Question

Ty 2 x R For the combined loading Shown here, make up values of Ty, E, F & FyrR and complete a stress analysis @ Pt A usi using Super position. Then complete a stress analyses in SOLIDWORKS AND compare the results Book a tine wlume to review your analysis on E Zoom

The Solidworks part is not needed just detailed calculations.

Answer 1

Given Information :

• For the given loading and the type of beam, Assume the values and give complete stress analysis of point A.
• The beam is of circular cross section.

Assumption :

The load  
य
is axial load applied in the negative Y direction.

Solution :

Assuming the values :

Ty = 10 KN

Fy = 10 KN

F= 0.2 m

E = 0.03 m = 30 mm

Analysing the stresses that are acting at point A :

Location of Point A: It is located at a distance of
F= 0.2 m
from the extension and on the Right Fiber of the beam.

At point A, the following streses will be acting :

1. Normal Compressive stress due to the force :
   य
   
2. Normal Compressive stress due to the force : $F_y$
3. Normal Bending stress due to the moment about Z axis developed by the eccentric force : $F_y$ equaling : $M_z=F_y*E$

The normal stress due to axial load is given by the formula

$\sigma_a=\frac{F}{A}$

Where
$\sigma_a=$ Axial Stress
$F=$ Axial Load
$A=$ Cross sectional area

From the equation of pure bending

$\frac{M}{I_N_A}=\frac{\sigma_b}{y}=\frac{E}{R}$

where
$M=$ Bending moment
$I_N_A=$ Area Moment of Inertia about Neutral Axis
$\sigma_b=$ Bending stress
$y=$ Distance between the neutral axis and the fiber at which bending stress is being calculated
$E=$ Young's Modulus of elasticity
$R=$ Radius of curvature of elastic curve

$\sigma_{b}=\frac{M*y}{I_{NA}}$

Calculating the stresses :

Normal Compressive stress due to  

य

Substituting values

$\sigma_{a1}=\frac{T_y}{\pi *R^2}$

Substitute the values :

$\sigma_{a1}=\frac{10*1000}{\pi *10^2}$

$\sigma_{a1}=31.83\ MPa$

Normal Compressive stress due to  $F_y$

$\sigma_{a2}=\frac{F_y}{\pi *R^2}$

Substiute the values :

$\sigma_{a2}=\frac{10*1000}{\pi *10^2}$

$\sigma_{a2}=31.83\ MPa$

Normal Bending stress due to the moment generated by the eccentricity of load  $F_y$

$\sigma_{b}=\frac{M*y}{I_{NA}}$

For circular cross-section,and for the point A situated at the periphery of the cross-section:

d y

$I_{NA}=\frac{\pi*d^4}{64}$

Substiute the values :

$\sigma_{b}=\frac{64*M*d}{2*\pi*d^4}$

$\sigma_{b}=\frac{32*M}{\pi*d^3}$

$\sigma_{b}=\frac{32*F_y*E}{\pi*d^3}$

$\sigma_{b}=\frac{32*10*1000*30}{\pi*20^3}$

$\sigma_{b}=381.97\ MPa$

Since point A is at the RIGHT Fiber of the beam, this fiber is in compression, Hence the normal stress due to Bending Moment generated by Eccentric Axial load is compressive in nature.

Now that all the stresses acting on the cross-section are normal to the cross-section, we can add them up.

$\sigma=\sigma_{a1}+\sigma_{a2}+\sigma_b$

$\sigma=31.83+3.83+381.97$

$\sigma=445.63\ MPa$