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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

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

Assignment: mechanics of solids 4

Shear modulus G=75 GPa,the twist angle is given as Θ=TL/J=(0.3
0.0710^8)/(0.025142867510^9) =0.083523 radians
In degrees Θ= 0.083523 180/ π =4.784 degrees Question 5 Internal moment m=42 N-m For a rectangular cross-section, the second moment of inertia I= (πd 4 )/64
b =0.15m, h=0.2m, the modulus S= (bh 2 )/6 =(0.150.2 2 )/6 =0.001,for circular section of
diameter d
S= (πd 3 )/32= (π0.075 3 )/32 =3.1076 *10^-6 m 3
Total modulus S =3.1076 *10^-6 +0.001, S= 0.0010031 m 3
S= (I2)/h, second moment of inertia I= (sh)/2= (0.00100310.2)/2 =0.000100311 m 2 Maximum tensile stress= internal moment/second moment of inertia=42/0.000100311 =418697.849687 N/m. Maximum Compressive stress=-418697.849687 N/m. Question 6 E=125 Gpa the second moment of inertia I= = =18.75 m 2 Mass=42kg and force F=422.59.8=1029 N Maximum normal stress σ max =10291.5/18.75=82.32 Nm

Assignment: mechanics of solids 5

The maximum normal strain ε max = σ max /E =82.32/125*10^9 =0.65856 * 10^-9

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