Question: Pumpjack frame problem D = 1 6 0 m , d = 0 . 1 6 m , A = 2 . 2 m ,

Pumpjack frame problem
D=160m, d=0.16m, A=2.2m, B=2.24m, P=3m, R=0.8m, H=5m. Weight of oil=26.8kN.
Please explain and justify your answers and steps with equations you have used to solve these questions.
Section-The frame
The aim of this section is to design the frame members I and J. The frame should be able to withstand the weight of the beam and the forces resulting from the operation of the pumpjack, including the weight of the crude oil and the forces due to the motor.
The frame members are made of mild steel. They may have different lengths but have the same profile and cross-sectional area.
1. Derive an expression for the displacement of the pivot point (i.e. where the bearing is) as a function of the force at the bearing (Fx,Fy), and the properties of the frame members I and J.
Your expression should have the following parameters: lengths of I and JLI and LJ; angles of the members to the horizontal 1 and J); the cross-section area a; the elastic modulus E of mild steel(200Gpa). Provide expressions for the axial forces along members I and J.
2. In the following, use Fx=-10kN and Fy=-40.6kN. The lengths of the members I and J are LI=7.5m and LJ=11.5m, respectively. Given the axial forces, determine the (uniform) cross-sectional area of the members such that they only deform (compress or elongate) within the elastic limit.
3. Given the compressive axial forces along the members, calculate the second moment of area that is necessary to prevent buckling of the members. Suggest a suitable cross-sectional profile made of 5mm thick steel plates. You may decide to weld several of these together, in which case consider them as a single block of steel. You may find that you need to use a higher cross-section area to make sure the frame does not buckle under load.
Pumpjack frame problem D = 1 6 0 m , d = 0 . 1 6

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