## 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.

# Principal planes,

For point B:
Within point in a stressed body, there exist three mutually perpendicular planes and the
resultant stress is always normal stress. The mutually perpendicular planes are known as
principal planes, and the resultant normal stresses acting on them are called principal
stresses.

Surname 2

For point A:

Surname 3

Question 2
For pure shear to occur, the stresses σ 1 and σ 1 must be equal in magnitude and opposite in sign
σ 1 = Pr/t and σ 2 = Pr/2t-F/A= Pr/2t-F/2πrt
Pr/t=- (Pr/2t-F/2πrt) collecting like terms together
Pr/t+ Pr/2t= F/2πrt solving for F, we get F=3π Pr 2 =3* π*0.1 2 *12000000Pa
F=360 kN
Required thickness is given by solving for t in equations above. Using the allowable normal
stress we have 110 MPa= Pr/2t-F/2πrt, therefore,
t= (P*r 2 *π-F)/220πr = (12000000*0.1 2 * π-190000)/ (220000000* π*0.1) = 0.00271 m

Surname 4

Question 3
Pipe diameter, outer =110mm=0.11m, inner diameter=90mm=0.09m
Wind pressure =1.5 kPa
Solution
Area of the sign board=2*1=2m 2 , resultant force W= wind pressure * area
W=21.5=3KN Torque T=resultant force W distance b, T=31.05=3.15 kN-M Bending moment M=W height=33.5=10.5 kN-M Shear force V=W =3 KN Bending moment M produces a tensile stress σ a at point A and tensile stress σ a = (Md 2 /2)/ [π(d 2 4 -d 1 4 )/64]= (10.5 0.11/2)/ [π(0.11 4 -0.09 4 )/64]=0.5775/(3.9662610^-
6)=145603.163 MPa
Torque produces shear stresses ҭ 1 at point A and B
Ҭ torsion == (T*d 2 /2)/ [π (d 2 4 -d 1 4 )/32]=(3.15 *0.11/2)/ [π(0.11 4 -0.09 4 )/32]= 0.17325/(7.932522*10^-
6)=21840.4689
Maximum plane shear stress Ҭ max =√ [( σ x – σ y ) 2 /2 + (Ҭ xy ) 2 ]
At point A the shear stress is zero

Surname 5
At point B σ x =0, σ y = σ a and Ҭ xy = Ҭ 1 =21840.4689 Pa, therefore, Maximum plane shear stress
Ҭ max =√ [(0 – 145603.163) 2 /2 + (21840.4689) 2 ]= √(1.10771510^10) =105248.04 Pa At point C, σ y = σ x =0 and Ҭ xy = Ҭ 1 + Ҭ 2, Ҭ 2 =2V/A= (23000)/2=3000 N/m 2 , hence Maximum plane
shear stress Ҭ max =√ [(0 – 0) 2 /2 + (3000+21840.4689) 2 ] =24840.4689 Pa
Question four
Solution
Tensile stress = tensile force P/ area A=34000/[ (π*0.05 2 )/4]
=34000/0.001963496=17.316 MPa
Shear torsion= Tr/I p = (34000 *0.05/2)/[(0.05) 4 π/32]=850/(6.13592210^-7)=1385.285 MPa
Principal stresses are σ 1,2 = [( σ x + σ y ) 2 /2 ±√[( σ x -σ y ) 2 /2+ (ҭ xy ) 2 ]
σ 1,2 = (0 +17.316)/2 ±√[(( 0 -17.316)/2) 2 + (1385.285) 2 ]
σ 1 = 8.658+√1919089.49=1393.97 MPa
σ 2 = -1376.6541 MPa
Maximum in plane shear stress
= √ [((0 -17.316)/2) 2 + (1385.285) 2 ] = 1385.31 MPa