Mathematical problems Assignment Help

Mathematical problems

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Question 1

• 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.

For point A:

Question 2

For pure shear to occur, the stresses σ₁ and σ₁ must be equal in magnitude and opposite in sign

σ₁= Pr/t and σ₂= 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 Pr²=3* π*0.1²*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*r²*π-F)/220πr = (12000000*0.1²* π-190000)/ (220000000* π*0.1) = 0.00271 m

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=2m², resultant force W= wind pressure * area

W=2*1.5=3KN

Torque T=resultant force W* distance b, T=3*1.05=3.15 kN-M

Bending moment M=W* height=3*3.5=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 = (M*d₂/2)/ [π(d₂⁴-d₁⁴)/64]= (10.5 *0.11/2)/ [π(0.11⁴-0.09⁴)/64]=0.5775/(3.96626*10^-6)=145603.163 MPa

Torque produces shear stresses ҭ₁ at point A and B

Ҭ_torsion == (T*d₂/2)/ [π (d₂⁴-d₁⁴)/32]=(3.15 *0.11/2)/ [π(0.11⁴-0.09⁴)/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

At point B σ_x=0, σ_y= σ_a and Ҭ_xy= Ҭ₁=21840.4689 Pa, therefore, Maximum plane shear stress

Ҭ_max =√ [(0 – 145603.163)²/2 + (21840.4689)²]= √(1.107715*10^10) =105248.04 Pa

At point C, σ_y= σ_x=0 and Ҭ_xy= Ҭ₁+ Ҭ_2, Ҭ₂ =2V/A= (2*3000)/2=3000 N/m², 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*0.05²)/4]

=34000/0.001963496=17.316 MPa

Shear torsion= Tr/I_p = (34000 *0.05/2)/[(0.05)⁴*π/32]=850/(6.135922*10^-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]

σ₁= 8.658+√1919089.49=1393.97 MPa

σ₂= -1376.6541 MPa

Maximum in plane shear stress

= √ [((0 -17.316)/2)²+ (1385.285)²] = 1385.31 MPa

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