Question

Determine the average normal stress developed in rod AB if the load has a mass of 55 kg . The diameter of rod AB is 8 mm.

Answer 1

Answer:

The normal stress is 10.7[MPa]

Explanation:

The normal stress can be calculated with the following equation:

$S_{norm} =\frac{F}{A} \\where:\\F= force [Newtons]\\A=area [m^2]\\S_{norm} = Normal stress [\frac{N}{m^{2} }] or [Pa]$

The area of the rod can be calculated using the equation:

$A=\frac{\pi }{4}*d^{2}  \\d=8[mm]=0.008[m]\\A=\frac{\pi }{4}*(0.008)^{2}  \\A=5.02*10^{-5} [m^{2} ]$

The force is the result of the mass multiplied by the gravity.

$F=55[kg]*9.81[m/s^{2} ] = 539.6[N]\\\\S_{norm} = 539.6/5.02*10^{-5} \\S_{norm} = 10.7*10^{6}[Pa] = 10.7[MPa]$

Answer 2

The average normal stress developed in rod AB if the load has a mass of 55 kg  should be  10.7[MPa].

Calculation of the average normal stress:

First we need to determine the area of the rod i.e.

$A = \pi /4 * d^2$

Here d = 8mm = 0.008 m

So

$= \pi /4 *(0.008)^2\\\\= 5.02*10^-5m^2$

Now the average normal stress should be

But before that force should be

= 55*9.81

= 539.6N

So, the average should be

$= 539.6/5.02*10^-5\\\\= 107*10^6$

= 10.7 M Pa

hence, The average normal stress developed in rod AB if the load has a mass of 55 kg  should be  10.7[MPa].

Learn more about force here: https://brainly.com/question/24848856