Question

a steel plate of width 120mm and of thickness 20mm is bent into a circular arc of radius 10m. ditermine the maximum stress induced and the bending moment which will produce the maximum stress. Take E =2×10*5 N/mm2​

Answer 1

Answer:

1200 mm john cena, undertaker the dedrman

Answer 2

Answer:

The maximum stress, σ, induced is $200N/mm^{2}$.

The bending moment, M, is $1.6 \times 10^{6}N-mm$.

Explanation:

Given,

The width of the steel plate, b = $120mm$

The plate's thickness, t = $20mm$

The radius of the circular arc, r = $10m$ = $10^{4}mm$

The Young's modulus, E = $2\times 10^{5} N/mm^{2}$

The maximum stress, σ =?

The bending moment, M =?

As we know,

• The maximum stress can be calculated by using the equation of bending theory given below:
• $\frac{\sigma}{y} = \frac{E}{r}$       -------equation (1)

Here,

• σ = The maximum stress
• y = The perpendicular distance
• E = The modulus elasticity
• r = The arc's radius

Now, we have to calculate the perpendicular axis (y).

• The perpendicular axis is half of the thickness.
• $y = \frac{t}{2}$ = $\frac{20}{2}$ = $10mm$

After putting the given values in the equation (1), we get:

• $\frac{\sigma}{10} = \frac{2 \times 10^{5} }{10^{4} }$
• σ = $200N/mm^{2}$

Now, the bending moment can be calculated by the equation given below:

• $\frac{M}{I} = \frac{\sigma}{y}$    -------equation (2)

Here,

• M = The bending moment
• I = The moment of inertia of the cross-section
• σ = The maximum stress
• y = The perpendicular distance

For this, we have to calculate the moment of inertia of the cross-section (I) by the formula given below:

• $I = \frac{bt^{3} }{12}$ = $\frac{120 \times (20)^{3}}{12}$ = $8\times 10^{4}mm^{4}$

After putting the values in equation (2), we get:

• $\frac{M}{8\times10^{4} } = \frac{200}{10}$
• M = $1.6 \times 10^{6}N-mm$

Hence, the bending moment, M = $1.6 \times 10^{6}N-mm$

And, the maximum stress, σ = $200N/mm^{2}$.