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 8.4.14P

A beam with a wide-flange cross section (see figure) has the following dimensions: h = 120 mm, r = 10 mm, h = 300 mm, and /ij = 260 mm. The beam is simply supported with span length L = 3,0 im A concentrated load P = 120 kN acts at the midpoint of the span.

At across section located 1.0 m from the left-hand support, determine the principal stresses tr, and tr2and the maximum shear stress Tmax at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axisChapter 8, Problem 8.4.14P, A beam with a wide-flange cross section (see figure) has the following dimensions: h = 120 mm, r =

(a).

Expert Solution
Check Mark
To determine

Find principal stresses and maximum shear stress at top of beam.

Answer to Problem 8.4.14P

Principal stressesσ1=82.64 MPa ,σ2=0 MPa

Maximum shear stressτmax=41.32 MPa

Explanation of Solution

Given Information:

Beam lengthL=3 m

Point loadP=120 kN

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 8.4.14P , additional homework tip  1

So bending moment at pointC is

  M=P2×L3=1202×33=60 kN-m

Shear force at pointC is

  V=P2=60 kN

Moment of inertia,

  I=bh312bh1312+th1312 about the neutral z axis.I=120×300312120×260312+10×260312I=108886666.7 mm4I=1.089×104 m4

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

So bending stress at top,

σx=MyIσx=60×0.151.089×104σx=82644.63 kPa =82.64 MPa

And shear stress at that point,

τ=VQItτ=0 MPa

For this situation no stress iny direction soσy=0

Principal normal stresses are given by following equation,

σ1,2=σx+σy2±(σxσy2)2+τxy2σ1,2=82.64+02±(82.6402)2+02σ1,2=41.32±41.32σ1=41.32+41.32=82.64 MPaσ2=41.3241.32=0 MPa

Maximum shear stress,

τmax=(σxσy2)2+τxy2τmax=(82.6402)2+02τmax=41.32 MPa

Conclusion:

Hence we get,

Principal stresses

σ1=82.64 MPa ,σ2=0 MPa

Maximum shear stressτmax=41.32 MPa

(b).

Expert Solution
Check Mark
To determine

Find principal stresses and maximum shear stress at top of web.

Answer to Problem 8.4.14P

Principal stresses

σ1=76.13 MPa ,σ2=4.5 MPa

Maximum shear stressτmax=40.32 MPa

Explanation of Solution

Given Information:

Beam lengthL=3 m

Point loadP=120 kN

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 8.4.14P , additional homework tip  2

So bending moment at pointC is

  M=P2×L3=1202×33=60 kN-m

Shear force at pointC is

  V=P2=60 kN

Moment of inertia,

  I=bh312bh1312+th1312 about the neutral z axis.I=120×300312120×260312+10×260312I=108886666.7 mm4I=1.089×104 m4

First moment of area of flange,

  Q=A1×y1=120×3002602×(2602+3002604)=336000 mm3=3.36×104 m3

So bending stress at top of web,

σx=MyIσx=60×0.131.089×104σx=71625.344 kPa =71.63 MPa

And shear stress at that point,

τxy=VQItτxy=60×3.36×1041.089×104×0.01τxy=18.51 MPa

For this situation no stress iny direction soσy=0

Principal normal stresses are given by following equation,

σ1,2=σx+σy2±(σxσy2)2+τxy2σ1,2=71.63+02±(71.6302)2+18.512σ1,2=35.815±40.315σ1=35.815+40.315=76.13 MPaσ2=35.81540.315=4.5 MPa

Maximum shear stress,

τmax=(σxσy2)2+τxy2τmax=(71.6302)2+18.512τmax=40.32 MPa

Conclusion:

Hence we get,

Principal stresses

σ1=76.13 MPa ,σ2=4.5 MPa

Maximum shear stressτmax=40.32 MPa

(c).

Expert Solution
Check Mark
To determine

Find principal stresses and maximum shear stress at neutral axis.

Answer to Problem 8.4.14P

Principal stresses

σ1=23.17 MPa ,σ2=23.17MPa

Maximum shear stressτmax=23.17 MPa

Explanation of Solution

Given Information:

Beam lengthL=3 m

Point loadP=120 kN

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 8.4.14P , additional homework tip  3

So bending moment at pointC is

  M=P2×L3=1202×33=60 kN-m

Shear force at pointC is

  V=P2=60 kN

Moment of inertia,

  I=bh312bh1312+th1312 about the neutral z axis.I=120×300312120×260312+10×260312I=108886666.7 mm4I=1.089×104 m4

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

  Q=A1×y1+A2×y2Q=120×3002602×(2602+3002604)+10×2602×(2604)Q=420500 mm3=4.205×104 m3

So bending stress at neutral axis,

σx=MyIσx=60×01.089×104σx=0 MPa

And shear stress at that point,

τxy=VQItτxy=60×4.205×1041.089×104×0.01τxy=23.17 MPa

For this situation no stress iny direction soσy=0

Principal normal stresses are given by following equation,

σ1,2=σx+σy2±(σxσy2)2+τxy2σ1,2=0+02±(002)2+23.172σ1,2=0±23.17σ1=0+23.17=23.17 MPaσ2=023.17=23.17 MPa

Maximum shear stress,

τmax=(σxσy2)2+τxy2τmax=(002)2+23.172τmax=23.17 MPa

Conclusion:

Hence we get,

Principal stresses

σ1=23.17 MPa ,σ2=23.17MPa

Maximum shear stressτmax=23.17 MPa

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