材料力学课堂教学软件（英文版）-范钦珊-chap.7 Composite Forces.pptVIP

Problem:The cylindrical vessel of average diameter 1800mm and wall thickness 14mm is subjected to internal pressure. Knowing that the material of the vessel is steel and the yield stress is σs= 400MPa. For the safety factor ns = 6.0, refer to the criterions below, determine the maximum internal pressure which may be applied to the vessel. Back to the first page Problem 8 The Third Kind of Exercise: The Strength Design under Complex Stress-State σt σm Solution： Determine the normal stress on a general element of the vessel Application of the maximum shearing stress criterion By using the maximum shearing stress criterion we have ＝10 .37 at. So, the maximum internal pressure can be obtained: Back to the first page Problem 8 The Third Kind of Exercise: The Strength Design under Complex Stress-State Application of the strain-energy density corresponding to the distortion criterion : σt σm Solution：Determine the normal stress on a general element of the vessel The strain-energy density corresponding to the distortion criterion is: Back to the first page Problem 8 The Third Kind of Exercise: The Strength Design under Complex Stress-State Application of the strain-energy density corresponding to the distortion criterion : σt σm ＝11 .97 at. So, the maximum internal pressure can be obtained: Back to the first page Problem 8 The Third Kind of Exercise: The Strength Design under Complex Stress-State Problem:The solid shaft ABCD and the loading are shown in the figure below. Knowing that ［σ］ = 120MPa, determine the smallest permissible diameter of the solid shaft ABCD. Back to the first page Problem 9 The Third Kind of Exercise: The Strength Design under Complex Stress-State T = FPr x z y 150 150 A B C D Draw out the sketch for calculation T N N Back to the first page FP MB=T FC FA FP MB=T FC FA sketch for calculation Problem 9 The Third Kind of Exercise: The Strength Design under Complex Stress-State Mmax Diagram of the internal forces Back to the first page Bendin