Explanation Given information: The copper hollow circular tube acts as a column. The right end of the tube is fixed and left end is free. The outer diameter ( d o ) of the tube is 35 mm . The thickness ( t ) of the tube is 7 mm . The Young’s modulus of the copper as E c u = 120 GPa . The yield stress of the copper as σ Y = 750 MPa . The length of the copper tube is 2 m . The load on the column is denoted by P . The eccentricity ( e ) of the loading is 14 mm . Calculation: Consider the inner diameter of the tubular column as d i . Calculate the inner diameter of the tubular column using the relation: d i = d o − 2 t (1) Substitute 35 mm for d o and 7 mm for t in Equation (1). d i = 35 − 2 × 7 = 35 − 14 = 21 mm Calculate the cross-sectional area ( A ) of the circular hollowtube using the relation: A = π 4 [ ( d o ) 2 − ( d i ) 2 ] (2) Substitute 35 mm for d o and 21 mm for d i in Equation (2): A = π 4 ( 35 2 − 21 2 ) = π 4 ( 1225 − 441 ) = 615.75 mm 2 × ( 1 m 2 10 6 mm 2 ) = 0.61575 × 10 − 3 m 2 Calculate the moment of inertia ( I ) using the relation: I = π 64 [ ( d 0 ) 4 − ( d i ) 4 ] (3) Substitute 35 mm for d o , 21 mm for d i in Equation (3): I = π 64 [ ( 35 ) 4 − ( 21 ) 4 ] = π 64 [ 1 , 500 , 625 − 194 , 481 ] = π 64 [ 1 , 306 , 144 ] = 64115.19 mm 4 × ( 1 m 4 10 12 mm 4 ) I = 64.1152 × 10 − 9 m 4 Calculate the value of radius of gyration ( r ) using a relation: r = I A (4) Substitute 64.1152 × 10 − 9 m 4 for I and 0.61575 × 10 − 3 m 2 for A in Equation (4). r = 64.1152 × 10 − 9 m 4 0.61575 × 10 − 3 m 2 = 1.04125 × 10 − 4 m 2 = 0.01020 m Consider the buckling of column: The coppercolumn is fixed at one end and free at another end, thus the value of effective length factor ( K ) is 2. The Young’s modulus ( E ) of the copper is 120 GPa . Show the unit conversion of the Young’s modulus as follows: E = 120 GPa × ( 10 9 N m 2 1 GPa ) = 120 × 10 9 N m 2 Show the expression for Euler’s formula as follows: P cr = π 2 E I y ( K L ) y 2 (5) Here, E is the Young’s modulus, I is the moment of inertia, K is the effective length factor, L is the length of the member. Substitute 2 m for L , 120 × 10 9 N m 2 for E , 64.1152 × 10 − 9 m 4 for I , and 2 for K in Equation (5). P c r = π 2 ( 120 × 10 9 ) ( 64.1152 × 10 − 9 ) ( 2 × 2 ) 2 = 75935 16 = 18 , 983.75 N × 1 kN 1 , 000 N = 4 , 745.937 kN Thus, the load on the column based on the buckling criteria is 4 , 745.937 kN _ . Check the validity of Euler’s formula: Consider the yielding stress of the steel, σ Y = 50 ksi

9th Edition

ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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 13.5, Problem 13.50P

To determine

The load supported by the column without failure.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 13.5, Problem 13.49P

Chapter 13.5, Problem 13.51P

Students have asked these similar questions

A reservoir at 300 ft elevation has a 6-in.-diameter discharge pipe located 50 ft below the surface. The pipe is 600 ft long and drops in elevation to 150 ft where the flow discharges to the atmosphere. The pipe is made of riveted steel with a roughness height of 0.005 ft. Determine the flow rate without a head loss Determine the flow rate with the pipe friction head loss. (hints: Since the velocity is not known for part b and the Reynolds number and friction factor depend on velocity, you will need to iterate to find the solution. A good first guess is the velocity from part (a))

Air at T₁-24°C, p₁-1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate of 3 kg/min and mixes with a saturated moist air stream entering at T₂-7°C, p2-1 bar. A single mixed stream exits at T3-17°C, p3-1 bar. Neglect kinetic and potential energy effects Step 1 Your answer is correct. Determine mass flow rate of the moist air entering at state 2, in kg/min. m2 = 2.1 Hint kg/min Using multiple attempts will impact your score. 5% score reduction after attempt 2 Step 2 Determine the relative humidity of the exiting stream. Փ3 = i % Attempts: 1 of 3 used

25 mm Brass core E = 105 GPa 0 = 20.9 x 10 °C PROBLEM 2.49 The aluminum shell is fully bonded to the brass core and the assembly is unstressed at a temperature of 15°C. Considering only axial deformations, determine the stress in the aluminum when the temperature reaches 195°C. 60 mm Aluminum shell E = 70 GPa a = 23.6 × 10°C

Chapter 13 Solutions

Mechanics Of Materials, Si Edition

Ch. 13.3 - A 50-in long steel rod has a diameter of 1 in....Ch. 13.3 - A 12-ft wooden rectangular column has the...Ch. 13.3 - The A992 steel column can be considered pinned at...Ch. 13.3 - A steel pipe is fixed supported at its ends. If it...Ch. 13.3 - Determine the maximum force P that can be...Ch. 13.3 - The A992 steel rod BC has a diameter of 50 mm and...Ch. 13.3 - 13-1. Determine the critical buckling load for the...Ch. 13.3 - The column consists of a rigid member that is...Ch. 13.3 - The aircraft link is made from an A992 steel rod....Ch. 13.3 - Rigid bars AB and BC are pin connected at B. If...

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.