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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


Area of the sign board=2*1=2m2, resultant force W= wind pressure * area


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 σa at point A and tensile stress

σa = (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

Ҭtorsion == (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 Ҭmax=√ [( σx – σy)2/2 + (Ҭxy)2]

At point A the shear stress is zero

At point B σx=0, σy= σa and Ҭxy= Ҭ1=21840.4689 Pa, therefore, Maximum plane shear stress

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

At point C, σy= σx=0 and Ҭxy= Ҭ1+ Ҭ2, Ҭ2 =2V/A= (2*3000)/2=3000 N/m2, hence Maximum plane shear stress   Ҭmax =√ [(0 – 0)2/2 + (3000+21840.4689)2] =24840.4689 Pa

Question four


Tensile stress = tensile force P/ area A=34000/[ (π*0.052)/4]

=34000/0.001963496=17.316 MPa

Shear torsion= Tr/Ip = (34000 *0.05/2)/[(0.05)4*π/32]=850/(6.135922*10^-7)=1385.285 MPa

Principal stresses are σ1,2  = [( σx + σy)2/2 ±√[( σxy)2/2+ (ҭxy)2]

σ1,2  = (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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