   # 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

#### Solutions

Chapter
Section
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. 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.

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:

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) Position Influence line ordinate of SC (kN/kN) 0 A 13 4 B 0 8 C− −13 8 C+ 23 16 D 0 19 E −14 22 F −12

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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