(a) Aluminum foil used for storing food weighs about 0.3 grams per square inch. How many atoms of aluminum are contained in one
square inch of the foil? (b) Using the densities and atomic weights given in Appendix A, calculate and compare the number of atoms per cubic centimeter in (i) lead and (ii) lithium.
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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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