Question

H7102 PRINCIPLES AND APPLICATIONS OF STRENGTH OF MATERIALS Q2. Figure Q2 shows a cylindrical pressure vessel made by welding semi-spherical caps to a cylindrical body. The cylindrical part has inner diameter Di = 1.8 m and the wall thickness, t, is uniform and equal to 6 mm. The internal pressure is p = 1.5 MPa. a) Determine the maximum and minimum direct stresses in the cylinder wall. What is their direction? [4 marks] b) What is the value, and in which direction, does the maximum shear stress, in the plane of the cylindrical wall occur? Are there shear stresses in other directions, larger than this one? [5 marks] c) If the weld can withstand a maximum tensile force of 787500 N per metre of length, what is the minimum internal pressure in the vessel at which the vessel fails? [5 marks] d) What is the value of a force F, applied in the direction of the axis of the cylinder which would make the direct stress in longitudinal direction, in the cylinder wall, one quarter the value of the hoop stress? [6 marks] 12.5 kytlmm?,9.8H 84 1-8 m= 18c0 mm

Could you solve the above questions in the photo?

Answer 1

a Giren data Pressure (P) = 15 mpn = 1.5x 100 Pa Imer Diameter (Di) = 1.8m Thickness (t) = 0 mm = 6X20-2 m Di = 300 < 20 The cylinder is thin cylinder . Radial stress To P* Di 2xt - 1-58106 1.8 2 * 6x 103 = 225 x 106 Pa = 225 mea Longitudinal stress 7 = r = 112.5 MPa . Stress block - 225 M PA =TY • 112.5 min = x = Tol VE 9 Maximum and minimum direct stresses A > + cu Tunz - Goiena. (1930) + zin] 24 2-0

Putting all the value » 2 168 75 56.25 .: Maximum direct stress Di = 168.75+ 56.25 = 225 MPa on Minimum direet stress a = 168.75 -56.25 = 112.5 Men 6 Maximum shear stress > Z-[(=)=zul]" 2 max . = 56.25 MPa and the angle of maximum shear stress pline 85 tone [ Zao

© Total force in the weld Pressure x cs area F 2 2 Parimeter p = ext Dil ndi i. Pour unit length force Format - PR For limiting condition F = 787500 N/m - 787500 = Px1.8 as a P = 1.75 x 106 Pa : P = 1.75 MPa . Maximum allorable pressure is 1.75 MPa

@ we know longitudinal stress is already half of the hoop stress I radial stress For ħ radial stress there must be a compressive stress of applied longitudinally suppose the stress is to ~ Te-a=t&r ²5-re=th to a = ħor = 14 +225 = 56.25 MIN Cross sectional area of the wall of cylinder Ру - жxp; XL - TX1.8 X 6X 10-3 . Force required » F 2 Dot Aw = 56.25x 106 ITX 1-8 x 10-3 F = 318086.26 N [compressive]