Question

The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of 0.75 in2, Determine the magnitude of the force P needed to displace point B 0.10 in. downward.

Answer 1

0.029N

Step-by-step explanation:

Pressure = displacement × density of steel × acceleration due to gravity

displacement = 0.1in = 0.1×0.0254m = 0.00254m, density of steel = 7900kg/m^3, acceleration due to gravity = 9.8m/s^2

Pressure = 0.00254×7900×9.8 = 196.65N/m^2

Force(P) = pressure × area

Area of 1 pin-connected A-36 steel member = 0.75in^2 = 4.84×10^-6m^2

Area of 3 pin-connected A-36 steel member = 3×4.84×10^-6 = 1.452×10^-5m^2

Force(P) = 196.65 × 1.4652×10^-5 = 0.0029N