Explanation Given information: The effective length of the column is L e = 2.4 m . The allowable yield strength of the steel is σ Y = 250 MPa . The modulus of elasticity of the steel is E = 200 GPa . The centric load acting in the column is P = 180 kN . Calculation: Consider the thickness of the angle section as 9.5 mm. Refer to Appendix C “Properties of Rolled-Steel Shapes” in the textbook. For L 89 × 64 × 9.5 mm angle section; The cross sectional area of the angle ( A ) is A = 1 , 360 mm 2 . The minimum moment of inertia is I = I y = 0.454 × 10 6 mm 4 . The minimum radius of gyration is r = r y = 18.2 mm . 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 200 GPa for E and 250 MPa for σ Y L r = 4.71 200 GPa × 1,000 MPa 1 GPa 250 = 133.22 Find the ratio of effective length ( L e ) and the minimum radius of gyration ( r ) as follows; L e r = 2.4 m × 1,000 mm 1 m 18.2 = 131.87 < L r = 133.22 Find the effective stress ( σ e ) using the equation. σ e = π 2 E ( L e / r y ) 2 Substitute 200 GPa for E and 131.87 for L e / r y . σ e = π 2 × 200 GPa × 1,000 MPa 1 GPa ( 131.87 ) 2 = 113.514 MPa Find the critical stress ( σ cr ) using the relation. σ cr = [ 0.658 σ Y σ e ] σ Y Substitute 250 MPa for σ Y and 113.514 MPa for σ e . σ cr = [ 0.658 250 113.514 ] 250 = 99.45 MPa Calculate the allowable stress ( σ a l l ) using the relation. σ a l l = σ cr 1.67 Substitute 99.45 MPa for σ cr . σ a l l = 99.45 1.67 = 59.55 MPa Calculate the allowable load ( P all ) using the equation. P all = σ all A Substitute 59.55 MPa for σ all and 2 ( 1 , 360 ) mm 2 for A . P all = 59.55 × 2 ( 1 , 360 ) = 162 × 10 3 N × 1 kN 1,000 N = 162 kN The centric load is greater than the allowable load. Hence, the design is unsafe

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill 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 10.3, Problem 83P

To determine

Find the lightest angles that can be used.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.3, Problem 82P

Chapter 10.3, Problem 84P

Students have asked these similar questions

CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE 1. The truss shown is supported by hinge at A and cable at E.Given: H = 4m, S = 1.5 m, α = 75⁰, θ = 33⁰.Allowable tensile stress in cable = 64 MPa.Allowable compressive stress in all members = 120 MPaAllowable tensile stress in all members = 180 MPa1.Calculate the maximum permissible P, in kN, if the diameter of the cable is 20 mm.2.If P = 40 kN, calculate the required area (mm2) of member BC.3. If members have solid square section, with dimension 15 mm, calculate the maximum permissible P (kN) based on the allowable strength of member HI.ANSWERS: (1) 45.6 kN; (2) 83.71 mm2; (3) 171.76 kN

CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE 2: A wire 4 meters long is stretched horizontally between points 4 meters apart. The wire is 25 mm2 in cross-section with a modulus of elasticity of 200 GPa. A load W placed at the center of the wire produces a sag Δ.1.Calculate the tension (N) in the wire if sag Δ = 30 mm.2.Calculate the magnitude of W, in N, if sag Δ = 54.3 mm.3. If W is 60 N, what is the sag (in mm)?ANSWERS: (1) 562 N, (2) 100 N, (3) 45.8 N

CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE 4 : A cable and pulley system at D is used to bring a 230-kg pole (ACB) to a vertical position as shown. The cable has tensile force T and is attached at C. The length of the pole is 6.0 m, the outer diameter is d = 140 mm, and the wall thickness t = 12 mm. The pole pivots about a pin at A. The allowable shear stress in the pin is 60 MPa and the allowable bearing stress is 90 MPa. The diameter of the cable is 8 mm.1.Find the minimum diameter (mm) of the pin at A to support the weight of the pole in the position shown.2.Calculate the elongation (mm) of the cable CD.3.Calculate the vertical displacement of point C, in mm.ANSWERS: (1) 6 mm, (2) 1.186 mm, (3) 1.337 mm--

Chapter 10 Solutions

Mechanics of Materials, 7th Edition

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 lightest angles that can be used.

Solution Summary: The author calculates the slenderness ratio using the equation.

Concept explainers

Videos

Find the lightest angles that can be used.

Want to see the full answer?

Chapter 10 Solutions