Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 8, Problem 8.4.16P

A W 200 x 41.7 wide-flange beam (see Table F-l(b), Appendix F) is simply supported with a span length of 2.5 m (see figure). The beam supports a concentrated load of 100 kN at 0.9 m from support B. At a cross section located 0,7 m from the left-hand support, determine the principal stresses tr, and 2 and the maximum shear stress r n M J t at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis,

  Chapter 8, Problem 8.4.16P, A W 200 x 41.7 wide-flange beam (see Table F-l(b), Appendix F) is simply supported with a span

(a).

Expert Solution
Check Mark
To determine

To find: Values of maximum stress and principal shear stress at top of beam.

Answer to Problem 8.4.16P

Values of principal stress :

σ1=151.6 MPa ,σ2=0 MPa

Maximum shear stressτmax=75.8 MPa

Explanation of Solution

Given Information:

Beam lengthL=2.5 m

Point loadP=100 kN

Dimensions of beam,

h=200b=165t=7.24h1=176.4

Concept Used:

Bending stressσx=MyI

Shear stressτxy=VQIt

Values of principal normal stress :σ1,2=σx+σy2±(σxσy2)2+τxy2

Maximum shear stress:τmax=(σxσy2)2+τxy2

Mechanics of Materials (MindTap Course List), Chapter 8, Problem 8.4.16P , additional homework tip  1

From equilibrium:

RB×2.5100×1.6=0RB=64 kN

So, bending moment at pointC is :

  M=RB×0.9=57.6 kN-m

Shear force at pointC is

  V=RB=64 kN

Moment of inertia:

  I=bh312bh1312+th1312 about the neutral z axis.I=165×200312165×176.4312+7.24×176.4312I=37837529.34 mm4I=3.8×105 m4

First, moment of area at the top of beam shall be zero,Q=0 m3

So, bending stress at top:

σx=MyIσx=57.6×0.13.8×105σx=151579 kPa =151.6 MPa

And shear stress at that point:

τ=VQItτ=0 MPa

For this situation no stress iny direction soσy=0

Values of Principal and normal stress are given by following equation:

σ1,2=σx+σy2±(σxσy2)2+τxy2σ1,2=151.6+02±(151.602)2+02σ1,2=75.8±75.8σ1=75.8+75.8=151.6 MPaσ2=75.875.8=0 MPa

Maximum shear stress:

τmax=(σxσy2)2+τxy2τmax=(151.602)2+02τmax=75.8 MPa

Conclusion:

Hence, we get:

Values of principal stress :

σ1=151.6 MPa ,σ2=0 MPa

Maximum shear stressτmax=75.8 MPa

(b).

Expert Solution
Check Mark
To determine

To find: The values of principal stress and principal stress at top of web.

Answer to Problem 8.4.16P

Values of principal stress:

σ1=146.13 MPa ,σ2=12.43 MPa.

Maximum shear stress:τmax=79.28 MPa.

Explanation of Solution

Given Information:

Beam lengthL=2.5 m

Point loadP=100 kN

Dimensions of beam,

h=200b=165t=7.24h1=176.4

Concept Used:

Bending stressσx=MyI

Shear stressτxy=VQIt

Values of principal normal stress:σ1,2=σx+σy2±(σxσy2)2+τxy2

Maximum shear stressτmax=(σxσy2)2+τxy2

Mechanics of Materials (MindTap Course List), Chapter 8, Problem 8.4.16P , additional homework tip  2

From equilibrium:

RB×2.5100×1.6=0RB=64 kN

So, bending moment at pointC is :

  M=RB×0.9=57.6 kN-m

Shear force at pointC is

  V=RB=64 kN

Moment of inertia:

  I=bh312bh1312+th1312 about the neutral z axis.I=165×200312165×176.4312+7.24×176.4312I=37837529.34 mm4I=3.8×105 m4

First, moment of area of flange:

  Q=A1×y1=165×200176.42×(176.42+200176.44)Q=183212.7 mm3=1.832×104 m3

So, bending stress at top of web:

σx=MyIσx=57.6×0.08823.8×105σx=133692.6 kPa =133.7 MPa

And shear stress at that point ::

τxy=VQItτxy=64×1.832×1043.8×105×0.00724τxy=42.62 MPa

For this situation no stress iny direction soσy=0 .

Values of principal normal stress are given by following equation:

σ1,2=σx+σy2±(σxσy2)2+τxy2σ1,2=133.7+02±(133.702)2+42.622σ1,2=66.85±79.28σ1=66.85+79.28=146.13 MPaσ2=66.8579.28=12.43 MPa

Maximum shear stress,

τmax=(σxσy2)2+τxy2τmax=(133.702)2+42.622τmax=79.28 MPa

Conclusion:

Hence, we get:

Values of principal stress :σ1=146.13 MPa ,σ2=12.43 MPa

Maximum shear stressτmax=79.28 MPa

(c).

Expert Solution
Check Mark
To determine

To find: Values of principal stress and maximum stress at neutral axis.

Answer to Problem 8.4.16P

Values of principal stress:

σ1=49.17 MPa,σ2=49.17 MPa

Maximum shear stressτmax=49.17 MPa.

Explanation of Solution

Given Information:

Beam lengthL=2.5 m

Point loadP=100 kN

Dimensions of beam:

h=200b=165t=7.24h1=176.4

Concept Used:

Bending stressσx=MyI

Shear stressτxy=VQIt

Principal normal stressesσ1,2=σx+σy2±(σxσy2)2+τxy2

Maximum shear stressτmax=(σxσy2)2+τxy2

Mechanics of Materials (MindTap Course List), Chapter 8, Problem 8.4.16P , additional homework tip  3

From equilibrium:

RB×2.5100×1.6=0RB=64 kN

So, bending moment at pointC is :

  M=RB×0.9=57.6 kN-m

Shear force at pointC is

  V=RB=64 kN

Moment of inertia,

  I=bh312bh1312+th1312 about the neutral z axis.I=165×200312165×176.4312+7.24×176.4312I=37837529.34 mm4I=3.8×105 m4

First moment of area for the section above the neutral axis:

  Q=A1×y1+A2×y2Q=165×200176.42×(176.42+200176.44)+7.24×176.42×(176.44)Q=211373.55 mm3=2.114×104 m3.

So, bending stress at neutral axis:

σx=M(y=0)Iσx=0 MPa

And shear stress at that point:

τxy=VQItτxy=64×2.114×1043.8×105×0.00724τxy=49.17 MPa

For this situation no stress iny direction soσy=0

Values of principal stress are given by following equation:

σ1,2=σx+σy2±(σxσy2)2+τxy2σ1,2=0+02±(002)2+49.172σ1,2=0±49.17σ1=0+49.17=49.17 MPaσ2=049.17=49.17 MPa

Maximum shear stress:

τmax=(σxσy2)2+τxy2τmax=(002)2+49.172τmax=49.17 MPa

Conclusion:

Hence, we get:

Values of principal stress:σ1=49.17 MPa ,σ2=49.17 MPa

Maximum shear stressτmax=49.17 MPa

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