4-10| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition|

Summary

This video details the solution to problem 4-10 from R.C. Hibbeler's 'Mechanics of Materials 9th Edition', focusing on axial loading. The problem involves an assembly of brass, copper, and stainless steel rods, along with a rigid bar, to determine the magnitude of a load 'P' given a specific displacement.

Highlights

Introduction to the Problem00:00:15

The problem describes an assembly consisting of two 10mm diameter brass/copper rods (AB and CD), a 15mm diameter stainless steel rod (EF), and a rigid bar (G). Given a horizontal displacement of 0.45mm at the end of rod EF, the goal is to determine the magnitude of load 'P'.

Free Body Diagram and Internal Loading00:01:24

To solve, a free body diagram is drawn. By cutting through the rods, the internal forces are determined. The force in rod EF (P_EF) is found to be 4P. For rods AB and CD, the forces (P_AB and P_CD) are both equal to P, derived from balancing forces on the rigid bar.

Calculating Areas of the Rods00:03:00

The cross-sectional areas of the rods are calculated using the formula pi/4 * d^2. For rod EF (stainless steel), with a 15mm diameter, the area is 56.25 * pi * 10^-6 square meters. For rods AB and CD (brass/copper), with a 10mm diameter, the area is 25 * pi * 10^-6 square meters.

Applying the Displacement Equation00:03:57

The total displacement at the end of rod EF (0.45mm or 0.45 * 10^-3 meters) is the sum of the elongations in rod EF and rods AB/CD. The formula for displacement (delta = PL/AE) is used. The lengths are 450mm (EF) and 300mm (AB/CD). The Young's Modulus (E) for stainless steel is 193 GPa and for brass/copper is 101 GPa.

Solving for Load P00:06:51

By substituting all known values (displacements, forces in terms of P, lengths, areas, and Young's Moduli) into the displacement equation, the equation is solved for P. After calculations, the magnitude of P is found to be 4967 Newtons, or approximately 4.97 kilonewtons.