Acircular cross section, d = 25 mm, experiences a torque load, t = 25 n·m, and a shear force, v = 85 kn. calculate the largest shear stress within the cross section. b. 239 mpa a. 231 mpa c. 8.15 mpa d. 95.6 mpa

largest shear stress = 239 N/mm²

Explanation:

Given data

cross section d = 25 mm

torque load T = 25 N·m

shear force V = 85 kN

to find out

largest shear stress

solution

we know that maximum shear stress due to load = 16T / πd³

maximum shear stress = 16 ( 25) / π(25)³

maximum shear stress = 8.148 N/mm²    ........................1

maximum shear stress due to force = 14/3 × avg shear stress

maximum shear stress due to force = 14 × 85 ×  / (3 ×π/4×25²)

maximum shear stress due to force = 230.88 N/mm²    ..................2

so

largest shear stress add equation 1 and 2

largest shear stress = 8.148 + 230.88

largest shear stress = 239 N/mm²

a displacement distance is size of displ

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force