Explanation Given information: Set the angle of the moment from y - axis is θ = 30 ° . Explanation: Show the free-body diagram of the square cross section as in Figure 1. Determine the moment of inertia in y and z - direction of the square cross section using the formula. I z = I y = b d 3 12 Here, b is width of the square section and d is depth of the square section. Substitute 250 mm for b and 250 mm for d . I z = I y = 250 mm × 1 m 1,000 mm × ( 250 mm × 1 m 1,000 mm ) 3 12 = 0.325521 × 10 − 3 m 4 Show the free body diagram of the square cross section for resultant moments as in Figure 2. The resultant moment in y -direction is: M y = M cos θ = 850 × cos 30 ° = 736.12 N-m The resultant moment in z -direction is as follows: M z = M sin θ = 850 × sin 30 ° = 425 N-m Determine the resultant normal stress at any point on the cross section using the equation. σ = − M z y I z + M y z I y (1) Here, σ is the normal stress at the point. Tensile stress is positive and compressive stress is negative. Stress at point A : The distance in y -direction is y A = − 125 mm . The distance in z -direction is z A = − 125 mm . Thus, the y and z coordinates of point A is (–125 mm, –125 mm). Substitute 425 N-m for M z , –125 mm for y , 0.325521 × 10 − 3 m 4 for I z , 736.12 N-m for M y , –125 mm for z , and 0.325521 × 10 − 3 m 4 for I y in Equation (1). σ A = − 425 × ( − 125 mm × 1 m 1,000 mm ) 0.325521 × 10 − 3 + 736.12 × ( − 125 mm × 1 m 1,000 mm ) 0.325521 × 10 − 3 = 163 , 200 − 282 , 670 = − 119 , 470 N/m 2 × 1 kPa 10 3 N/m 2 = − 119.4 kPa ≈ 119 kPa Thus, the stress at point A is σ A = − 119 kPa _ . Stress at point B : The distance in y -direction is y B = − 125 mm . The distance in z -direction is z B = 125 mm . Thus, the y and z coordinates of point B is (–125 mm, 125 mm). Substitute 425 N-m for M z , –125 mm for y , 0.325521 × 10 − 3 m 4 for I z , 736.12 N-m for M y , 125 mm for z , and 0.325521 × 10 − 3 m 4 for I y in Equation (1). σ B = − 425 × ( − 125 mm × 1 m 1,000 mm ) 0

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

Videos

Question

Chapter 11.5, Problem 97P

To determine

Find the stress at each corner of the square cross section.

Sketch the stress distribution.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11.5, Problem 96P

Chapter 11.5, Problem 98P

Students have asked these similar questions

The hoist consists of a single rope and an arrangement of frictionless pulleys as shown. If the angle 0 = 59°, determine the force that must be applied to the rope, Frope, to lift a load of 4.4 kN. The three-pulley and hook assembly at the center of the system has a mass of 22.5 kg with a center of mass that lies on the line of action of the force applied to the hook. e ΘΕ B CC 10 BY NC SA 2013 Michael Swanbom Fhook Note the figure may not be to scale. Frope = KN HO Frope

Determine the tension developed in cables AB and AC and the force developed along strut AD for equilibrium of the 400-lb crate. x. 5.5 ft C 2 ft Z 2 ft D 6 ft B 4 ft A 2.5 ft

A block of mass m hangs from the end of bar AB that is 7.2 meters long and connected to the wall in the xz plane. The bar is supported at A by a ball joint such that it carries only a compressive force along its axis. The bar is supported at end B by cables BD and BC that connect to the xz plane at points C and D respectively with coordinates given in the figure. Cable BD is elastic and can be modeled as a linear spring with a spring constant k = 400 N/m and unstretched length of 6.34 meters. Determine the mass m, the compressive force in beam AB and the tension force in cable BC. Z D (c, 0, d) C (a, 0, b), A e B y f m BY NC SA x 2016 Eric Davishahl Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value a 8.1 m b 3.3 m C 2.7 m d 3.9 m e 2 m f 5.4 m The mass of the block is The compressive force in bar AB is The tension in cable S is N. kg.

Chapter 11 Solutions

Statics and Mechanics of Materials (5th Edition)

Ch. 11.2 - In each case, the beam is subjected to the...Ch. 11.2 - In each ease, express the shear and moment...Ch. 11.2 - In each ease, express the shear and moment...Ch. 11.2 - In each ease, express the shear and moment...Ch. 11.2 - In each ease, express the shear and moment...Ch. 11.2 - Prob. 5FP Ch. 11.2 - Prob. 6FP Ch. 11.2 - In each ease, draw the shear and moment diagrams...Ch. 11.2 - Prob. 8FP Ch. 11.2 - Prob. 1P

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 stress at each corner of the square cross section. Sketch the stress distribution.

Solution Summary: The author shows the free-body diagram of the square cross section for resultant moments as in Figure 1.