   # 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. P12.3

#### Solutions

Chapter
Section
Chapter 12, Problem 3P
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. P12.3

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 DEF (w) is 20 kN/m.

The horizontal distance of the point AB and DE (L1) is 8 m.

The horizontal distance of the point BC and EF (L2) is 5 m.

The vertical distance of the members AD, BE, and CF (lv) is 8 m.

Calculation:

Consider the girder DE.

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

lm=0.8L1

Substitute 8 m for L1.

lm=0.8×8=6.4m

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

le=0.1L1

Substitute 8 m for L1.

le=0.1×8=0.8m

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

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

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

VDE=wlm2

Substitute 20 kN/m for w and 6.4 m for lm.

VDE=20×6.42=64kN

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=0SDw×leVDE=0

Substitute 20 kN/m for w, 0.8 m for le, and 64 kN for VDE.

SD20×0.864=0SD=80kN()

Determine the moment at the left end.

Take moment about left end is equal to zero.

MD=0MDw×le×le2VDE×le=0

Substitute 20 kN/m for w, 0.8 m for le, and 64 kN for VDE.

MD20×0.8×0.8264×0.8=0MD=57.6kNm(Counterclockwise)

Consider upward direction is positive and clockwise moment is positive.

Determine the support reaction at the right end.

Apply the equations of equilibrium to the right end portion.

FY=0SEw×leVDE=0

Substitute 20 kN/m for w, 0.8 m for le, and 64 kN for VDE.

SE20×0.864=0SE=80kN()

Determine the moment at the right end.

Take moment about right end is equal to zero.

ME=0MEw×le×le2VDE×le=0

Substitute 20 kN/m for w, 0.8 m for le, and 64 kN for VDE.

ME20×0.8×0.8264×0.8=0ME=57.6kNm(Clockwise)

Determine the maximum bending moment at the middle of the girder using the relation.

Mmax=wlm28

Substitute 20 kN/m for w and 6.4 m for lm.

Mmax=20×6.428=102

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