## Why choose us?

We understand the dilemma that you are currently in of whether or not to place your trust on us. Allow us to show you how we can offer you the best and cheap essay writing service and essay review service.

# Mechanics of solids

Mechanics of solids

Assignment1: Q4 – To calculate angle of twist take length L=70mm, If you have assumed the length by
now, write the assumption in the begining of the solution

Assignment: mechanics of solids

Question 1

Assignment: mechanics of solids 2

Normal stress σ = F n /A (N/m 2 ) Where F n is normal component of force
A=is the cross-sectional area of the hand brake
From the data
F n =42 N and A=1.07510^-6 therefore, normal stress σ = 42/ (1.07510^-6) =39069767.4 N/m
or Pa
Young modulus E = stress σ/ strain ε
Therefore, Strain ε = stress σ/ Young modulus E, ε = 39069767.4/14010^9 ε=0.27 Question two Diameter 8mm and length 200mm.allowable torsional shear stress is 60Mpa. Solution D=0.008m, R=0.004m T/J=Gθ /L =ҭ/r .J=πD 4 /32 = (π0.008 4 )/32 =4.02285610^-10 M 4 The complete torsion equation is T/J=Gθ /L =ҭ/R The maximum permissible torque T= (ҭmaxJ)/R = (6010^64.022856*10^-
10)/0.004=6.034285 N-M
Question 3
P=42N, length=225mm,shear modulus G=120 Gpa, diameter d=15mm.

Assignment: mechanics of solids 3

Solution
D=0.015m, L=0.225m, T=Fd=420.225=9.45 Nm
G=120 Gpa, R=0.0075
J= (π0.015 4 )/32 =4.97209710^-6 m 4
A) Maximum shear stress( ҭmax)= (9.45 0.0075)/4.97209710^-6
Ҭmax =14254.5489 Pa
Θ= (TL)/J=(9.450.225)/(12010^94.97209710^-6) =2.12625/596651.64 Θ= 0.00000356 radians Θ= (0.00000356180)/ π =0.000204 degrees

Question 4
Diameter d=4.0mm, length L=70mm, d=0.004m,R=0.002,L=0.07m
Maximum shear stress is given by;
J= (π0.004 4 )/32 = 5.63210^-8/224=0.0251428610^-8 m 4 . Torque T= (ҬmaxJ)/R, Ҭmax = (TR)/J =(0.30.002)/0.02514286*10^-8
Ҭmax = 0.02386363 *10^8 Pa
(b)

Assignment: mechanics of solids 4

Shear modulus G=75 GPa,the twist angle is given as Θ=TL/J=(0.3