• Engineering
• erobz
In summary, the conversation discusses a problem involving normal and shear stresses in a pipe or other closed tubular geometry. The book solution for normal and shear stresses are given, along with the individual's calculations for the values. The individual also mentions using a single pipe wall thickness in the calculation for the first moment, which results in the book answer being obtained. The individual provides their worked solution for reference.
erobz
Homework Helper
Gold Member
Homework Statement
Determine the normal and shear stresses developed at the points of interest.
Relevant Equations
Strength of Materials

Normal Stress ##\sigma##:

## \sigma = \frac{F}{A}## - tensile/compressive
##\sigma = \frac{Mr}{I}## - bending

Shear Stress ##\tau##:

## \tau = \frac{Tr}{J}## - Torsion
## \tau = \frac{QV}{It}## -Shear Flow

The book solution for normal and shear stresses respectively are:

a) ##20.4~\text{MPa}, 14.34 ~\text{MPa} ## - I find both

b) ##-21.5~\text{MPa}, \boxed{19.98~\text{MPa}}## - I find the normal stress, but I'm not getting the book answer for the shear stress.

I'm getting Torsion + Shear Flow ##\approx 25.6 \text{MPa}##

I'm just brushing up for the heck of it...No professor to ask. Does anyone get solution in the book?

@erobz
I may be wrong, but for the cross-section shared by points a and b, my values are:

Fx = 0
Mx = 90 kN-mm

Fy = 1.5 kN
My = 108 kN-mm

Fz = 1.2 kN
Mz = 67.5 kN-mm

Would you mind to verify?

erobz and DeBangis21
Lnewqban said:
@erobz
I may be wrong, but for the cross-section shared by points a and b, my values are:

Fx = 0
Mx = 90 kN-mm

Fy = 1.5 kN
My = 108 kN-mm

Fz = 1.2 kN
Mz = 67.5 kN-mm

Would you mind to verify?

Sorry, I had some bad editing there. Yeah, those match my values.

Lnewqban
Never mind. I believe I found the problem. For the shear flow in a pipe (or other closed tubular geometry depending on position I suppose):

$$\tau = \frac{QV}{I (2t)}$$

Its either that or you only use a quarter of the pipe in ##Q## computation( I was using a half - with single wall thickness). To use a single pipe wall thickness ##t##, the first moment ##Q## is computed for one quarter of the pipe.Adding that to the Torsion gets me the book answer...

Last edited:
Lnewqban
Here is my worked solution if anyone is interested.

Last edited:
DeBangis21 and Lnewqban

1. What is normal stress and how is it different from shearing stress?

Normal stress is a type of stress that acts perpendicular to the surface of an object, while shearing stress acts parallel to the surface. Normal stress is caused by forces acting in opposite directions, while shearing stress is caused by forces acting in the same direction but with different magnitudes.

Combined loading refers to a situation where an object is subjected to both normal and shearing stresses simultaneously. This can result in a complex stress state, as the two types of stress can interact and affect each other. The overall stress on the object will depend on the magnitude and direction of the applied forces.

3. What are the equations used to calculate normal and shearing stresses in combined loading?

The normal stress is calculated using the equation σ = F/A, where σ is the stress, F is the force acting on the object, and A is the cross-sectional area. The shearing stress is calculated using the equation τ = F/A, where τ is the shearing stress, F is the force acting on the object, and A is the area of the surface being sheared.

4. How do you determine the maximum stress in a combined loading situation?

The maximum stress in a combined loading situation can be determined by using the Mohr's circle method. This involves plotting the normal and shearing stresses on a graph and finding the point where the stress state is at its maximum. The radius of the circle at this point represents the maximum stress.

Combined loading can lead to failure in an object if the stresses exceed the strength of the material. This can happen if the normal and shearing stresses are acting in a way that causes high levels of stress concentration, or if the stresses are applied repeatedly over time, leading to fatigue failure. It is important to carefully consider the stress state in combined loading situations to prevent failure.

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