Question 2.3: Two solid circular rods are welded to a plate at B to form a......
Two solid circular rods are welded to a plate at B to form a single rod, as shown in Fig. 1. Consider the 30-kN force at B to be uniformly distributed around the circumference of the collar at B and the 10 kN load at C to be applied at the centroid of the end cross section. Determine the axial stress in each portion of the rod.
Plan the Solution Since each segment of the rod satisfies the conditions for uniform axial stress, we can use Eq. 2.5 to calculate the two required axial stresses. First, however, we need to compute the force in each rod by using an appropriate free-body diagram and equation of equilibrium.
σ_i = \frac{F_i}{A_i} Axial- Stress Equation (2.5)
Free-body Diagrams: First we draw free-body diagrams that expose the rod forces F_1 (or F_{BC}) and F_2 (or F_{BC}). We show F_1 and F_2 positive in tension.
Equations of Equilibrium: From free-body diagram 1 (Fig. 2a),
\underrightarrow{+} \left(\sum{F} \right)_1 = 0 : –F_1 – 30 kN + 10 kN = 0, F_1 = -20 kN
and, from free-body diagram 2 (Fig. 2b),
\underrightarrow{+} \left(\sum{F} \right)_2 = 0 : –F_2 + 10 kN = 0, F_2 = 10 kN
A_1 = \frac{\pi }{4} d^2_1 =\frac{\pi}{4} (20 mm)² = 314.2 mm²
A_2 = \frac{\pi }{4} d^2_2 =\frac{\pi}{4} (15 mm)² = 176.7 mm²
Axial Stresses: Using Eq. 2.5, we obtain the axial stresses
σ_1 = \frac{F_1} {A_1}=\frac{20 kN} {314.2 mm^2} = -63.7 \frac{MN }{m^2}σ_1 = \frac{F_2} {A_2}=\frac{10 kN} {176.7 mm^2} = -56.6 \frac{MN }{m^2}
\begin{matrix} σ_1 = -63.7 MPa (63.7 MPa C) \\ σ_2 = 56.6 MPa (56.6 MPa T) \end{matrix}
Review the Solution In this problem we could ‘‘mentally’’ solve the equilibrium problems and ‘‘see’’ that AB is in compression and that BC is in tension.
