Solutions Manual
Mechanics of Materials - An Integrated Learning System
5th Edition
By
Timothy A. Philpot,
Jeffery S. Thomas
( All Chapters Included - 100% Verified Solutions )
1
,Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot
P1.1 A steel bar of rectangular cross section, 15 mm by 60 mm, is loaded by a compressive force of
110 kN that acts in the longitudinal direction of the bar. Compute the average normal stress in the bar.
Solution
The cross-sectional area of the steel bar is
A = (15 mm )( 60 mm ) = 900 mm 2
The normal stress in the bar is
F (110 kN )(1,000 N/kN )
σ= = = 122.222 MPa = 122.2 MPa Ans.
A 900 mm 2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright2Act without the permission of the copyright owner is unlawful.
,Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot
P1.2 A circular pipe with outside diameter of 4.5 in. and wall thickness of 0.375 in. is subjected to an
axial tensile force of 42,000 lb. Compute the average normal stress in the pipe.
Solution
The outside diameter D, the inside diameter d, and the wall thickness t are related by
D = d + 2t
Therefore, the inside diameter of the pipe is
d = D − 2t = 4.5 in. − 2 ( 0.375 in.) = 3.75 in.
The cross-sectional area of the pipe is
π π
A=
4
(D 2
− d2) =
4
( 4.5 in.) − ( 3.75 in.) = 4.8597 in.2
2
2
The average normal stress in the pipe is
F 42,000 lb
σ= = = 8,642.6 psi = 8,640 psi Ans.
A 4.8597 in.2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright3Act without the permission of the copyright owner is unlawful.
, Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot
P1.3 A circular pipe with an outside diameter of 80 mm is subjected to an axial compressive force of
420 kN. The average normal stress may not exceed 130 MPa. Compute the minimum wall thickness
required for the pipe.
Solution
From the definition of normal stress, solve for the minimum area required to support a 420 kN load
without exceeding a normal stress of 130 MPa
F F ( 420 kN )(1,000 N/kN )
σ= ∴ Amin ≥ = = 3, 230.77 mm 2
A σ 130 N/mm 2
The cross-sectional area of the pipe is given by
π
A= (D2 − d 2 )
4
Set this expression equal to the minimum area and solve for the maximum inside diameter d
π
( 80 mm ) − d 2 ≥ 3, 230.77 mm 2
2
4
4
( 80 mm ) − d 2 ≥ ( 3, 230.77 mm 2 )
2
π
∴ d max ≤ 47.8169 mm
The outside diameter D, the inside diameter d, and the wall thickness t are related by
D = d + 2t
Therefore, the minimum wall thickness required for the aluminum tube is
D − d 80 mm − 47.8169 mm
tmin ≥ = = 16.092 mm = 16.09 mm Ans.
2 2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright4Act without the permission of the copyright owner is unlawful.
Mechanics of Materials - An Integrated Learning System
5th Edition
By
Timothy A. Philpot,
Jeffery S. Thomas
( All Chapters Included - 100% Verified Solutions )
1
,Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot
P1.1 A steel bar of rectangular cross section, 15 mm by 60 mm, is loaded by a compressive force of
110 kN that acts in the longitudinal direction of the bar. Compute the average normal stress in the bar.
Solution
The cross-sectional area of the steel bar is
A = (15 mm )( 60 mm ) = 900 mm 2
The normal stress in the bar is
F (110 kN )(1,000 N/kN )
σ= = = 122.222 MPa = 122.2 MPa Ans.
A 900 mm 2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright2Act without the permission of the copyright owner is unlawful.
,Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot
P1.2 A circular pipe with outside diameter of 4.5 in. and wall thickness of 0.375 in. is subjected to an
axial tensile force of 42,000 lb. Compute the average normal stress in the pipe.
Solution
The outside diameter D, the inside diameter d, and the wall thickness t are related by
D = d + 2t
Therefore, the inside diameter of the pipe is
d = D − 2t = 4.5 in. − 2 ( 0.375 in.) = 3.75 in.
The cross-sectional area of the pipe is
π π
A=
4
(D 2
− d2) =
4
( 4.5 in.) − ( 3.75 in.) = 4.8597 in.2
2
2
The average normal stress in the pipe is
F 42,000 lb
σ= = = 8,642.6 psi = 8,640 psi Ans.
A 4.8597 in.2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright3Act without the permission of the copyright owner is unlawful.
, Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot
P1.3 A circular pipe with an outside diameter of 80 mm is subjected to an axial compressive force of
420 kN. The average normal stress may not exceed 130 MPa. Compute the minimum wall thickness
required for the pipe.
Solution
From the definition of normal stress, solve for the minimum area required to support a 420 kN load
without exceeding a normal stress of 130 MPa
F F ( 420 kN )(1,000 N/kN )
σ= ∴ Amin ≥ = = 3, 230.77 mm 2
A σ 130 N/mm 2
The cross-sectional area of the pipe is given by
π
A= (D2 − d 2 )
4
Set this expression equal to the minimum area and solve for the maximum inside diameter d
π
( 80 mm ) − d 2 ≥ 3, 230.77 mm 2
2
4
4
( 80 mm ) − d 2 ≥ ( 3, 230.77 mm 2 )
2
π
∴ d max ≤ 47.8169 mm
The outside diameter D, the inside diameter d, and the wall thickness t are related by
D = d + 2t
Therefore, the minimum wall thickness required for the aluminum tube is
D − d 80 mm − 47.8169 mm
tmin ≥ = = 16.092 mm = 16.09 mm Ans.
2 2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright4Act without the permission of the copyright owner is unlawful.
