Explanation Given information: The effective length of the column is L e = 20 ft . The allowable yield strength of the steel is σ Y = 36 ksi . The modulus of elasticity of the steel is E = 29 × 10 6 psi . The centric load acting in the column is P = 65 kips . Calculation: The outer dimension of the tube is d o = 6 in . . Find the inner dimension of the tube ( d i ) using the relation. d i = d o − 2 t Substitute 6 in. for d o and 1 2 in . for t . d i = 6 − 2 t Find the cross sectional area ( A ) using the equation. A = d o 2 − d i 2 Substitute 6 in. for d o and ( 6 − 2 t ) for d i . A = 6 2 − ( 6 − 2 t ) 2 = 36 − 36 + 24 t − 4 t 2 = 24 t − 4 t 2 Find the moment of inertia ( I ) of the cross section using the equation. I = d o 4 12 − d i 4 12 Substitute 6 in. for d o and ( 6 − 2 t ) for d i . I = 6 4 12 − ( 6 − 2 t ) 4 12 = 1 12 ( 6 4 − ( 6 − 2 t ) 4 ) Find the minimum radius of gyration ( r ) using the relation. r = I A Substitute 1 12 ( 6 4 − ( 6 − 2 t ) 4 ) for I and 24 t − 4 t 2 for A . r = 1 12 ( 6 4 − ( 6 − 2 t ) 4 ) 24 t − 4 t 2 = 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 Find the slenderness ratio L r using the equation. L r = 4.71 E σ Y Here, the modulus of elasticity of the material is E and the allowable yield strength is σ Y Substitute 29 × 10 6 psi for E and 36 ksi for σ Y L r = 4.71 29 × 10 6 36 ksi × 1,000 psi 1 ksi = 133.68 Find the ratio of effective length ( L e ) and the minimum radius of gyration ( r ) as follows; L e r = 20 ft × 12 in . 1 ft 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 = 240 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 (1) Consider 240 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 < L r = 133.68 Find the effective stress ( σ e ) using the equation. σ e = π 2 E ( L e / r y ) 2 Substitute 29 × 10 6 psi for E and 240 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 for L e / r y . σ e = π 2 × 29 × 10 6 psi × 1 ksi 10 3 psi ( 240 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 ) 2 = 4

7th Edition

ISBN: 9780100257061

Author: BEER

Publisher: YUZU

See similar textbooks

Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

Videos

Question

Chapter 10.3, Problem 82P

To determine

Find the thickness of the lightest tube.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.3, Problem 81P

Chapter 10.3, Problem 83P

Students have asked these similar questions

3. A cast-iron tube is used to support a compressive load, knowing that the maximum allowable change in length is 0.025%. a. Determine the maximum normal stress in the tube b. Find the minimum wall thickness for a load of 7 KN if the outside diameter is 50 mm.

Three wires are used to suspend the plate shown. Aluminum wires of 18-in. diameter are used at A and Bwhile a steel wire of 112-in. diameter is used at C. Knowing that the allowable stress for aluminum (Ea510.4 3 106 psi) is 14 ksi and that the allowable stress for steel (Es529 3 106 psi) is 18 ksi, determine the maximum load P that can be applied

An annular washer distributes the load P applied to a steel rod to a timber support. The rod's diameter is 22 mm, and the washer's inner diameter is 25 mm, which is larger than the hole's permissible outer diameter. Knowing that the axial normal stress in the steel rod is 35 MPa and the average bearing stress between the washer and the timber must not exceed 5 MPa, examine the smallest allowed outer diameter, d, of the washer. %3D %3D +22 mm P Figure 4

Chapter 10 Solutions

EBK MECHANICS OF MATERIALS

Ch. 10.1 - Knowing that the spring at A is of constant k and...Ch. 10.1 - Two rigid bars AC and BC are connected by a pin at...Ch. 10.1 - 10.3 and 10.4 Two rigid bars AC and BC are...Ch. 10.1 - 10.3 and 10.4 Two rigid bars AC and BC are...Ch. 10.1 - The steel rod BC is attached to the rigid bar AB...Ch. 10.1 - The rigid rod AB is attached to a hinge at A and...Ch. 10.1 - The rigid bar AD is attached to two springs of...Ch. 10.1 - A frame consists of four L-shaped members...Ch. 10.1 - Determine the critical load of a pin-ended steel...Ch. 10.1 - Determine the critical load of a pin-ended wooden...

Ch. 10.1 - A column of effective length L can be made by...Ch. 10.1 - A compression member of 1.5-m effective length...Ch. 10.1 - Determine the radius of the round strut so that...Ch. 10.1 - Determine (a) the critical load for the square...Ch. 10.1 - A column with the cross section shown has a...Ch. 10.1 - A column is made from half of a W360 216...Ch. 10.1 - A column of 22-ft effective length is made by...Ch. 10.1 - A single compression member of 8.2-m effective...Ch. 10.1 - Knowing that P = 5.2 kN, determine the factor of...Ch. 10.1 - Members AB and CD are 30-mm-diameter steel rods,...Ch. 10.1 - The uniform brass bar AB has a rectangular cross...Ch. 10.1 - A 1-in.-square aluminum strut is maintained in the...Ch. 10.1 - A 1-in.-square aluminum strut is maintained in the...Ch. 10.1 - Column ABC has a uniform rectangular cross section...Ch. 10.1 - Column ABC has a uniform rectangular cross section...Ch. 10.1 - Column AB carries a centric load P of magnitude 15...Ch. 10.1 - Each of the five struts shown consists of a solid...Ch. 10.1 - A rigid block of mass m can be supported in each...Ch. 10.2 - An axial load P = 15 kN is applied at point D that...Ch. 10.2 - An axial load P is applied to the 32-mm-diameter...Ch. 10.2 - The line of action of the 310-kN axial load is...Ch. 10.2 - Prob. 32P Ch. 10.2 - An axial load P is applied to the 32-mm-square...Ch. 10.2 - Prob. 34P Ch. 10.2 - Prob. 35P Ch. 10.2 - Prob. 36P Ch. 10.2 - Solve Prob. 10.36, assuming that the axial load P...Ch. 10.2 - The line of action of the axial load P is parallel...Ch. 10.2 - Prob. 39P Ch. 10.2 - Prob. 40P Ch. 10.2 - The steel bar AB has a 3838-in. square cross...Ch. 10.2 - For the bar of Prob. 10.41, determine the required...Ch. 10.2 - A 3.5-m-long steel tube having the cross section...Ch. 10.2 - Prob. 44P Ch. 10.2 - An axial load P is applied to the W8 28...Ch. 10.2 - Prob. 46P Ch. 10.2 - A 100-kN axial load P is applied to the W150 18...Ch. 10.2 - A 26-kip axial load P is applied to a W6 12...Ch. 10.2 - Prob. 49P Ch. 10.2 - Axial loads of magnitude P = 84 kN are applied...Ch. 10.2 - An axial load of magnitude P = 220 kN is applied...Ch. 10.2 - Prob. 52P Ch. 10.2 - Prob. 53P Ch. 10.2 - Prob. 54P Ch. 10.2 - Axial loads of magnitude P = 175 kN are applied...Ch. 10.2 - Prob. 56P Ch. 10.3 - Using allowable stress design, determine the...Ch. 10.3 - Prob. 58P Ch. 10.3 - Prob. 59P Ch. 10.3 - A column having a 3.5-m effective length is made...Ch. 10.3 - Prob. 61P Ch. 10.3 - Bar AB is free at its end A and fixed at its base...Ch. 10.3 - Prob. 63P Ch. 10.3 - Prob. 64P Ch. 10.3 - A compression member of 8.2-ft effective length is...Ch. 10.3 - A compression member of 9-m effective length is...Ch. 10.3 - A column of 6.4-m effective length is obtained by...Ch. 10.3 - A column of 21-ft effective length is obtained by...Ch. 10.3 - Prob. 69P Ch. 10.3 - Prob. 70P Ch. 10.3 - Prob. 71P Ch. 10.3 - Prob. 72P Ch. 10.3 - Prob. 73P Ch. 10.3 - For a rod made of aluminum alloy 2014-T6, select...Ch. 10.3 - Prob. 75P Ch. 10.3 - Prob. 76P Ch. 10.3 - A column of 4.6-m effective length must carry a...Ch. 10.3 - A column of 22.5-ft effective length must carry a...Ch. 10.3 - Prob. 79P Ch. 10.3 - A centric load P must be supported by the steel...Ch. 10.3 - A square steel tube having the cross section shown...Ch. 10.3 - Prob. 82P Ch. 10.3 - Prob. 83P Ch. 10.3 - Two 89 64-mm angles are bolted together as shown...Ch. 10.3 - Prob. 85P Ch. 10.3 - Prob. 86P Ch. 10.3 - Prob. 87P Ch. 10.3 - Prob. 88P Ch. 10.4 - An eccentric load is applied at a point 22 mm from...Ch. 10.4 - Prob. 90P Ch. 10.4 - Prob. 91P Ch. 10.4 - Solve Prob. 10.91 using the interaction method and...Ch. 10.4 - A column of 5.5-m effective length is made of the...Ch. 10.4 - Prob. 94P Ch. 10.4 - A steel compression member of 9-ft effective...Ch. 10.4 - Prob. 96P Ch. 10.4 - Two L4 3 38-in. steel angles are welded together...Ch. 10.4 - Solve Prob. 10.97 using the interaction method...Ch. 10.4 - A rectangular column is made of a grade of sawn...Ch. 10.4 - Prob. 100P Ch. 10.4 - Prob. 101P Ch. 10.4 - Prob. 102P Ch. 10.4 - Prob. 103P Ch. 10.4 - Prob. 104P Ch. 10.4 - A steel tube of 80-mm outer diameter is to carry a...Ch. 10.4 - Prob. 106P Ch. 10.4 - Prob. 107P Ch. 10.4 - Prob. 108P Ch. 10.4 - Prob. 109P Ch. 10.4 - Prob. 110P Ch. 10.4 - Prob. 111P Ch. 10.4 - Prob. 112P Ch. 10.4 - Prob. 113P Ch. 10.4 - Prob. 114P Ch. 10.4 - Prob. 115P Ch. 10.4 - A steel column of 7.2-m effective length is to...Ch. 10 - Determine (a) the critical load for the steel...Ch. 10 - Prob. 118RP Ch. 10 - Prob. 119RP Ch. 10 - (a) Considering only buckling in the plane of the...Ch. 10 - Member AB consists of a single C130 3 10.4 steel...Ch. 10 - The line of action of the 75-kip axial load is...Ch. 10 - Prob. 123RP Ch. 10 - Prob. 124RP Ch. 10 - A rectangular column with a 4.4-m effective length...Ch. 10 - Prob. 126RP Ch. 10 - Prob. 127RP Ch. 10 - Prob. 128RP

Knowledge Booster

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.

Find the thickness of the lightest tube.

Solution Summary: The author calculates the thickness of the lightest tube using the equation.

Concept explainers

Videos

Find the thickness of the lightest tube.

Want to see the full answer?

Chapter 10 Solutions