Mathematical problems

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**SAMPLE ANSWER**

**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 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

**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=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*d_{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/(3.96626*10^-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

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}]= √(1.107715*10^10) =105248.04 Pa

At point C, σ_{y}= σ_{x}=0 and Ҭ_{xy}= Ҭ_{1}+ Ҭ_{2, }Ҭ_{2 }=2V/A= (2*3000)/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/(6.135922*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}]

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

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