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Mechanics of Materials: An Integrated Learning System, 4

th Ed. Timothy A. Philpot

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 Copyright Act without the permission of the copyright owner is unlawful.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

( )( )

2 15 mm 60 mm 900 mmA== The normal stress in the bar is ( )( ) 2 110 kN 1,000 N/kN 122.222 MPa 900 m 12 m 2.2 MPa F A

= = = =

Ans.

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Mechanics of Materials: An Integrated Learning System, 4

th Ed. Timothy A. Philpot

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 Copyright Act without the permission of the copyright owner is unlawful.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 2D d t=+ Therefore, the inside diameter of the pipe is

• )2 4.5 in. 2 0.375 in. 3.75 in.d D t= − = − =

The cross-sectional area of the pipe is

( ) ( )( )

22

• 2 2

4.5 in. 3.75 in. 4.8597 in.44 A D d



= − = − =



The average normal stress in the pipe is 2 42,000 lb 8,642.6 psi 4.8597 in.8,640 psi F A

= = = =

Ans.

• / 4

Mechanics of Materials: An Integrated Learning System, 4

th Ed. Timothy A. Philpot

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 Copyright Act without the permission of the copyright owner is unlawful.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 ( )( ) 2 min2 420 kN 1,000 N/kN 3,230.77 mm 130 N/mm FF A A  

=   = =

The cross-sectional area of the pipe is given by 22 () 4 A D d  =− Set this expression equal to the minimum area and solve for the maximum inside diameter d

• )

( ) ( )

2 22 2 22 max 80 mm 3,230.77 mm 4 4 80 mm 3,230.77 mm 47.8169 mm d d d  

 −



−



The outside diameter D, the inside diameter d, and the wall thickness t are related by 2D d t=+ Therefore, the minimum wall thickness required for the aluminum tube is min 80 mm 47.8169 mm 16.092 mm 22 16.09 mm Dd t

−−

 = = = Ans.

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