Question

You are part of a team that is designing a structural component. The applied pressure loading will be obtained by another team member using CFD analysis. The structure can be idealised as a simply supported beam. The tensile strength of the material will be used to size the structure. While you are waiting for the CFD calculations, you decide to perform an initial calculation on the structure with a uniformly distributed load (as shown below). y 3 B х 2a AO a tk B L= 150 mm E = 69.0 GPa, o allow(tension) = 305 MPa, tallow= 150 MPa , w = 28 kN/m in order to evaluate the initial design, you will: a) Determine the critical dimension, a, based on the assumed loading. Make sure that your design can withstand the tensile and shear loads. b) For your given dimension of a, determine the maximum deflection.

Answer 1

Given Information :

• 
  E = 69 G Pa
  
• 
  w 28 kN/m
  
• 
  O allow 305 MPa
  
• 
  Tallow 150 M Pa
  
• 
  L = 150 mm = 0.15 m
  

Solution :

(a) Critical Dimension :

Ans :

Let the reactions at the supports be :

R & Rb

Using the static equilibrium equations :

ΣF, = 0

Rg + R =w* L

R2 + Rp = 28 * 0.15 = 4.2 kN

Taking moment about point A :

Μ = 0

R; * 0.15 - w * 0.15 * 0.15 2 = 0

* 0.15 – 28 * 0.15 * 0.15 2 = 0

Rp = 2.1 kN

Therefore :

R = 2.1 kN

Finding the position and the value of Maximum Bending moment along the beam :

Writing the Moment equation of an arbitrary section taken at a distance of x from the hinged support at A:

W * M = Ra*2 2

In order to find the position of maximum bending moment, differentiate the above equation with respect to x and equate it to zero.

W* C DM. dc d -(Ra * I dac 2

W* 2 * 2 0 = Ra 2

Ra C

2.1 T= 28

r= 0.075 m

Therefore at a distance of from the hinged support, maximum bending moment occurs.

r= 0.075 m

Substitute the value to find the maximum bending moment :

W * M = Ra*2 2

28 *0.0752 Мmax = 2.1 + 0.075 -

Mmar 0.07875 kN - m

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

M *у оъ INA

Substitute the values for maximum bending stress :

2a y = a 2

a* 2a4 INA = (2a) 12 3

M *у оъ INA

M ka О. 2a4 3

M + 3 оъ 2a3

0.07875 * 10 * 108 * 3 305 20

a = 728 mm

Therefore considering bending stress, the critical dimension is :
a = 728 mm

Finding the position and the value of Maximum Shear force, along the length of the beam:

Writing the Shear force equation for the given loading, taking an arbitrary section at a distance x from the hinged support at point A:

Ra TT W* C

As it can be seen that the shear force equation is linear, the maximum shear force is at ;

T

Vimax = 2.1 kN

Finding the critical dimension based on Maximum shear stress :

Since the given section is a rectangle :

Γω Tmar * Ταυg

| cs Tmar k

Substitute the values :

2.1 * 1000 150 یه ای ad

a = 3.240 mm

Therefore considering shear loads, the critical dimension is :
a = 3.240 mm

(b) Maximum deflection :

Ans :

Considering the critical dimension :
a = 728 mm

Calculating the Area moment of Inertia :

1 a * (2a) 12

I 7.28 * (2 * 7.28) 12

I = 1872.55 mm

Writing the Deflection equation :

EI *y' = Mon

0% α- EI * y" = R * 1 - 2

Integrating the equation :

W* EI *y = Ra* +C 6

Integrating the equation :

W* C EI *y = Ra* + Cl2 + C 6 24

Finding the constants of integation using the boundary conditions :

r = 0 + y = 0

$EI*0=R_a*\frac{0^3}{6}-\frac{w*0^4}{24}+C_1*0+C_2$

$C_2=0$

L + y = 0

$0=2.1*1000*\frac{0.15^3}{6}-\frac{28*1000*0.15^4}{24}+C_1*0.15$

$C_1*0.15=-0.590625$

$C_1=-3.9375$

The Deflection equation is :

$EI*y=2100*\frac{x^3}{6}-\frac{28000*x^4}{24}-3.9375x$

The Maximum Deflection is :

Substitute the values :

$y_{max}=\frac{5}{384*1872.55*69*10^3*1000}*28000*(150)^4$

Therefore the maximum deflection is :  .