Engineering Normal and Shearing Stress - Combined Loading

AI Thread Summary
The discussion focuses on calculating normal and shear stresses under combined loading conditions. The original poster presents two sets of stress values, noting discrepancies with book solutions, particularly for shear stress. A participant confirms their calculations and suggests that the issue lies in the method used for computing shear flow in a pipe, specifically regarding the first moment of area (Q). They clarify that using a quarter of the pipe for Q computation, rather than half, aligns their results with the book's answers. The conversation concludes with an offer to share the worked solution for further reference.
erobz
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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
1679773212754.png


Here is my combined loading:

1679774155412.png


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?
 
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@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?
 
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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?
1679780434749.png


Sorry, I had some bad editing there. Yeah, those match my values.
 
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...
 
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Here is my worked solution if anyone is interested.

1679796068512.png

1679796114607.png


1679840078687.png
 
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