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For the beam of Problem 8.10, determine the maximum positive and negative shears and the maximum positive and negative bending moments at point C due to a concentrated live load of 150 kN, a uniformly distributed live load of 50 kN/m, and a uniformly distributed dead load of 25 kN/m. FIG. P8.10, P8.11

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Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

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Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 9, Problem 6P
Textbook Problem
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For the beam of Problem 8.10, determine the maximum positive and negative shears and the maximum positive and negative bending moments at point C due to a concentrated live load of 150 kN, a uniformly distributed live load of 50 kN/m, and a uniformly distributed dead load of 25 kN/m.

Chapter 9, Problem 6P, For the beam of Problem 8.10, determine the maximum positive and negative shears and the maximum

FIG. P8.10, P8.11

To determine

Find the maximum positive and negative shears and the maximum positive and negative bending moments at point C.

Explanation of Solution

Given Information:

The concentrated live load (P) is 150 kN.

The uniformly distributed live load (wL) is 50 kN/m.

The uniformly distributed dead load (wD) is 25 kN/m.

Calculation:

Apply a 1 kN unit moving load at a distance of x from left end A.

Sketch the free body diagram of beam as shown in Figure 1.

Refer Figure 1.

Find the equation of support reaction (By) at B using equilibrium equation:

Take moment about point D.

Consider moment equilibrium at point D.

Consider clockwise moment as positive and anticlockwise moment as negative.

Sum of moment at point D is zero.

ΣMD=0By(12)1(16x)=012By=16xBy=43x12        (1)

Find the equation of support reaction (Dy) at D using equilibrium equation:

Apply vertical equilibrium equation of forces.

Consider upward force as positive (+) and downward force as negative ().

By+Dy=1

Substitute 43x12 for By.

43x12+Dy=1Dy=143+x12Dy=x1213        (2)

Influence line for the shear at point C.

Apply 1 kN load at just left of C.

Find the equation of shear force at C of portion AB (0x8m).

Sketch the free body diagram of the section AC as shown in Figure 2.

Refer Figure 2.

Apply equilibrium equation of forces.

Consider upward force as positive (+) and downward force as negative ().

ΣFy=0

BySC1=0SC=By1

Substitute 43x12 for By.

SC=(43x12)1=13x12

Apply 1 kN load at just right of C.

Find the equation of shear force at C of portion CF (8mx22m).

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

Refer Figure 3.

Apply equilibrium equation of forces.

Consider upward force as positive (+) and downward force as negative ().

ΣFy=0

BySC=0SC=By

Substitute 43x12 for By.

SC=(43x12)=43x12

Thus, the equations of the influence line for SC as follows

SC=13x12, 0x8m        (3)

SC=43x12, 8mx22m        (4)

Find the value of influence line ordinate of shear force SC at various points of x using the Equations (3) and (4) and summarize the value as in Table 1.

x (m)PositionInfluence line ordinate of SC(kN/kN)
0A13
4B0
8C13
8C+23
16D0
19E14
22F12

Draw the influence lines for the shear force at point C using Table 1 as shown in Figure 4.

Refer Figure 4.

The maximum positive ILD ordinate at point C is 23kN/kN.

The maximum negative ILD ordinate at point C is 13kN/kN.

Find the positive area (A1) of the influence line diagram of shear force at point C.

A1=12(LAB)(13)+12(LCD)(23)

Here, LAB is the length of beam between A to B and LCD is the length of beam between C to D.

Substitute 4 m for LAB and 8 m for LCD.

A1=12(4)(13)+12(8)(23)=3.333m

Find the negative area (A2) of the influence line diagram of shear force at point C.

A2=12(LBC)(13)+12(LDF)(12)

Here, LBC is the length of beam between B to C and LDF is the length of beam between E to F.

Substitute 4 m for LBC and 3 m for LDF.

A2=12(4)(13)+12(6)(12)=0.6671.5=2.167m

Find the maximum positive shear at point C using the equation.

Maximum positive SC=P(maximum positive ILD ordinate at C)+wL(A1)+wD(A1+A2)

Substitute 150 kN for P, 23kN/kN for maximum positive ILD ordinate at C, 50 kN/m for wL, 3.333m for A1, 25 kN/m for wD, and 2.167m for A2.

Maximum positive SC=150(23)+50(3.333)+25(3.3332.167)=100+166.67+29

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