Normal and Shearing Stress - Combined Loading

[erobz]

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

Here is my combined loading:

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?

 

[Lnewqban]

@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]

@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.

 

[erobz]

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...

 

[erobz]

Here is my worked solution if anyone is interested.