
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 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 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
Solution
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 ±√[( σx -σy)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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