(a) Explanation Given information: Outer diameter of the drill pipe is 3 in. Wall thickness is 0.25 in. Weight of the drill pipe is 50 lb / ft . Calculation: Let N be the normal force and T be the torque of the drill pipe. Sketch the free body diagram of the drill pipe as shown in Figure 1. Refer Figure 1. The value of normal force is N = − 2 , 500 lb and torque is T = 800 lb ⋅ ft . Outer diameter of the pipe ( d 0 ) is 1.5 in. Thickness of the pipe ( t ) is 0.25 in. Find the inner diameter of the pipe ( d i ) as shown below: d i = d 0 − 2 t = 3 in . − 2 ( 0.25 in . ) = 2.5 in . Find the area ( A ) of the pipe. A = π 4 ( ( 3 in . ) 2 − ( 2.5 in . ) 2 ) = 2.1598 in 2 Find the polar moment of inertia ( J ) of the pipe: J = π 32 ( ( 3 in . ) 4 − ( 2.5 in . ) 4 ) = 4.1172 in 4 Find the normal stress ( σ ) due to axial stress using the relation: σ = N A = − 2 , 500 lb 2.1598 in 2 = − 1 , 157.5 psi The outer radius of the pipe is c = 1.5 in . Find the shear stress ( τ ) due to torsional shear stress using the relation: τ = T c J = ( 800 lb ⋅ ft ) ( 1.5 in . ) 4 (b) To determine The maximum in-plane shear stress at a point on its surface at section a .

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ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

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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 …

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Chapter 9.3, Problem 9.42P

(a)

To determine

The principal stresses.

(b)

To determine

The maximum in-plane shear stress at a point on its surface at section a.

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Chapter 9.3, Problem 9.41P

Chapter 9.3, Problem 9.43P

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MECHANICS OF MATERIALS-TEXT

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