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
bartleby

Videos

Textbook Question
Book Icon
Chapter 8, Problem 8.4.15P

A beam with a wide-flange cross section (see figure) has the following dimensions: b = 5 in., t = 0.5 in,, ft = 12 in., and /?, = 10.5 in. The beam is simply supported with span length L = 10 ft and supports a uniform load q = 6 kips/fL

Calculate the principal stresses *rl and and the maximum shear stress t__ at a cross section located

[|] JA

3 ft from the left-hand support at each of the following locations: (a) the bottom of the beam, (b) the bottom of the web, and (c) the neutral axisChapter 8, Problem 8.4.15P, A beam with a wide-flange cross section (see figure) has the following dimensions: b = 5 in., t =

(a).

Expert Solution
Check Mark
To determine

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

Answer to Problem 8.4.15P

Principal stresses

σ1=6.8 ksi ,σ2=0 ksi

Maximum shear stressτmax=6.8 ksi

Explanation of Solution

Given Information:

Beam lengthL=10 ft=120 in

Uniform loadq=6 kips/ft

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 8.4.15P , additional homework tip  1

So bending moment at pointC is

  M=qL2×1q2=6×102×162=27 kip-ft=324 kip-in

Shear force at pointC is

  V=qL2q=6×1026=24 kip

Moment of inertia,

  I=bh312bh1312+th1312 about the neutral z axis.I=5×123125×10.5312+0.5×10.5312I=285.9 in4

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

So bending stress at top,

σx=MyIσx=324×6285.9σx=6.8 ksi

And shear stress at that point,

τ=VQItτ=0 ksi

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=6.8+02±(6.802)2+02σ1,2=3.4±3.4σ1=3.4+3.4=6.8 ksiσ2=3.43.4=0 ksi

Maximum shear stress,

τmax=(σxσy2)2+τxy2τmax=(6.802)2+02τmax=6.8 ksi

Conclusion:

Hence we get,

Principal stresses

σ1=6.8 ksi ,σ2=0 ksi

Maximum shear stressτmax=6.8 ksi

(b).

Expert Solution
Check Mark
To determine

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

Answer to Problem 8.4.15P

Principal stresses

σ1=7.6 ksi ,σ2=1.65 ksi

Maximum shear stressτmax=4.626 ksi

Explanation of Solution

Given Information:

Beam lengthL=10 ft=120 in

Uniform loadq=6 kips/ft

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 8.4.15P , additional homework tip  2

So bending moment at pointC is

  M=qL2×1q2=6×102×162=27 kip-ft=324 kip-in

Shear force at pointC is

  V=qL2q=6×1026=24 kip

Moment of inertia,

  I=bh312bh1312+th1312 about the neutral z axis.I=5×123125×10.5312+0.5×10.5312I=285.9 in4

First moment of area of flange,

  Q=A1×y1=5×1210.52×(10.52+1210.54)=21.1 in3

So bending stress at top of web,

σx=MyIσx=324×5.25285.9σx=5.95 ksi

And shear stress at that point,

τxy=VQItτxy=24×21.1285.9×0.5τxy=3.5425 ksi

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=5.95+02±(5.9502)2+3.54252σ1,2=2.975±4.626σ1=2.975+4.626=7.6 ksiσ2=2.9754.626=1.65 ksi

Maximum shear stress,

τmax=(σxσy2)2+τxy2τmax=(5.9502)2+3.54252τmax=4.626 ksi

Conclusion:

Hence we get,

Principal stresses

σ1=7.6 ksi ,σ2=1.65 ksi

Maximum shear stressτmax=4.626 ksi

(c).

Expert Solution
Check Mark
To determine

Find principal stresses and maximum shear stress at neutral axis.

Answer to Problem 8.4.15P

Principal stresses

σ1=4.7 ksi ,σ2=4.7 ksi

Maximum shear stressτmax=4.7 ksi

Explanation of Solution

Given Information:

Beam lengthL=10 ft=120 in

Uniform loadq=6 kips/ft

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 8.4.15P , additional homework tip  3

So bending moment at pointC is

  M=qL2×1q2=6×102×162=27 kip-ft=324 kip-in

Shear force at pointC is

  V=qL2q=6×1026=24 kip

Moment of inertia,

  I=bh312bh1312+th1312 about the neutral z axis.I=5×123125×10.5312+0.5×10.5312I=285.9 in4

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

  Q=A1×y1+A2×y2Q=5×1210.52×(10.52+1210.54)+0.5×10.52×(10.54)Q=28 in3

So bending stress at neutral axis,

σx=M(y=0)Iσx=0 ksi

And shear stress at that point,

τxy=VQItτxy=24×28285.9×0.5τxy=4.7 ksi

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+4.72σ1,2=0±4.7σ1=0+4.7=4.4 ksiσ2=04.7=4.7 ksi

Maximum shear stress,

τmax=(σxσy2)2+τxy2τmax=(002)2+4.72τmax=4.7 ksi

Conclusion:

Hence we get,

Principal stresses

σ1=4.7 ksi ,σ2=4.7 ksi

Maximum shear stressτmax=4.7 ksi

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 8 Solutions

Mechanics of Materials (MindTap Course List)

Ch. 8 - A spherical stainless-steel tank having a diameter...Ch. 8 - Solve the preceding problem if the diameter is 480...Ch. 8 - : A hollow, pressurized sphere having a radius r =...Ch. 8 - A fire extinguisher tank is designed for an...Ch. 8 - Prob. 8.3.2PCh. 8 - A scuba t a n k (see fig ure) i s bci ng d e...Ch. 8 - A tall standpipc with an open top (see figure) has...Ch. 8 - An inflatable structure used by a traveling circus...Ch. 8 - A thin-walled cylindrical pressure vessel of a...Ch. 8 - A strain gage is installed in the longitudinal...Ch. 8 - A circular cylindrical steel tank (see figure)...Ch. 8 - A cylinder filled with oil is under pressure from...Ch. 8 - Solve the preceding problem if F =90 mm, F = 42...Ch. 8 - A standpipe in a water-supply system (see figure)...Ch. 8 - A cylindrical tank with hemispherical heads is...Ch. 8 - : A cylindrical tank with diameter d = 18 in, is...Ch. 8 - A pressurized steel tank is constructed with a...Ch. 8 - Solve the preceding problem for a welded Tank with...Ch. 8 - A wood beam with a cross section 4 x 6 in. is...Ch. 8 - Prob. 8.4.2PCh. 8 - A simply supported beam is subjected to two point...Ch. 8 - A cantilever beam with a width h = 100 mm and...Ch. 8 - A beam with a width h = 6 in. and depth h = 8 in....Ch. 8 - Beam ABC with an overhang BC is subjected to a...Ch. 8 - A cantilever beam(Z, = 6 ft) with a rectangular...Ch. 8 - Solve the preceding problem for the following...Ch. 8 - A simple beam with a rectangular cross section...Ch. 8 - An overhanging beam ABC has a guided support at A,...Ch. 8 - Solve the preceding problem if the stress and...Ch. 8 - A cantilever wood beam with a width b = 100 mm and...Ch. 8 - . A cantilever beam (width b = 3 in. and depth h =...Ch. 8 - A beam with a wide-flange cross section (see...Ch. 8 - A beam with a wide-flange cross section (see...Ch. 8 - A W 200 x 41.7 wide-flange beam (see Table F-l(b),...Ch. 8 - A W 12 x 35 steel beam is fixed at A. The beam has...Ch. 8 - A W 360 x 79 steel beam is fixed at A. The beam...Ch. 8 - A W 12 X 14 wide-flange beam (see Table F-l(a),...Ch. 8 - A cantilever beam with a T-section is loaded by an...Ch. 8 - Beam A BCD has a sliding support at A, roller...Ch. 8 - , Solve the preceding problem using the numerical...Ch. 8 - A W 12 x 35 steel cantilever beam is subjected to...Ch. 8 - A W 310 x 52 steel beam is subjected to a point...Ch. 8 - A solid circular bar is fixed at point A. The bar...Ch. 8 - A cantilever beam with a width h = 100 mm and...Ch. 8 - Solve the preceding problem using the following...Ch. 8 - A cylindrical tank subjected to internal...Ch. 8 - A cylindrical pressure vessel having a radius r =...Ch. 8 - A pressurized cylindrical tank with flat ends is...Ch. 8 - A cylindrical pressure vessel with flat ends is...Ch. 8 - The tensional pendulum shown in the figure...Ch. 8 - The hollow drill pipe for an oil well (sec figure)...Ch. 8 - Solve the preceding problem if the diameter is 480...Ch. 8 - . A segment of a generator shaft with a hollow...Ch. 8 - A post having a hollow, circular cross section...Ch. 8 - A sign is supported by a pole of hollow circular...Ch. 8 - A sign is supported by a pipe (see figure) having...Ch. 8 - A traffic light and signal pole is subjected to...Ch. 8 - Repeat the preceding problem but now find the...Ch. 8 - A bracket ABCD having a hollow circular cross...Ch. 8 - A gondola on a ski lift is supported by two bent...Ch. 8 - Beam A BCD has a sliding support at A, roller...Ch. 8 - A double-decker bicycle rack made up of square...Ch. 8 - A semicircular bar AB lying in a horizontal plane...Ch. 8 - Repeat Problem 8.5-22 but replace the square tube...Ch. 8 - An L-shaped bracket lying in a horizontal plane...Ch. 8 - A horizontal bracket ABC consists of two...Ch. 8 - , An arm A BC lying in a horizontal plane and...Ch. 8 - A crank arm consists of a solid segment of length...Ch. 8 - A moveable steel stand supports an automobile...Ch. 8 - A mountain bike rider going uphill applies a force...Ch. 8 - Determine the maximum tensile, compressive, and...Ch. 8 - Prob. 8.5.32PCh. 8 - A plumber's valve wrench is used to replace valves...Ch. 8 - A compound beam ABCD has a cable with force P...Ch. 8 - A steel hanger bracket ABCD has a solid, circular...
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Mechanics of Materials Lecture: Beam Design; Author: UWMC Engineering;https://www.youtube.com/watch?v=-wVs5pvQPm4;License: Standard Youtube License