# Principle stresses and maximum shearing stress in a cylindrical shaft. Pl help

• taureau20
In summary: In summary: In summary, the homework statement states that the principle stresses are found to be 78.09 MPa and -148.135 MPa.

## Homework Statement

Nevermind the rough picture, the shaft is cylindrical. Knowing that the post has a diameter of 60mm, determine the principle stresses and the maximum shearing stress at point K.

## Homework Equations

These are equations for your reference:

$\sigma = My/I$
$\tau = Tc/J$
Where M is bending moment, T is twisting torque. J is polar moment of inertia. c is radius of cylindrical shaft. y is distance of point from neutral axis.

## The Attempt at a Solution

I find:

J = pi/2 . c^4 = 1.27e-6 m^4; I = J/2 = 6.36e-7 m^4

$\tau_K = Tc/J = M_yc/J = 1.125e3\cdot0.03/ 1.27e-6 = 26.56MPa$

$(\sigma_K)_y= M_zc/I = -2.7e3\cdot0.03/ 6.36e-7 = -127.32 MPa$

$(\sigma_K)_z = 0$ right??

Because of the shearing force, $P_z, \tau = 3/2 \cdot P/A = 3/2 \cdot 7.5e3/(\pi /2 \times .03^2) = 7.965 MPa$
So, $\tau_{total}= 26.57MPa + 7.965MPa = 34.535MPa$
Now, because of $M_x$ there won't be any normal longitudinal stress at K because K lies on neutral axis, right? So the only normal longitudinal stress is due to $M_z$.

Now, because of P_y there will be compressive stress = $P_y/A = -18e3/(\pi / 2 \times .03^2) = -12.73MPa$
This I add with the normal longitudinal stress $\sigma_y=-127.36MPa$ calculated earlier.
So, $(\sigma_y)_{total} = -127.36 - 12.73MPa = - 140.09MPa$

So this is the diagram for the stress distribution on an element at point K:

So the principle stresses are found to be:

$\tau_{max} = \sqrt{((\sigma_y-\sigma_z)/2)^2 + (\tau_{yz})^2} = \sqrt{((-140.09-0)/2)^2 + 34.535^2} = 78.09 MPa.$$And, \sigma_{max,min} = \sigma_{ave} \pm \tau_{max} = (-140.09/2 \pm 78.09) MPa = -148.135MPa, 8.045MPa$The answers in the back of the text are : 6.45MPa, -140.0MPa for 73.3MPa. I am not sure which value is for which stress. I guess the first and last values are for the principle stresses and the middle value is for the maximum shearing stress.

Last edited:
taureau20: Excellent work. Double-check the given question in your textbook and see if you erroneously omitted or supplied a negative sign on Py, Pz, My, or Mz in your above diagram. Secondly, check your formula for cross-sectional area; it is currently incorrect. See if this resolves the problem. Also, per the international standard (ISO 31-0), always leave a space between a numeric value and its following unit symbol; e.g., 73.3 MPa, not 73.3MPa.

O Christ! How can I write area = pi/2 r^2 when its pi.r^2!

## 1. What are principle stresses in a cylindrical shaft?

Principle stresses refer to the maximum and minimum stresses that occur in a specific direction on a cylindrical shaft. These stresses are important in determining the structural integrity and strength of the shaft.

## 2. How are principle stresses calculated in a cylindrical shaft?

Principle stresses can be calculated using the formula σ1= (σx + σy)/2 ± √((σx-σy)^2/4 + τxy^2), where σ1 is the maximum principle stress, σx and σy are the normal stresses in the x and y directions, and τxy is the shear stress in the xy direction.

## 3. What is the significance of principle stresses in a cylindrical shaft?

The principle stresses help in determining the failure criterion of the shaft and predicting its failure under different loading conditions. They also provide insight into the direction of the stresses and help in designing safe and efficient shafts.

## 4. What is the maximum shearing stress in a cylindrical shaft?

The maximum shearing stress is the highest stress that occurs in a direction perpendicular to the principle stresses. It can be calculated using the formula τmax = (σmax - σmin)/2, where σmax and σmin are the maximum and minimum principle stresses.

## 5. How does the maximum shearing stress affect the strength of a cylindrical shaft?

The maximum shearing stress is an important factor in determining the strength and failure of a cylindrical shaft. If the shearing stress exceeds the ultimate shear strength of the material, it can lead to failure and damage of the shaft. Therefore, it is crucial to consider the maximum shearing stress in the design and analysis of cylindrical shafts.