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Abhinav R.
Tutor for last 3 years and pursuing PhD From IIT(ISM) Dhanbad
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Mechanical Engineering
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Question:

A cylinder (500 mm internal diameter and 20 mm wall thickness) with closed ends is subjected simultaneously to an internal pressure of 0-60 MPa, bending moment 64000 Nm and torque 16000 Nm. Determine the maximum tensile stress and shearing stress in the wall.

Abhinav R.
Answer:

Given: d = 500 mm = 0·5 m; t = 20 mm = 0·02 m; p = 0·60 MPa = 0.6 MN/m2; M = 64000 Nm = 0·064 MNm; T= 16000 Nm = 0·016 MNm. Maximum tensile stress: First let us determine the principle stresses σ1 and σ2 assuming this as a thin cylinder. We know, σ1 = (p d)/2t =(0.6×0.5)/(2×0.02) =7.5 MN/m2 σ2 = (p d)/4t =(0.6×0.5)/(4×0.02) =3.75 MN/m2. Next consider effect of combined bending moment and torque on the walls of the cylinder. Then the principal stresses σ1 ' and σ2 ' are given by σ1 ' = 16/(π×d^3 ) [M+√((M^2 )+T^2)] σ2 ' = 16/(π×d^3 ) [M-√((M^2 )+T^2)] ∴ σ1' =16/(π×(0.5)^3 ) [0.064+√((0.064)^2 +(0.016)^2)] = 5.29MN/m^2 and σ2' =16/(π×(0.5)^3 ) [0.064 - √((0.064)^2 +(0.016)^2)] = -0.08MN/m^2 σ1 (net)= σ1+ σ1'= 12.79 MN/m^2 σ2 (net)= σ2+ σ2'= 3.67 MN/m^2 Maximum shearing stress τ_max= (σ1 (net) – σ2 (net))/2 = (12.79-3.67 )/2 = 4.56 MN/m^2

Manufacturing Engineering
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Question:

In orthogonal turning of medium carbon steel. The specific machining energy is 2.0 J/mm3. The cutting velocity, feed and depth of cut are 120 m/min, 0.2 mm/rev and 2 mm respectively. The main cutting force in Newton is .

Abhinav R.
Answer:

The energy consumption per unit volume of material removal, commonly known as specific energy e = Power(W)/MRR( mm3/s) = Fc/(1000 fd) (Fc= cutting force) or, 2.0 = Fc/1000*0.2*2 ⇒Fc = 800 N

Industrial Engineering
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Question:

following data refers to the activities of a project , where, node 1 refers to the start and node 5 refers to the end of the project. Activity Duration (days) 1-2 2 2-3 1 4-3 3 1-4 3 2-5 3 3-5 2 4-5 4 The critical path (CP) in the network is

Abhinav R.
Answer:

(B) 1-4-3-5 Subject : Operations Research Topic : Pert and Cpm Explanation : 1 - 2 - 5 = 5 1 - 2 - 3 - 5 = 5 1 - 4 - 3 - 5 = 8 1 - 4 - 5 = 7 A critical path is a contrinuous path from the source node to the sink node such that a delay in any activity results in a corresponding delay in whole process so here critical path is 1 - 4 - 3 - 5 .

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