Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition)
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Chapter 10, Problem 10.1P
To determine

Determine the moments at the supports and draw bending moment diagram.

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Answer to Problem 10.1P

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  1

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  2

Explanation of Solution

Given information:

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  3

For AB,

In the member AB fixed end is subjected to the udl load with 25kN/m of the first half length.

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  4

Determine the fixed end moment for beam span AB, shown in the figure below, using the following formula.

MFAB=abwx×dx×x× (lx)2l2=11wl2192

MFAB=11×25×62192=51.56kNm

And,

MFBA=blwx×dx×x2× (lx)l2=5wl2192

MFBA=5×25×62192=23.44kNm

In the member BC, where B is roller support and C is fixed end support. It is subjected to point loads of with 15 kN, 15 kN and 15 kN.

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  5

Determine the fixed end moment for beam span BC, shown in the figure below, using the following formula.

MFBC=5Pl16

Where, P is the concentrated load acting on the beam BC.

MFBC=5×15×816=37.5kNm

By using slope deflection formula, calculate the moment at the end of the beam

MAB=MFAB+2EIl(2θA+θB+ 3Δl)MAB=51.56+2EI6(2×0+θB+ 3×06)=51.56+2EI6θB (1)

MBA=MFBA+2EIl(2θB+θA+ 3Δl)MAB=23.44+2EI6(2θB+0+ 3×06)=23.44+4EI6θB (2)

Similarly for the beam BC,

MBC=MFBC+2EIl(2θB+θC 3Δl)MBC=37.5+2EI8(2×θB+0 3×06)=37.5+4EI8θB (3) MCB=MFCB+2EIl(2θC+θB 3Δl)MCB=37.5+2EI8(2×0+θB 3×06)=37.5+2EI8θB (4)

Since the moment equilibrium at support B.

MB=0

MBA+MBC=0

23.44+4EI6θB37.5+4EI8θB=0θB=12.05EI

Now substitute the value of θB in equation 2, 3, 4 and 5, we get

MAB=51.56+2EI6( 12.05 EI)=47.54kN.mMBA=23.44+4EI6( 12.05 EI)=31.47kN.mMBC=37.5+2EI4( 12.05 EI)=31.47kN.mMCB=37.5+2EI8( 12.05 EI)=40.51kN.m

Now determine the shear reaction at the end, of span AB and BC.

For span AB

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  6

Moment about A

MA=031.47RB×6+25×3×3247.54=0RB1=16.07

And,

RA+RB=25×3=75kNSubstituteRBRA=7516.07=58.93kN

For Span BC

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  7

Moment about B

MB=040.51RC×8+15×6+15×6+15×631.47=0RC=23.63kN

And

RB+RC=15+15+15=45kNSubstituteRBRB2=4523.63=21.37kN

Shear force diagram for the whole beam.

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  8

At point A = 58.93kN

At right of point D = 58.93kN75kN=16.07kN

At point B = 21.37kN

At right of point B = 21.37kN

At point E = 21.37kN15kN=6.37kN

At right of point E = 6.37kN

At point F = 6.37kN15kN=8.63kN

At right of point F = 8.63kN

At point G 8.63kN15kN=23.63kN

At right of point F = 23.63kN

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  9

The maximum Positive sagging moment in span AB, where the shear force is equal to zero.

58.9325x=0x=2.3572m

The maximum positive sagging moment is determine by the below formula.

Mmax1=wl22MA=25× 2.35722247.54=21.91kN.m

Bending moment at point E.

ME=31.47+21.27×2=11.27kNm

Bending moment at point F.

MF=31.47+21.27×415×2=24.01kNm

Bending moment at point G.

MG=31.47+21.27×415×415×2=6.75kNm

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition), Chapter 10, Problem 10.1P , additional homework tip  10

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