Question

A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is applied to the top side of the plate. A normal tensile force of 20kN is applied to the right side of the plate. The elastic modulus, E, is 115 GPa for titanium. If the left and bottom edges of the plate are xed, calculate the normal strain and elongation of both the TOP and RIGHT side of the plate. Report your answer with proper units and signicant digits.

Answer 1

Answer:

For the Top Side

- Strain ε  = 0.00021739

- Elongation is 0.00260868 cm

For The Right side

- Strain ε  = 0.00021739

-Elongation is 0.00347826 cm

Explanation:

Given the data in the question;

Length of the squared titanium plate = 12 cm by 12 cm = 0.12 m by 0.12 m

Thickness = 5 mm = 0.005 m

Force to the Top F$_t$ = 15 kN = 15000 Newton

Force to the right F$_r$ = 20 kN = 20000 Newton

elastic modulus, E = 115 GPa = 115 × 10⁹ pascal

Now, For the Top Side;

- Strain = σ/E = F$_t$  / AE

we substitute

= 15000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 15000 / 69000000

Strain ε  = 0.00021739

- Elongation

Δl = ε × l

we substitute

Δl = 0.00021739 ×  12 cm

Δl = 0.00260868 cm

Hence, Elongation is 0.00260868 cm

For The Right side

- Strain = σ/E = F$_r$  / AE

we substitute

Strain = 20000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 20000 / 69000000

Strain ε = 0.000289855

- Elongation

Δl = ε × l

we substitute

Δl = 0.000289855×  12 cm

Δl = 0.00347826 cm

Hence, Elongation is 0.00347826 cm