Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4.7, Problem 120P

a)

To determine

Show that the maximum compressive stresses are in the ratio 4:5:7:9.

a)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The load act on the point of the bars is P.

Calculation:

At the point A:

Show the cross-sectional diagram of the square bar as in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 4.7, Problem 120P , additional homework tip  1

Here, a is sides of the square bar.

Refer to Figure 1.

The maximum compressive stress of the square bar (σA)s is as follows:

(σA)s=PAsPecAIs=PAs(1+AsecAIs) (1)

Here, e is the eccentricity of the load and cA is the distance between the centroid from extreme fibre.

The cross-sectional area of the square bar (As) is a2.

The eccentricity of the load (e) is a2.

The distance between the centroid from extreme fibre (cA) is a2.

The moment of inertia (Is) of the section of the square bar is a412.

Calculate the maximum compressive stress of the square bar (σA)s:

Substitute a2 for As, a2 for e, a2 for cA, and a412 for Is in Equation (1).

(σA)s=PAs(1+a2×a2×a2a412)=PAs12a4P4a4As=PAs3PAs=4PAs

Show the cross-sectional diagram of the circular bar as in Figure 2.

Mechanics of Materials, 7th Edition, Chapter 4.7, Problem 120P , additional homework tip  2

Here, c is radius of the circular bar.

Refer to Figure 2.

The maximum compressive stress of the circular bar (σA)c is as follows:

(σA)c=PAcPecAIc=PAc(1+AcecAIc) (2)

The cross-sectional area of the circular bar (Ac) is as follow:

Ac=πc2=a2

The eccentricity of the load (e) is aπ.

The distance between the centroid from extreme fibre (cA) is aπ.

The moment of inertia (Ic) of the section of the circular bar is as follows:

Ic=π4c4=π4×(a4π)=π4×a4π2=a44π2

Calculate the maximum compressive stress of the circular bar (σA)c:

Substitute a2 for Ac, aπ for e, aπ for cA, and a44π2 for Ic in Equation (2).

(σA)c=PAc(1+a2×aπ×aπa44π2)=PAc4π2a4Pa4π2Ac=PAc4PAc=5PAc

Show the cross-sectional diagram of the diamond shape bar as in Figure 3.

Mechanics of Materials, 7th Edition, Chapter 4.7, Problem 120P , additional homework tip  3

Here, a is side of the diamond shape bar.

Refer to Figure 3.

The maximum compressive stress of the diamond shape bar (σA)d is as follows:

(σA)d=PAdPecAId=PAd(1+AdecAId) (3)

The cross-sectional area of the diamond shape bar (Ad) is as follow:

Ad=a2

The eccentricity of the load (e) is 2a2.

The distance between the centroid from extreme fibre (cA) is 2a2.

The moment of inertia (Id) of the section of the diamond shape bar is as follows:

Id=a412

Calculate the maximum compressive stress of the diamond shape bar (σA)d:

Substitute a2 for Ad, 2a2 for e, 2a2 for cA, and a412 for Id in Equation (3).

(σA)d=PAd(1+a2×2a2×2a2a412)=PAd12×2a4P4a4Ad=PAd6PAd=7PAd

Show the cross-sectional diagram of the triangular bar as in Figure 4.

Mechanics of Materials, 7th Edition, Chapter 4.7, Problem 120P , additional homework tip  4

Here, s is side of the triangular bar.

Refer to Figure 4.

The maximum compressive stress of the triangular bar (σA)t is as follows:

(σA)t=PAtPecAIt=PAt(1+AtecAIt) (4)

The cross-sectional area of the triangular bar (At) is as follow:

At=12×s×3s2=3s24

The distance between the centroid from extreme fibre (cA) is as follows:

cA=23×3s2=3s3×33=3s33=s3

The eccentricity of the load (e) is s3.

The moment of inertia (It) of the section of the triangular bar is as follows:

It=136×s×(3s2)3=s36×33s38=3s496

Calculate the maximum compressive stress of the triangular bar (σA)t:

Substitute 3s24 for At, s3 for e, s3 for cA, and 3s496 for It in Equation (4).

(σA)t=PAt(1+3s24×s3×s33s496)=PAt96s4P3×4s4At=PAt8PAt=9PAt

Calculate the maximum compressive stresses are in the ratio:

A=(σA)s:(σA)c:(σA)d:(σA)t

Substitute (4PAs) for (σA)s, (5PAc) for (σA)c, (7PAd) for (σA)d, and (9PAt) for (σA)t.

A=4PAs:5PAc:7PAd:9PAt=4As:5Ac:7Ad:9At

The four bars shown have the same cross-sectional area.

A=4:5:7:9

Hence the maximum compressive stresses are in the ratio 4:5:7:9 is proved.

b)

To determine

Show that the maximum tensile stresses are in the ratio 2:3:5:3.

b)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The load act on the point of the bars is P.

Calculation:

At the point B:

Refer to Figure 1.

The maximum tensile stress of the square bar (σB)s is as follows:

(σB)s=PAs+PecBIs=PAs(AsecBIs1) (5)

Here, the e is the eccentricity of the load and cB is the distance between the centroid from extreme fibre.

The cross-sectional area of the square bar (As) is a2.

The eccentricity of the load (e) is a2.

The distance between the centroid from extreme fibre (cB) is a2.

The moment of inertia (Is) of the section of the square bar is a412.

Calculate the maximum tensile stress of the square bar (σB)s:

Substitute a2 for As, a2 for e, a2 for cB, and a412 for Is in Equation (5).

(σB)s=PAs(a2×a2×a2a4121)=12a4P4a4AsPAs=3PAsPAs=2PAs

Refer to Figure 2.

The maximum tensile stress of the circular bar (σB)c is as follows:

(σB)c=PAc+PecBIc=PAc(AcecBIc1) (6)

The cross-sectional area of the circular bar (Ac) is as follow:

Ac=πc2=a2

The eccentricity of the load (e) is aπ.

The distance between the centroid from extreme fibre (cB) is aπ.

The moment of inertia (Ic) of the section of the circular bar is as follows:

Ic=π4c4=π4×(a4π)=π4×a4π2=a44π2

Calculate the maximum tensile stress of the circular bar (σB)c:

Substitute a2 for Ac, aπ for e, aπ for cB, and a44π2 for Ic in Equation (6).

(σB)c=PAc(a2×aπ×aπa44π21)=4π2a4Pa4π2AcPAc=4PAcPAc=3PAc

Refer to Figure 3.

The maximum tensile stress of the diamond shape bar (σB)d is as follows:

(σB)d=PAd+PecBId=PAd(AdecBId1) (7)

The cross-sectional area of the diamond shape bar (Ad) is as follow:

Ad=a2

The eccentricity of the load (e) is 2a2.

The distance between the centroid from extreme fibre (cB) is 2a2.

The moment of inertia (Id) of the section of the diamond shape bar is as follows:

Id=a412

Calculate the maximum tensile stress of the diamond shape bar (σB)d:

Substitute a2 for Ad, 2a2 for e, 2a2 for cB, and a412 for Id in Equation (7).

(σB)d=(a2×2a2×2a2a4121)=12×2a4P4a4AdPAd=6PAdPAd=5PAd

Refer to Figure 4.

The maximum tensile stress of the triangular bar (σB)t is as follows:

(σB)t=PAt+PecBIt=PAt(AtecBIt1) (8)

The cross-sectional area of the triangular bar (At) is as follows:

At=12×s×3s2=3s24

The distance between the centroid from extreme fibre (cB) is as follows:

cB=13×3s2=3s3×2×33=3s3×23=s23

The eccentricity of the load (e) is s3.

The moment of inertia (It) of the section of the triangular bar is as follows:

It=136×s×(3s2)3=s36×33s38=3s496

Calculate the maximum tensile stress of the triangular bar (σB)t:

Substitute 3s24 for At, s3 for e, s23 for cB, and 3s496 for It in Equation (8).

(σB)t=PAt(3s24×s3×s233s4961)=96s4P3×4×2s4AtPAt=4PAtPAt=3PAt

Calculate the maximum tensile stresses are in the ratio:

A=(σA)s:(σA)c:(σA)d:(σA)t

Substitute (2PAs) for (σA)s, (3PAc) for (σA)c, (5PAd) for (σA)d, and (3PAt) for (σA)t.

B=2PAs:3PAc:5PAd:3PAt=2As:3Ac:5Ad:3At

The four bars shown have the same cross-sectional area.

B=2:3:5:3

Hence the maximum tensile stresses are in the ratio 2:3:5:3 is proved.

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!
Students have asked these similar questions
5.86 The cast iron inverted T-section supports two concentrated loads of magnitude P. The working stresses are 48 MPa in tension, 140 MPa in compression, and 30 MPa in shear. (a) Show that the neutral axis of the cross section is located at d ¼ 48:75 mm and that the moment of inertia of the cross-sectional area about this axis is I ¼ 11:918  106 mm4. (b) Find the maximum allowable value of P.
Prob.4: [2.37] The 1.5 m concrete post is reinforced with six steel bars, each with 28 mm diameter. Knowing the E, = 200 GPa and Ec = 200 GPa, determine the normal stresses in the steel and concrete when a 1550 kN axial centric force P is applied to the post. 450 mm 1.5 m
M= 500 Nm PROBLEM 4.2 Knowing that the couple shown acts in the vertical plane, determine the stress at (a) point A, and (b) point B. [Ans. (a) -116.4 MPa (b) -87.3 MPa] 30 mm 40 mm Fig. P4.2 E231 hp

Chapter 4 Solutions

Mechanics of Materials, 7th Edition

Ch. 4.3 - 4.9 through 4.11 Two vertical forces are applied...Ch. 4.3 - Knowing that a beam of the cross section shown is...Ch. 4.3 - Knowing that a beam of the cross section shown is...Ch. 4.3 - Solve Prob. 4.13, assuming that the beam is bent...Ch. 4.3 - Knowing that for the extruded beam shown the...Ch. 4.3 - The beam shown is made of a nylon for which the...Ch. 4.3 - Solve Prob. 4.16, assuming that d = 40 mm.Ch. 4.3 - Knowing that for the beam shown the allowable...Ch. 4.3 - 4.19 and 4.20 Knowing that for the extruded beam...Ch. 4.3 - 4.19 and 4.20 Knowing that for the extruded beam...Ch. 4.3 - Straight rods of 6-mm diameter and 30-m length are...Ch. 4.3 - A 900-mm strip of steel is bent into a full circle...Ch. 4.3 - Straight rods of 0.30-in. diameter and 200-ft...Ch. 4.3 - A 60-Nm couple is applied to the steel bar shown,...Ch. 4.3 - (a) Using an allowable stress of 120 MPa,...Ch. 4.3 - A thick-walled pipe is bent about a horizontal...Ch. 4.3 - A couple M will be applied to a beam of...Ch. 4.3 - A portion of a square bar is removed by milling,...Ch. 4.3 - In Prob. 4.28, determine (a) the value of h for...Ch. 4.3 - For the bar and loading of Concept Application...Ch. 4.3 - Prob. 31PCh. 4.3 - It was assumed in Sec. 4.1B that the normal...Ch. 4.5 - 4.33 and 4.34 A bar having the cross section shown...Ch. 4.5 - 4.33 and 4.34 A bar having the cross section shown...Ch. 4.5 - 4.35 and 4.36 For the composite bar indicated,...Ch. 4.5 - Prob. 36PCh. 4.5 - 4.37 and 4.38 Wooden beams and steel plates are...Ch. 4.5 - 4.37 and 4.38 Wooden beams and steel plates are...Ch. 4.5 - 4.39 and 4.40 A copper strip (Ec = 105 GPa) and an...Ch. 4.5 - 4.39 and 4.40 A copper strip (Ec = 105 GPa) and an...Ch. 4.5 - 4.41 and 4.42 The 6 12-in. timber beam has been...Ch. 4.5 - 4.41 and 4.42 The 6 12-in. timber beam has been...Ch. 4.5 - 4.43 and 4.44 For the composite beam indicated,...Ch. 4.5 - Prob. 44PCh. 4.5 - Prob. 45PCh. 4.5 - Prob. 46PCh. 4.5 - A concrete slab is reinforced by 58-in.-diameter...Ch. 4.5 - Solve Prob. 4.47, assuming that the spacing of the...Ch. 4.5 - The reinforced concrete beam shown is subjected to...Ch. 4.5 - Prob. 50PCh. 4.5 - Knowing that the bending moment in the reinforced...Ch. 4.5 - A concrete beam is reinforced by three steel rods...Ch. 4.5 - The design of a reinforced concrete beam is said...Ch. 4.5 - For the concrete beam shown, the modulus of...Ch. 4.5 - 4.55 and 4.56 Five metal strips, each 0.5 1.5-in....Ch. 4.5 - 4.55 and 4.56 Five metal strips, each 0.5 1.5-in....Ch. 4.5 - The composite beam shown is formed by bonding...Ch. 4.5 - A steel pipe and an aluminum pipe are securely...Ch. 4.5 - The rectangular beam shown is made of a plastic...Ch. 4.5 - Prob. 60PCh. 4.5 - Knowing that M = 250 Nm, determine the maximum...Ch. 4.5 - Knowing that the allowable stress for the beam...Ch. 4.5 - Semicircular grooves of radius r must be milled as...Ch. 4.5 - Prob. 64PCh. 4.5 - A couple of moment M = 2 kNm is to be applied to...Ch. 4.5 - The allowable stress used in the design of a steel...Ch. 4.6 - The prismatic bar shown is made of a steel that is...Ch. 4.6 - Prob. 68PCh. 4.6 - Prob. 69PCh. 4.6 - Prob. 70PCh. 4.6 - The prismatic rod shown is made of a steel that is...Ch. 4.6 - Solve Prob. 4.71, assuming that the couples M and...Ch. 4.6 - 4.73 and 4.74 A beam of the cross section shown is...Ch. 4.6 - 4.73 and 4.74 A beam of the cross section shown is...Ch. 4.6 - 4.75 and 4.76 A beam of the cross section shown is...Ch. 4.6 - Prob. 76PCh. 4.6 - 4.77 through 4.80 For the beam indicated,...Ch. 4.6 - Prob. 78PCh. 4.6 - Prob. 79PCh. 4.6 - 4.77 through 4.80 For the beam indicated,...Ch. 4.6 - 4.81 through 4.83 Determine the plastic moment Mp...Ch. 4.6 - Prob. 82PCh. 4.6 - Prob. 83PCh. 4.6 - Determine the plastic moment Mp of a steel beam of...Ch. 4.6 - Determine the plastic moment Mp of the cross...Ch. 4.6 - Determine the plastic moment Mp of a steel beam of...Ch. 4.6 - Prob. 87PCh. 4.6 - Prob. 88PCh. 4.6 - Prob. 89PCh. 4.6 - Prob. 90PCh. 4.6 - Prob. 91PCh. 4.6 - Prob. 92PCh. 4.6 - Prob. 93PCh. 4.6 - Prob. 94PCh. 4.6 - Prob. 95PCh. 4.6 - Prob. 96PCh. 4.6 - Prob. 97PCh. 4.6 - Prob. 98PCh. 4.7 - Knowing that the magnitude of the horizontal force...Ch. 4.7 - A short wooden post supports a 6-kip axial load as...Ch. 4.7 - Two forces P can be applied separately or at the...Ch. 4.7 - A short 120 180-mm column supports the three...Ch. 4.7 - As many as three axial loads, each of magnitude P...Ch. 4.7 - Two 10-kN forces are applied to a 20 60-mm...Ch. 4.7 - Portions of a 1212-in. square bar have been bent...Ch. 4.7 - Knowing that the allowable stress in section ABD...Ch. 4.7 - A milling operation was used to remove a portion...Ch. 4.7 - A milling operation was used to remove a portion...Ch. 4.7 - The two forces shown are applied to a rigid plate...Ch. 4.7 - Prob. 110PCh. 4.7 - Prob. 111PCh. 4.7 - A short column is made by nailing four 1 4-in....Ch. 4.7 - A vertical rod is attached at point A to the cast...Ch. 4.7 - A vertical rod is attached at point A to the cast...Ch. 4.7 - Knowing that the clamp shown has been tightened...Ch. 4.7 - Prob. 116PCh. 4.7 - Three steel plates, each of 25 150-mm cross...Ch. 4.7 - A vertical force P of magnitude 20 kips is applied...Ch. 4.7 - The four bars shown have the same cross-sectional...Ch. 4.7 - Prob. 120PCh. 4.7 - An eccentric force P is applied as shown to a...Ch. 4.7 - Prob. 122PCh. 4.7 - Prob. 123PCh. 4.7 - Prob. 124PCh. 4.7 - A single vertical force P is applied to a short...Ch. 4.7 - The eccentric axial force P acts at point D, which...Ch. 4.9 - 4.127 through 4.134 The couple M is applied to a...Ch. 4.9 - 4.127 through 4.134 The couple M is applied to a...Ch. 4.9 - 4.127 through 4.134 The couple M is applied to a...Ch. 4.9 - 4.127 through 4.134 The couple M is applied to a...Ch. 4.9 - 4.127 through 4.134 The couple M is applied to a...Ch. 4.9 - 4.127 through 4.134 The couple M is applied to a...Ch. 4.9 - Prob. 133PCh. 4.9 - Prob. 134PCh. 4.9 - 4.135 through 4.140 The couple M acts in a...Ch. 4.9 - 4.135 through 4.140 The couple M acts in a...Ch. 4.9 - Prob. 137PCh. 4.9 - 4.135 through 4.140 The couple M acts in a...Ch. 4.9 - 4.135 through 44.140 The couple M acts in a...Ch. 4.9 - 4.135 through 4.140 The couple M acts in a...Ch. 4.9 - Prob. 141PCh. 4.9 - 4.141 through 4.143 The couple M acts in a...Ch. 4.9 - 4.141 through 4.143 The couple M acts in a...Ch. 4.9 - The tube shown has a uniform wall thickness of 12...Ch. 4.9 - Prob. 145PCh. 4.9 - Knowing that P = 90 kips, determine the largest...Ch. 4.9 - Knowing that a = 1.25 in., determine the largest...Ch. 4.9 - A rigid circular plate of 125-mm radius is...Ch. 4.9 - Prob. 149PCh. 4.9 - A beam having the cross section shown is subjected...Ch. 4.9 - Prob. 151PCh. 4.9 - Prob. 152PCh. 4.9 - Prob. 153PCh. 4.9 - Prob. 154PCh. 4.9 - Prob. 155PCh. 4.9 - Prob. 156PCh. 4.9 - Prob. 157PCh. 4.9 - Prob. 158PCh. 4.9 - A beam of unsymmetric cross section is subjected...Ch. 4.9 - Prob. 160PCh. 4.10 - For the curved bar shown, determine the stress at...Ch. 4.10 - For the curved bar shown, determine the stress at...Ch. 4.10 - Prob. 163PCh. 4.10 - Prob. 164PCh. 4.10 - The curved bar shown has a cross section of 40 60...Ch. 4.10 - Prob. 166PCh. 4.10 - Prob. 167PCh. 4.10 - Prob. 168PCh. 4.10 - The curved bar shown has a cross section of 30 30...Ch. 4.10 - Prob. 170PCh. 4.10 - Prob. 171PCh. 4.10 - Three plates are welded together to form the...Ch. 4.10 - 4.173 and 4.174 Knowing that the maximum allowable...Ch. 4.10 - Prob. 174PCh. 4.10 - Prob. 175PCh. 4.10 - Prob. 176PCh. 4.10 - Prob. 177PCh. 4.10 - Prob. 178PCh. 4.10 - Prob. 179PCh. 4.10 - Knowing that P = 10 kN, determine the stress at...Ch. 4.10 - Prob. 181PCh. 4.10 - Prob. 182PCh. 4.10 - Prob. 183PCh. 4.10 - Prob. 184PCh. 4.10 - Prob. 185PCh. 4.10 - Prob. 186PCh. 4.10 - Prob. 187PCh. 4.10 - Prob. 188PCh. 4.10 - Prob. 189PCh. 4.10 - Prob. 190PCh. 4.10 - For a curved bar of rectagular cross section...Ch. 4 - Two vertical forces are applied to a beam of the...Ch. 4 - Prob. 193RPCh. 4 - Prob. 194RPCh. 4 - Determine the plastic moment Mp of a steel beam of...Ch. 4 - In order to increase corrosion resistance, a...Ch. 4 - The vertical portion of the press shown consists...Ch. 4 - The four forces shown are applied to a rigid plate...Ch. 4 - Prob. 199RPCh. 4 - Prob. 200RPCh. 4 - Three 120 10-mm steel plates have been welded...Ch. 4 - A short length of a W8 31 rolled-steel shape...Ch. 4 - Two thin strips of the same material and same...
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
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Everything About COMBINED LOADING in 10 Minutes! Mechanics of Materials; Author: Less Boring Lectures;https://www.youtube.com/watch?v=N-PlI900hSg;License: Standard youtube license