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6th Edition

KASSIMALI + 1 other

Publisher: Cengage,

ISBN: 9781337630931

Chapter 9, Problem 7P

Textbook Problem

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For the beam of Problem 8.23, determine the maximum positive and negative shears and the maximum positive and negative bending moments at point *D* due to a concentrated live load of 30 k, a uniformly distributed live load of 3 k/ft, and a uniformly distributed dead load of 1 k/ft.

FIG. P8.23, P8.25

To determine

Draw the influence lines for the shear and bending moment at point *D*.

**Given Information:**

The concentrated live load (*P)* is 30 k.

The uniformly distributed live load

The uniformly distributed dead load

**Calculation:**

Use Muller-Breslau's principle.

Influence line for the shear at point *D*.

Cut the beam at *D* to obtain the released structure of the beam. Next apply a small relative displacement in positive direction of *D* of the portion *CD* downward by *D* of the portion *ED* upward by

Sketch the deflected shape of the released beam when the beam cut at *D* as shown in Figure 1.

The general shape of the influence line for shear *D* is the similar of deflected shape.

Sketch the general shape of the influence line for shear *D* as shown in Figure 2.

Find the numerical value of influence line ordinate at *D*.

Place the 1 k load first just to the left of *D* and then just to the right of *D*.

Sketch the free body diagram of beam as shown in Figure 3.

Find the vertical reaction *A*.

Consider moment at *B* from *A*.

Consider clockwise moment as positive and anticlockwise moment as negative.

Find the vertical reaction *G*.

Consider moment at *F* from *G*.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the vertical reactions

Consider moment at *B* from *G*.

Consider clockwise moment as negative and anticlockwise moment as positive.

Substitute 0 for

Consider moment at *F* from *A*.

Consider clockwise moment as positive and anticlockwise moment as negative.

Substitute 0 for

Solve Equations (1) and (2),

Find the shear at *D* *D* using the equation:

Substitute 0.5 k for

Thus, the value of influence line ordinate just left of *D* is ‑0.5 k/k.

Find the shear at *D* *D* using the equation:

Substitute 0.5 k for

Thus, the value of influence line ordinate just right of *D* is 0.5 k/k.

The numerical values of influence line ordinate in different points on the beam for shear *D* are determined using geometry (similar triangle) of the influence line and summarize the values as in Table 4.

x (ft) | Points | Influence line ordinate of |

0 | A | 0 |

20 | ||

40 | C | 0 |

70 | ||

70 | ||

100 | E | 0 |

115 | F | |

130 | G | 0 |

Sketch the influence line diagram for shear at point *D* as shown in Figure 4.

Refer Figure 4.

The maximum positive ILD ordinate is

The maximum negative ILD ordinate is

Find the positive area *D*.

Here, *A* to *C* and *D* to *E*.

Substitute 40 ft for

Find the negative area *D*.

Here, *C* to *D* and *E* to *G*.

Substitute 30 ft for

**Find the maximum positive shear at point D using the equation.**

Substitute 30 k for *P*,

Structural Analysis

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