical reaction at point A by resolving the vertical component of forces.

+Fy=0Ay2(30)+By=0Ay60+50=0Ay=10k

Find the horizontal reaction at point A by resolving the horizontal component of forces.

+Fx=0Ax+30Bx=0        (1)

Consider the section DEB.

Find the horizontal reaction at point B by taking moment about the hinge at D.

+MDDEB=0By(15)2(15)(152)Bx(20)=050(15)225Bx(20)=0Bx=26.25k

Substitute 26.25 k for Bx in Equation (1).

Ax+3026.25=0Ax=3.75k

Show the free-body diagram of the members and joints of the entire frame as in Figure 2.

Consider point A:

Resolve the vertical component of forces.

+FY=010AYAC=0AYAC=10k

Resolve the horizontal component of forces.

+FX=0AXAC3.75=0AXAC=3.75k

Consider the member AC:

Resolve the vertical component of forces.

+FY=0AYACCYAC=010CYAC=0CYAC=10k

Resolve the horizontal component of forces.

+FX=0AXAC+CXAC=03.75+CXAC=0CXAC=3.75k

Take moment about the point C.

+MCAC=0MCACAXAC(20)=0MCAC3.75(20)=0MCAC=75k-ft

Consider the point C:

Resolve the vertical component of forces.

+FY=0CYACCYCE=010CYCE=0CYCE=10k

Resolve the horizontal component of forces

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