Normal Stress, sigma = normal force / area at the section sigma = Psin 60 degree

Question

Normal Stress, sigma = normal force / area at the section sigma = Psin 60 degree / (A0 / sin 60 degree) sigma = P sin 60 degree sin 60 degree / A0 A0 = 3 times 5 in2 A0 = 15 in2 sigma = 1. 4 sin 60 degree sin 60 degree / 15 = 0. 07 ksi sigma = 70 psi Shear stress, tau = shear force / shear area at the section tau = P cos 60 degree /(A0 / sin 60 degree) tau = P cos 60 degree sin 60 degree / A0 = 1. 4cos60 degree sin60 degree / 15 = 0. 0404 ksi tau = 40. 4 psi The 1. 4-kip load P is supported by two wooden members of uniform cross section that are joined by simple glued scarf spilce shown. Determine the normal and sharing stresses in the glued splice. Fig. P1. 29 Given that, The load P = 14 kips From the figure. The normal force at the section = P sin 60 degree the shear force at the section=p cos 60 degree

Explanation / Answer

The area of the section is A0/sin(60)=15/(sqrt(3)/2))=30/sqrt(2) which you are right it is is bigger than 15 (though it is not 30). But note that the author doesn't say the area is 15, but it moves the sin(60) to the numerator

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