• New York University

• ME-UY MISC

 Name: Quiz 4 10/9/2019 The linkage is made of two pin-connected A-36 steel members, each having a cross-sectional area of 1.5 in2. Determine the magnitude of the force P needed to displace point A 0.025 in. downward. (20 points) Solution: Static Equilibrium +↑∑ Fy=2FAC cos( tan−1 1.5 2 )−P=0→ FAC= 5 8 P δ AC= FAC LAC AE = 5 8 P(12√2 2 +1.52 ) 1.5(29×106 ) =4.31034×10−7 P∈¿ We also have, δ AC=LAC−LA ' C=(√2 2 +1.52 ) 12−√(2×12−0.025 ) 2 +(1.5×12) 2=30−29.98=0.02∈¿ Set to two δ AC to be equal to each other and solve for P, 4.31034×10−7 P=0.05→P=46400 lbs Alternatively, Using the relationship between the sides and angles for triangles we have, 1.5 2 tan−1 ¿ ¿ ¿ c 2−a 2−b 2 +2abcosγ=0→LAC ' 2 −LAC 2 −δA2 +2LAC (δA)cos ¿ LAC−δ AC ¿ 2 LAC ' 2 =¿ Inserting the given numerical values for the various quantities and solving for P we obtain,