   # 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

#### Solutions

Chapter
Section
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.

### Explanation of Solution

Given Information:

The concentrated live load (P) is 30 k.

The uniformly distributed live load (wL) is 3 k/ft.

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 SD by moving end D of the portion CD downward by Δ1 and end D of the portion ED upward by Δ2 to obtain the deflected shape.

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 (SD) at D is the similar of deflected shape.

Sketch the general shape of the influence line for shear (SD) at 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 (Ay) at A.

Consider moment at B from A.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMB=0Ay×20=0Ay=0

Find the vertical reaction (Gy) at G.

Consider moment at F from G.

Consider clockwise moment as negative and anticlockwise moment as positive.

ΣMF=0Gy×20=0Gy=0

Find the vertical reactions Cy and Ey.

Consider moment at B from G.

Consider clockwise moment as negative and anticlockwise moment as positive.

ΣMB=0Gy(110)+Ey(80)1(50)+Cy(20)=0

Substitute 0 for Gy.

(Gy)(110)+Ey(80)1(50)+Cy(20)=080Ey+20Cy=50        (1)

Consider moment at F from A.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMF=0Ay(115)+Cy(75)1(45)+Ey(15)=0

Substitute 0 for Ay.

(0)(115)+Cy(75)1(45)+Ey(15)=075Cy+15Ey=45        (2)

Solve Equations (1) and (2),

Cy=0.5kEy=0.5k

Find the shear at D (SD,L) when unit load 1 k placed just left of D using the equation:

SD,L=Ey

Substitute 0.5 k for Ey.

SD,L=0.5k

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

Find the shear at D (SD,R) when unit load 1 k placed just right of D using the equation:

SD,R=Cy

Substitute 0.5 k for Cy.

SD,R=0.5k

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 SD at 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 SD (k/k) 0 A 0 20 B 13 40 C 0 70 D− −12 70 D+ 12 100 E 0 115 F −14 130 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 12k/k.

The maximum negative ILD ordinate is 12k/k.

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

A1=12(LAC)(13)+12(LDE)(12)

Here, LAC is the length of beam between A to C and LDE is the length of beam between D to E.

Substitute 40 ft for LAC and 30 ft for LDE.

A1=12(40)(13)+12(30)(12)=6.667+7.5=14.167ft

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

A2=12(LCD)(12)+12(LEG)(14)

Here, LCD is the length of beam between C to D and LEG is the length of beam between E to G.

Substitute 30 ft for LCD and 30 ft for LEG.

A2=12(30)(12)+12(30)(14)=7.53.75=11.25ft

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

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

Substitute 30 k for P, 12k/k for maximum positive ILD ordinate, 3 k/ft for wL, 14.167ft for A1, 1 k/ft for wD, and 11

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