Consider a bar with a variable cross section supporting a load P, as shown in the figure below. The bar is fixed at one end and carries the load P at the other end. Designate the width of the bar at the top by W1 , at the bottom by W2 , its thickness by t, and its length by L. The bar's modulus of elasticity is E. Analytically determining how much the bar will deflect at various points along its length when it is subjected to the load P. Neglect the weight of the bar in the following analysis, assuming that the applied load is considerably larger than the weight of the bar. Use Hooke’s law to show that the force can be related to the extension δ of a uniform bar with cross section area A as follows: F = (AE/L)*δ where E is the Young’s modulus, and L is the length of the bar. Use the above knowledge, approximate the original problem (a) as a five-element system shown in (b), and then (c), find the displacements of node 1 to 5. Values for the parameters: E=10.4E6 lb/in^2, W1 = 2 in, W2 =1 in, thickness t=0.125 in, L=10 in, and P=1000 lb
Consider a bar with a variable cross section supporting a load P, as shown in the figure below. The bar is fixed at one end and carries the load P at the other end. Designate the width of the bar at the top by W1 , at the bottom by W2 , its thickness by t, and its length by L. The bar's modulus of elasticity is E. Analytically determining how much the bar will deflect at various points along its length when it is subjected to the load P. Neglect the weight of the bar in the following analysis, assuming that the applied load is considerably larger than the weight of the bar.
Use Hooke’s law to show that the force can be related to the extension δ of a uniform bar with cross section area A as follows: F = (AE/L)*δ
where E is the Young’s modulus, and L is the length of the bar. Use the above knowledge, approximate the original problem (a) as a five-element system shown in (b), and then (c), find the displacements of node 1 to 5. Values for the parameters: E=10.4E6 lb/in^2, W1 = 2 in, W2 =1 in, thickness t=0.125 in, L=10 in, and P=1000 lb.
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