[amp_mcq option1=”$$\\frac{2}{3}$$” option2=”$$\\frac{3}{2}$$” option3=”$$\\frac{3}{4}$$” option4=”$$\\frac{4}{3}$$” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\frac{3}{2}}$.<\/p>\n
The maximum shear stress in a circular beam occurs at the neutral axis, and the average shear stress occurs at the outer edge of the beam. The ratio of maximum shear stress to average shear stress is given by the following equation:<\/p>\n
$$\\frac{\\tau_{max}}{\\tau_{avg}} = \\frac{3}{2}$$<\/p>\n
This equation can be derived by considering the equilibrium of a small element of the beam. The shear stress on the top surface of the element is $\\tau_{avg}$, and the shear stress on the bottom surface of the element is $\\tau_{max}$. The shear force on the element is equal to the difference between the shear stresses on the top and bottom surfaces, multiplied by the area of the element. The equilibrium equation for the element is then:<\/p>\n
$$\\tau_{avg} A – \\tau_{max} A = 0$$<\/p>\n
Solving for $\\tau_{max}$, we get:<\/p>\n
$$\\tau_{max} = \\frac{3}{2} \\tau_{avg}$$<\/p>\n
Therefore, the ratio of maximum shear stress to average shear stress is $\\frac{3}{2}$.<\/p>\n
Option A is incorrect because it is the ratio of the maximum shear stress to the minimum shear stress. The minimum shear stress occurs at the outer edge of the beam, and the maximum shear stress occurs at the neutral axis. The ratio of the maximum shear stress to the minimum shear stress is therefore always greater than 1.<\/p>\n
Option B is incorrect because it is the ratio of the average shear stress to the minimum shear stress. The minimum shear stress occurs at the outer edge of the beam, and the average shear stress occurs at the neutral axis. The ratio of the average shear stress to the minimum shear stress is therefore always less than 1.<\/p>\n
Option C is incorrect because it is the ratio of the maximum shear stress to the average stress. The average stress is equal to the product of the Young’s modulus and the strain. The maximum shear stress is always greater than the average stress.<\/p>\n
Option D is incorrect because it is the ratio of the average stress to the minimum stress. The minimum stress is equal to the product of the Young’s modulus and the strain. The average stress is always greater than the minimum stress.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\frac{2}{3}$$” option2=”$$\\frac{3}{2}$$” option3=”$$\\frac{3}{4}$$” option4=”$$\\frac{4}{3}$$” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[639],"tags":[],"class_list":["post-5639","post","type-post","status-publish","format-standard","hentry","category-theory-of-structures","no-featured-image-padding"],"yoast_head":"\n