Question 1.1: Load Resultants at a Section of a Piping An L-shaped pipe as......
Load Resultants at a Section of a Piping
An L-shaped pipe assembly of two perpendicular parts AB and BC is connected by an elbow at B and bolted to a rigid frame at C. The assembly carries a vertical load P_A, a torque T_A at A, as well as its own weight (Figure 1.3a). Each pipe is made of steel of unit weight w and nominal diameter d.
Find
What are the axial force, shear forces, and moments acting on the cross-section at point O?
Given
a=0.6 m , b=0.48 m , d=63.5 mm \left(2.5 \text { in.), } P_A=100 N , T_A=25 N \cdot m , w=5.79 lb / ft \right. (see Table A.4).
Assumption
The weight of the pipe assembly is uniformly distributed over its entire length.
See Figure 1.3 and Equation (1.5).
\begin{array}{rccc} \Sigma F_x=0 & \Sigma F_y=0 & \Sigma F_z=0 \\ \Sigma M_x=0 & \Sigma M_y=0 & \Sigma M_z=0 \end{array} (1.5)
Using the conversion factor from Table A.1, w=5.79 (N/m)/(0.0685) =84.53 N/m. Thus, the weights of the pipes AB and BO are equal to
W_{A B}=(84.53)(0.6)=50.72 N , \quad W_{B O}=(84.53)(0.48)=40.57 N
Free-body: Part ABO. We have six equations of equilibrium for the 3D force system of six unknowns (Figure 1.3b). The first three from Equation (1.5) results in the internal forces on the pipe at point O as follows:
\begin{array}{ll} \Sigma F_x=0: & F=0 \\ \Sigma F_y=0: & V_y-50.72-40.57-100=0: \quad V_y=191.3 N \\ \Sigma F_z=0: & V_z=0 \end{array}
Applying the last three from Equation (1.5), the moments about point O are found to be
\begin{array}{ll} \Sigma M_x=0: & T+(50.72)(0.3)+100(0.6)=0, \quad T=-75.2 N \cdot m \\ \Sigma M_y=0: & M_y=0 \\ \Sigma M_z=0: & M_z-25-100(0.48)-(50.72)(0.48)-(40.57)(0.24)=0, \\ & M_z=107.1 N \cdot m \end{array}
Comment: The negative value calculated for T means that the torque vector is directed opposite to that indicated in Figure 1.3b.
| TABLE A.1 Conversion Factors: SI Units to US Customary Units |
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| Quantity | SI Unit | US Equivalent |
| Acceleration | m/s^2 (meter per square second) | 3.2808 ft/s² |
| Area | m^2 (square meter) | 10.76 ft² |
| Force | N (newton) | 0.2248 lb |
| Intensity of force | N/m (newton per meter) | 0.0685 lb/ft |
| Length | m (meter) | 3.2808 ft |
| Mass | kg (kilogram) | 2.2051 lb |
| Moment of a force, torque | N · m (newton meter) | 0.7376 lb · ft |
| Moment of inertia of a plane area | m^4 (meter to fourth power) | 2.4025× 10^6 in.^4 |
| Moment of inertia of a mass | kg · m^2 (kilogram meter squared) | 0.7376 ft · s² |
| Power | W (watt) | 0.7376 ft · lb/s |
| kW (kilowatt) | 1.3410 hp | |
| Pressure or stress | Pa (pascal) | 0. 145× 10^{−3} psi |
| Specific weight | kN/m^3 (kilonewton per cubic meter) | 3.684×10^{−3}lb/in.^ 3 |
| Velocity | m/s (meter per second) | 3.2808 ft/s |
| Volume | m^3 (cubic meter) | 35.3147 ft³ |
| Work or energy | J (joule, newton meter) | 0.7376 ft · lb |
| Notes: 1 mile, mi=5280 ft=1609 m; 1 kilogram, kg=2.20946 lb=9.807 N; 1 joule, J=1 N · m; 1 inch, in.=25.4 mm; 1 foot, ft=12 in.=304.6 mm; 1 acceleration of gravity, g=9. 8066 m/ s²=32.174 ft/s². | ||