To determine Plot the distribution of normal stress acting along section a - a . Explanation Given information: The thickness of the plate is 10 mm. The horizontal force ( P ) is 80 kN. The load P acts along the centreline of the thickness such that distance d is 50 mm. Calculation: Show the free body diagram of cut section at a-a as shown in Figure 1. Here, N is the normal reaction at the fixed end of post, M is the moment at fixed end of the post, d is diameter of the post, and P is the load. Apply the equilibrium force in the x direction, which is equal to zero. Refer Figure 1: ∑ F x = 0 N − 80 = 0 N = 80 kN Take a moment about O is equal to zero. ∑ M c = 0 M − ( 80 × 0.05 ) = 0 M − 4 = 0 M = 4 kN ⋅ m Determine the area for rectangular section ( A ) using the following relation. A = b × d (1) Here, t is the thickness of the plate and d is the depth of the plate. Substitute 10 mm for b and 200 mm for d in Equation (1). A = 10 × 200 = 2 , 000 mm 2 × ( 1 m 2 10 6 mm 2 ) = 0.002 m 2 Determine the moment of inertia ( I ) using the following relation: I = 1 12 b d 3 (2) Substitute 10 mm for b and 200 mm for d in Equation (2). I = 1 12 × 10 × 200 3 = 6.667 × 10 6 mm 4 × 10 − 12 m 1 mm = 6.667 × 10 − 6 m 4 Determine the normal stress at point A ( σ A ) developed in the combination of axial and bending stress as shown below: σ A = N A − M y I (3) Here, N is the normal force, M is the moment of cylindrical post, A is the area of cross section, I is the moment of inertia, and y is the distance. Substitute 80 kN for N , 6.667 × 10 − 6 m 4 for I , 0.002 m 2 for A , 0.1 m for y, and 4 kN ⋅ m for M in Equation (3). σ A = 80 kN ( 10 3 N 1 kN ) 0.002 − 4 kN ⋅ m ( 10 3 N ⋅ m kN ⋅ m ) × 0.1 6.667 × 10 − 6 = 80 × 10 3 0

9th Edition

ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

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 8.2, Problem 8.32P

To determine

Plot the distribution of normal stress acting along section a-a.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.2, Problem 8.31P

Chapter 8.2, Problem 8.33P

Students have asked these similar questions

I don't know how to solve this

I am not able to solve this question. Each part doesn't make sense to me.

Exercises Find the solution of the following Differential Equations 1) y" + y = 3x² 3) "+2y+3y=27x 5) y"+y=6sin(x) 7) y"+4y+4y = 18 cosh(x) 9) (4)-5y"+4y = 10 cos(x) 11) y"+y=x²+x 13) y"-2y+y=e* 15) y+2y"-y'-2y=1-4x³ 2) y"+2y' + y = x² 4) "+y=-30 sin(4x) 6) y"+4y+3y=sin(x)+2 cos(x) 8) y"-2y+2y= 2e* cos(x) 10) y+y-2y=3e* 12) y"-y=e* 14) y"+y+y=x+4x³ +12x² 16) y"-2y+2y=2e* cos(x)

Chapter 8 Solutions

MECHANICS OF MATERIALS-TEXT

Ch. 8.1 - If it is subjected to an internal pressure of p =...Ch. 8.1 - If it is subjected to an internal pressure of p =...Ch. 8.1 - The thin-walled cylinder can be supported in one...Ch. 8.1 - If the inner diameter of the tank is 22 in., and...Ch. 8.1 - Prob. 8.5P Ch. 8.1 - 8–6. If the flow of water within the pipe in Prob....Ch. 8.1 - A boiler is constructed of 8-mm-thick steel plates...Ch. 8.1 - 88. The steel water pipe has an inner diameter of...Ch. 8.1 - The steel water pipe has an inner diameter of 12...Ch. 8.1 - The A-36-steel band is 2 in. wide and is secured...

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.