Question

(i) a circular rod which the diameter d=7cm is subjected to the forces as shown in the figure, f=4kn, a=0.6m the permissible stress is σ=120mpa try to check the strength of the rod with the third strength theory calculate the bending moment and (the following part is scribbled and a chart is present with axes and a curve)

Explanation:

Step1: Analyze the Load and Supports

The rod has two concentrated loads \( F = 4\,\text{kN} \) and two moments \( Fa \) at supports \( A \) and \( B \). First, find the reactions at \( A \) and \( B \). By equilibrium, vertical forces: \( R_A + R_B = 2F \). Moment about \( A \): \( Fa + F(a) + F(2a) - R_B(3a) = 0 \)? Wait, no, the diagram shows \( A \) has a moment \( Fa \) (counterclockwise) and \( B \) has a moment \( Fa \) (clockwise), and two downward forces at \( C \) and \( D \) (each \( F \)), separated by \( a \) from \( A \), \( a \) between \( C \) and \( D \), and \( a \) from \( D \) to \( B \). So total length from \( A \) to \( B \) is \( 3a \).

Equilibrium of moments about \( A \): \( \sum M_A = 0 \)
\( Fa (\text{clockwise from } B) + F(a) + F(2a) - R_B(3a) = 0 \)? Wait, no, the support \( A \) has a moment \( Fa \) (counterclockwise), support \( B \) has a moment \( Fa \) (clockwise). The forces at \( C \) (distance \( a \) from \( A \)) and \( D \) (distance \( 2a \) from \( A \)) are downward \( F \) each.

So \( \sum M_A = 0 \):
\( -Fa (\text{counterclockwise at } A) + F(a) + F(2a) - Fa (\text{clockwise at } B) - R_B(3a) = 0 \)? Wait, maybe better to consider shear force and bending moment diagrams.

Alternatively, the shear force between \( A \) and \( C \): let's define sections. From \( A \) to \( C \) (length \( a \)): shear force \( V = R_A \). From \( C \) to \( D \) (length \( a \)): \( V = R_A - F \). From \( D \) to \( B \) (length \( a \)): \( V = R_A - 2F + R_B \). But by vertical equilibrium, \( R_A + R_B = 2F \), so \( R_A - 2F + R_B = 0 \).

Now, bending moment: at \( A \), \( M_A = Fa \) (counterclockwise, so bending moment at \( A \) is \( Fa \) (positive if we take sagging as positive? Wait, maybe the moment at \( A \) is a reaction moment. Wait, the diagram shows \( A \) has a moment \( Fa \) (curved arrow counterclockwise) and \( B \) has a moment \( Fa \) (curved arrow clockwise), and two downward forces at \( C \) and \( D \).

Let's recast: the rod is a beam with fixed moments at \( A \) ( \( M_A = Fa \), counterclockwise) and \( B \) ( \( M_B = Fa \), clockwise), and two concentrated loads \( F \) at \( C \) ( \( x = a \)) and \( D \) ( \( x = 2a \) ), where \( a = 0.6\,\text{m} \), \( F = 4\,\text{kN} \).

First, find shear force \( V(x) \):

- From \( A \) ( \( x=0 \)) to \( C \) ( \( x=a \)): \( V(x) = R_A \). But since there are no vertical reactions (wait, the supports at \( A \) and \( B \) are shown as "mm" (maybe pin or roller? But with moments). Wait, maybe it's a beam with end moments and two concentrated loads. Let's calculate bending moment \( M(x) \):

At \( x = 0 \) ( \( A \)): \( M(0) = Fa \) (from the end moment).

At \( x = a \) ( \( C \)): \( M(a) = Fa + R_A \cdot a - F \cdot 0 \) (wait, no, if \( R_A = 0 \)? Wait, vertical forces: sum of vertical forces is \( 2F \) downward, but if there are no vertical reactions (since supports are moments only? No, the diagram has \( A \) and \( B \) as supports with moments and maybe shear? Wait, the original problem says "circular rod" subjected to forces, so maybe it's a shaft under bending and torsion? Wait, no, the forces are vertical, so bending. Wait, the third strength theory is for combined stress (torsion and bending), but here maybe only bending? Wait, the diagram shows moments \( Fa \) (maybe torsion? Wait, the curved arrow at \( A \) is a moment, maybe torsion. Wait, the problem says "third strength theory", which is \( \sigma_r = \sqrt{\sigma^2 + 4\tau^2} \leq [\sigma] \) or \( \sigma_r = \sigma_1 - \sigma_3 \leq [\sigma] \). Wait, maybe the rod is under bending and torsion. Wait, the moment \( Fa \) could be torsion, and the vertical forces cause bending.

Wait, let's clarify: diameter \( d = 7\,\text{cm} = 0.07\,\text{m} \), \( F = 4\,\text{kN} = 4000\,\text{N} \), \( a = 0.6\,\text{m} \), \( [\sigma] = 120\,\text{MPa} = 120 \times 10^6\,\text{Pa} \).

First, calculate bending moment and torque (torsion moment).

Assume the moments \( Fa \) are torsional moments (torque \( T = Fa \)), and the vertical forces \( F \) cause bending moment \( M \).

Torque \( T = Fa = 4000\,\text{N} \times 0.6\,\text{m} = 2400\,\text{N·m} \).

Bending moment: the maximum bending moment from the two vertical forces. The beam between \( A \) and \( B \) has two loads at \( a \) and \( 2a \) from \( A \). The bending moment at the midpoint ( \( x = 1.5a \)): \( M = F \times 0.5a = 4000 \times 0.3 = 1200\,\text{N·m} \)? Wait, no, if the beam is simply supported? No, the supports have moments. Wait, maybe it's a cantilever? No, two supports.

Alternatively, maybe the rod is under pure bending and torsion. Let's proceed with third strength theory:

Third strength theory (von Mises for plane stress, or maximum shear stress theory): \( \sigma_{r3} = \frac{1}{W} \sqrt{M^2 + T^2} \times 32 / \pi d^3 \)? Wait, for a circular shaft, the bending stress \( \sigma = \frac{32M}{\pi d^3} \), shear stress \( \tau = \frac{16T}{\pi d^3} \). Then third strength theory: \( \sigma_{r3} = \sqrt{\sigma^2 + 4\tau^2} \leq [\sigma] \).

First, calculate \( M \) (bending moment) and \( T \) (torque).

Assume \( T = Fa = 4000\,\text{N} \times 0.6\,\text{m} = 2400\,\text{N·m} \).

Bending moment: from the two vertical forces \( F \) at \( C \) and \( D \), the maximum bending moment is at the midpoint between \( C \) and \( D \), which is \( x = 1.5a \), so \( M = F \times 0.5a = 4000 \times 0.3 = 1200\,\text{N·m} \).

Now, calculate \( \sigma \) (bending stress) and \( \tau \) (shear stress):

\( \sigma = \frac{32M}{\pi d^3} \), \( d = 0.07\,\text{m} \), so \( d^3 = (0.07)^3 = 3.43 \times 10^{-4}\,\text{m}^3 \).

\( \sigma = \frac{32 \times 1200}{\pi \times 3.43 \times 10^{-4}} \approx \frac{38400}{1.077 \times 10^{-3}} \approx 35.7 \times 10^6\,\text{Pa} = 35.7\,\text{MPa} \).

\( \tau = \frac{16T}{\pi d^3} = \frac{16 \times 2400}{\pi \times 3.43 \times 10^{-4}} \approx \frac{38400}{1.077 \times 10^{-3}} \approx 35.7\,\text{MPa} \) (wait, no, \( T = Fa = 2400\,\text{N·m} \), so \( \tau = \frac{16 \times 2400}{\pi (0.07)^3} \approx \frac{38400}{1.077 \times 10^{-3}} \approx 35.7\,\text{MPa} \)? Wait, no, \( d = 7\,\text{cm} = 0.07\,\text{m} \), so \( d^3 = 0.07^3 = 3.43 \times 10^{-4}\,\text{m}^3 \). Then \( \tau = \frac{16 \times 2400}{\pi \times 3.43 \times 10^{-4}} \approx \frac{38400}{1.077 \times 10^{-3}} \approx 35.7\,\text{MPa} \).

Now, third strength theory: \( \sigma_r = \sqrt{\sigma^2 + 4\tau^2} = \sqrt{(35.7)^2 + 4 \times (35.7)^2} = \sqrt{5 \times (35.7)^2} = 35.7 \times \sqrt{5} \approx 35.7 \times 2.236 \approx 79.8\,\text{MPa} \leq 120\,\text{MPa} \), so the rod is safe.

But wait, maybe the bending moment is different. Let's recalculate bending moment. If the two forces \( F \) are at \( x = a \) and \( x = 2a \), and the supports at \( x = 0 \) and \( x = 3a \) have moments \( M_A = Fa \) and \( M_B = Fa \).

The bending moment equation:

For \( 0 \leq x \leq a \): \( M(x) = M_A + V \cdot x \). But \( V = 0 \) (no vertical reaction), so \( M(x) = Fa \).

For \( a \leq x \leq 2a \): \( M(x) = Fa - F(x - a) \). At \( x = 1.5a \), \( M = Fa - F(0.5a) = 0.5Fa \).

For \( 2a \leq x \leq 3a \): \( M(x) = Fa - F(x - a) - F(x - 2a) + M_B \). At \( x = 3a \), \( M = Fa - F(2a) - F(a) + Fa = 0 \).

Wait, this is confusing. Maybe the moments \( Fa \) are torsional (torque), and the vertical forces cause bending. Let's proceed with the given data:

\( d = 7\,\text{cm} = 0.07\,\text{m} \), \( F = 4\,\text{kN} = 4000\,\text{N} \), \( a = 0.6\,\text{m} \), \( [\sigma] = 120\,\text{MPa} \).

Torque \( T = Fa = 4000 \times 0.6 = 2400\,\text{N·m} \).

Bending moment \( M = F \times a = 4000 \times 0.6 = 2400\,\text{N·m} \)? Wait, no, if there are two forces, maybe \( M = F \times a \) (single force). Wait, the diagram shows two forces at \( C \) and \( D \), each \( F \), so the bending moment between \( C \) and \( D \) is \( M = F \times a \) (distance between the two forces is \( a \), so the moment is \( F \times a \)).

Now, calculate the section modulus \( W = \frac{\pi d^3}{32} \approx \frac{\pi (0.07)^3}{32} \approx \frac{\pi \times 3.43 \times 10^{-4}}{32} \approx 3.37 \times 10^{-5}\,\text{m}^3 \).

Bending stress \( \sigma = \frac{M}{W} \), shear stress \( \tau = \frac{T}{W_t} \), where \( W_t = \frac{\pi d^3}{16} \approx 6.74 \times 10^{-5}\,\text{m}^3 \).

Then, third strength theory: \( \sigma_r = \sqrt{\sigma^2 + 4\tau^2} \).

Calculate \( \sigma = \frac{2400}{3.37 \times 10^{-5}} \approx 71.2 \times 10^6\,\text{Pa} = 71.2\,\text{MPa} \).

\( \tau = \frac{2400}{6.74 \times 10^{-5}} \approx 35.6 \times 10^6\,\text{Pa} = 35.6\,\text{MPa} \).

Then \( \sigma_r = \sqrt{(71.2)^2 + 4 \times (35.6)^2} = \sqrt{5069.44 + 4 \times 1267.36} = \sqrt{5069.44 + 5069.44} = \sqrt{10138.88} \approx 100.7\,\text{MPa} \leq 120\,\text{MPa} \), so the rod is safe.

Step2: Check Strength with Third Theory

The third strength theory (maximum shear stress theory) for combined bending and torsion is \( \sigma_r = \sigma_1 - \sigma_3 \), where \( \sigma_1 = \frac{\sigma}{2} + \sqrt{(\frac{\sigma}{2})^2 + \tau^2} \), \( \sigma_3 = \frac{\sigma}{2} - \sqrt{(\frac{\sigma}{2})^2 + \tau^2} \), so \( \sigma_r = 2\sqrt{(\frac{\sigma}{2})^2 + \tau^2} = \sqrt{\sigma^2 + 4\tau^2} \).

Substituting \( \sigma = \frac{32M}{\pi d^3} \), \( \tau = \frac{16T}{\pi d^3} \), we get \( \sigma_r = \frac{32}{\pi d^3} \sqrt{M^2 + T^2} \).

Calculate \( M^2 + T^2 = (2400)^2 + (2400)^2 = 2 \times (2400)^2 \), so \( \sqrt{M^2 + T^2} = 2400\sqrt{2} \).

Then \( \sigma_r = \frac{32 \times 2400\sqrt{2}}{\pi (0.07)^3} \approx \frac{32 \times 2400 \times 1.414}{3.1416 \times 3.43 \times 10^{-4}} \approx \frac{110592}{1.077 \times 10^{-3}} \approx 102.7 \times 10^6\,\text{Pa} = 102.7\,\text{MPa} \leq 120\,\text{MPa} \).

Thus, the rod satisfies the third strength theory.

Answer:

The rod is safe as the equivalent stress \( \sigma_r \approx 102.7\,\text{MPa} \) (or corrected calculation) is less than the permissible stress \( 120\,\text{MPa} \).