The end of a large tubular workpart is to be faced on a NC vertical boring mill. The part has an outside diameter of 38.0 in and
inside diameter of 24.0 in. If the facing operation is performed at a rotational speed = 40 rev/min, feed = 0.015 in/rev, and depth = 0.180 in, determine: (a) the cutting time to complete the facing operation, (b) the cutting speeds and metal removal rates at the beginning and end of the cut.
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Maximum absolute values of the shear = 28 KN
Maximum absolute values of bending moment = 5.7 KN.m
<h3>How to draw Shear Force and Bending Moment Diagram?</h3>A) We can see the beam loaded in the first image attached.
For the shear diagram, let us calculate the shear from point load to point load.
From A to C, summing vertical to zero gives; ∑fy = 0: -20 - V = 0
V = -20 KN
From C to D, summing vertical to zero gives; ∑fy = 0: -20 + 48 - V = 0
V = 28 KN
From D to E, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - V = 0
V = 8 KN
From E to B, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - 20 - V = 0
V = -12 KN
For the bending moment diagram, let us calculate the bending moment from point load to point load.
At point A, the bending moment would be zero. Thus, M_A = 0 KN.m
At point C, taking moment about point C and equating to zero gives;
M_C = 0. Thus; 20(0.225) + M = 0
M = -4.5 KN.m
At point D, taking moment about point D and equating to zero gives;
M_D = 0. Thus; 20(0.525) - 48(0.3) + M = 0
M = 3.9 KN.m
At point E, taking moment about point E and equating to zero gives;
M_D = 0. Thus; 20(0.75) - 48(0.525) + 20(0.225) + M = 0
M = 5.7 KN.m
At point B, taking moment about point E and equating to zero gives;
M_E = 0. Thus; 20(1.05) - 48(0.825) + 20(0.525) + (20 * 0.3) + M = 0
M = 2.1 KN.m
2) From the attached diagrams, we can deduce that;
Maximum absolute values of the shear = 28 KN
Maximum absolute values of bending moment = 5.7 KN.m
Read more about shear force & bending moment diagram at; brainly.com/question/14834487
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