# Find the maximum stress at a point in a machine member

A machine member having a diameter of 50 mm and a length of 250 mm is supported at one end as a cantilever subjected to different forces as shown in the figure. Find the maximum stress at points A and B.

I know to find axial stress will be equal to 15kn/area of crossection and bending moment due to 3kn force can be found as (3kn x 250 x 25)/moment of inertia. I'm not sure will it be polar moment of inertia or area moment of inertia and then how will I be able to achieve maximum stress. kindly looking for help, thanks in advance.

Assuming

• x is the horizontal axis parallel to P =15kN (+x to the right, -x to the left)
• y is the vertical axis on the plane (+y upwards, -y downwards)
• z is the axis perpendicular to the x-y (coming out of the image).

Also

• Q= 3kN is the load at the end of the beam
• $$M_t=1 [kNm]$$ is the torsional load

Then there are three stresses at each point A, B

• an axial stress which is common for both $$\sigma_{xx} = \frac{15 kN}{A_m}$$

where $$A_m$$ is the cross-sectional area of the member. The direction is horizontal

• a normal stress due to bending

$$\sigma_{xx,b}= \frac{Q\cdot L}{I_{zz}}\frac{d}{2}= \frac{32\cdot Q\cdot L}{\pi d^3}$$

where $$I_{xx}=\frac{\pi d^4}{64}$$ the second moment of area.

The direction for the stress is positive at A, and negative at B.

• a shear stress due to the shear load $$\tau_{xy}=0$$

• a shear stress due to the torsional load: $$\tau_{xz}=\frac{M_t}{J_{p}}\frac{d}{2}$$

where $$J_{p}=\frac{\pi d^4}{32}$$ the polar moment of area.

$$\tau_{xz}=\frac{16\cdot M_t}{\pi d^3}$$

to sum up:

type A B
Tensile normal $$4\cdot \frac{15 kN }{\pi d^2}$$ $$4\cdot \frac{15 kN}{\pi d^2}$$
Bending normal $$\frac{32\cdot Q\cdot L}{\pi d^3}$$ $$- \frac{32\cdot Q\cdot L}{\pi d^3}$$
Shear shear xy $$0$$ $$0$$
Torsional shear xz $$-\frac{16\cdot M_t}{\pi d^3}$$ $$\frac{16\cdot M_t}{\pi d^3}$$

Then for example for A,

• the total normal stresses will be:

$$4\cdot \frac{15 kN }{\pi d^2} + \frac{32\cdot Q\cdot L}{\pi d^3}$$

• the magnitude of the shear stresses will be

$$\sqrt{\left(0\right)^2 + \left(-\frac{16\cdot M_t}{\pi d^3}\right)^2 } =\frac{16\cdot M_t}{\pi d^3}$$

The maximum shear stress due to vertical load is

$$\tau_{max} = \frac{VQ_{max}}{I_c b}=\frac{4V}{3A}=\frac{4*3kN}{3\pi25^2}$$ And ot happens at the horizintal axis passing through the centroid.

The max shear stress due to torsion happens at the surface of the bar.

$$\tau_{max} = \frac{2t_{max}}{\pi r^3}=\frac{2kN*1000mm/m}{\pi *25^3}$$

So the maximum shear stress is the sum of the two stresses and happens along the two sides of the bar.

As for the bending moment:

M=3kN*250mm=750kNmm $$I_x = \pi r^4 / 4 \rightarrow S_X=I/c= \pi r^3/4$$

$$\sigma =m/S=750*4kNmm/(\pi*250^3)$$

This has to be added and subtracted from the normal tension of $$\frac{15kN}{\pi 25^2}$$

You are not alone, many people are confused about the moment of inertia and polar moment of inertia of the circular shape, $$I, I_p$$, and their uses.

Let's review two equations:

$$\sigma_b = \dfrac {My}{I}$$, where $$\sigma_b$$ is the normal stress induced by the bending moment $$M$$, and $$I$$ is the moment of inertia. For circle, $$I = \dfrac {\pi d^4}{64}$$, where $$d$$ is the diameter of the circle.

$$\tau = \dfrac {TL}{J}$$, where $$\tau$$ is the shear stress induced by the torsional force $$T$$, and $$J$$ is the polar moment of inertia, sometimes written as $$I_p$$. For circular shape, $$J$$ or $$I_p = \dfrac {\pi d^4}{32}$$

The cantilever beam in the example subjects to combined stresses of "tension", "bending", "shear" and "torsion". Therefore,

$$\sigma_{A,B} = \dfrac {F_t}{A} \pm \dfrac {My}{I_x}$$ - $$Normal Stress$$

$$\tau_{A,B} = \sqrt (F_y/A)^2 + (TL/I_p)^2$$ - $$Shear Stress$$

Note: The concept above ($$"M" vs "I"$$; $$"T" vs "I_p"$$) applies to all shapes but differs in the form of the polar moment of inertia ($$I_p$$ or $$J$$).

For a circular shaft, the maximum and minimum stresses always fall on the perimeter at the opposite points (extreme points). The locations depend on the direction of the bending moment and the nature of the axial force, whether it is a tensile or compressive force. You can stop at finding and expressing the shear stress and normal stress separately for most of the practical concerns, but occasionally (especially for academic works), you might need to consider the maximum point stress by combining the shear stress and normal stress using the method of "square root of squares, $$c = \sqrt{a^2 + b^2}$$).

Note, the maximum/minimum stress locations stated above do not hold for other shapes.