   # 12.1 through 12.5 Draw the approximate shear and bending moment diagrams for the girders of the frames shown in Figs. P12.1 through P12.5. FIG. P12.2

#### Solutions

Chapter
Section
Chapter 12, Problem 2P
Textbook Problem
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## 12.1 through 12.5 Draw the approximate shear and bending moment diagrams for the girders of the frames shown in Figs. P12.1 through P12.5. FIG. P12.2

To determine

Draw the shear and bending moment diagrams for the girders of given frame.

### Explanation of Solution

Given information:

The uniformly distributed load acting along the girder EF and CD (w) is 1.5 k/ft.

The horizontal distance of the member EF and CD (L) is 20 ft.

The vertical distance of the members AC, CE, BD, and DF (lv) is 12 ft.

Calculation:

The span length and loads for the two girders of the frame EF and CD are same; therefore the approximate shear and bending moment diagrams for the girders will also be the same.

Determine the span for the middle portion of the girder using the relation.

lm=0.8L

Substitute 20 ft for L.

lm=0.8×20=16ft

Determine the span for the two end portion of the girder using the relation.

le=0.1L

Substitute 20 ft for L.

le=0.1×20=2ft

Draw the statically determinate girder portion as in Figure (1).

Consider the equilibrium of the simply supported middle portion of the girder.

Determine the vertical reactions at the end portion using the relation.

V=wlm2

Substitute 1.5 k/ft for w and 16 ft for lm.

V=1.5×162=12k

Consider the equilibrium conditions of the end portions of the girder.

Consider upward direction is positive and counter clockwise moment is positive.

Determine the support reaction at the left end.

Apply the equations of equilibrium to the left end portion.

FY=0SLw×leV=0

Substitute 1.5 k/ft for w, 2 ft for le, and 12 k for V.

SL1.5×212=0SL=15k()

Determine the moment at the left end.

Take moment about left end is equal to zero.

ML=0MLw×le×le2V×le=0

Substitute 1

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