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01 Solutions 46060



5/6/10



2:43 PM



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•1–101. The 200-mm-diameter aluminum cylinder supports a compressive load of 300 kN. Determine the average normal and shear stress acting on section a–a. Show the results on a differential element located on the section.



300 kN



a



Referring to the FBD of the upper segment of the cylinder sectional through a–a shown in Fig. a, +Q©Fx¿ = 0;



Na - a - 300 cos 30° = 0



+a©Fy¿ = 0;



Va - a - 300 sin 30° = 0



30



Na - a = 259.81 kN



a



Va - a = 150 kN



0.1 Section a–a of the cylinder is an ellipse with a = 0.1 m and b = m. Thus, cos 30° 0.1 b = 0.03628 m2. A a - a = pab = p(0.1)a cos 30°



A sa - a B avg =



259.81(103) Na - a = = 7.162(106) Pa = 7.16 MPa Aa - a 0.03628



Ans.



A ta - a B avg =



150(103) Va - a = = 4.135(106) Pa = 4.13 MPa Aa - a 0.03628



Ans.



d



The differential element representing the state of stress of a point on section a–a is shown in Fig. b



1–102. The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determine the average normal stress in the shank, the average shear stress along the cylindrical area of the plate defined by the section lines a–a, and the average shear stress in the bolt head along the cylindrical area defined by the section lines b–b.



8 mm



a 7 mm



b



8 kN



18 mm b a



P = ss = A



8 (103)



= 208 MPa



Ans.



(tavg)a =



8 (103) V = = 4.72 MPa A p (0.018)(0.030)



Ans.



(tavg)b =



8 (103) V = = 45.5 MPa A p (0.007)(0.008)



Ans.



p 4



(0.007)2



68



30 mm