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This example illustrates the usage of a faientice theory to identify the strength of a mechanical element or component. The instance may also work out any kind of confusion existing between the phrases stamina of a device component, strength of a material, and also stamina of a component at a suggest. A specific force F used at D near the finish of the 15-in lever displayed in Fig. 5–16, which is quite similar to a socket wrench, outcomes in specific stresses in the cantilevered bar OABC. This bar (OABC) is of AISI 1035 steel, forged and heat-treated so that it has a minimum (ASTM) yield stamina of 81 kpsi. We presume that this component would certainly be of no worth after yielding. Hence the pressure F forced to initiate yielding deserve to be related to as the strength of the component part. Find this pressure.


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Solution


We will assume that lever DC is strong sufficient and thus not a part of the problem. A 1035 steel, heat-treated, will have actually a reduction in location of 50 percent or more and also therefore is a ductile material at normal temperatures. This additionally suggests that anxiety concentration at shoulder A need not be thought about. A stress element at A on the peak surface will be subjected to a tensile bfinishing tension and a torsional tension. This allude, on the 1-in-diameter section, is the weakest area, and governs the stamina of the assembly. The two stresses are

sigma_x=fracMI / c=frac32 Mpi d^3=frac32(14 F)pileft(1^3 ight)=142.6 F

au_z x=fracT rJ=frac16 Tpi d^3=frac16(15 F)pileft(1^3 ight)=76.4 F

Employing the distortion-power concept, we uncover, from Eq. (5–15), that

sigma^prime=left(sigma_x^2-sigma_x sigma_y+sigma_y^2+3 au_x y^2 ight)^1 / 2 (5–15)

sigma^prime=left(sigma_x^2+3 au_z x^2 ight)^1 / 2=left<(142.6 F)^2+3(76.4 F)^2 ight>^1 / 2=194.5 F

Equating the von Mises anxiety to S_y, we solve for F and also get 

F=fracS_y194.5=frac81000194.5=416 lbf

In this example the toughness of the material at allude A is S_y=81 message kpsi. The stamina of the assembly or component is F = 416 lbf.

Let us use the MSS theory for comparichild. For a allude undergoing plane stress with only one nonzero normal stress and also one shear stress and anxiety, the 2 nonzero primary stresses will certainly have actually oppowebsite indications, and also for this reason the maximum shear tension is obtained from the Mohr’s circle in between them. From Eq. (3–14)

au_1, au_2=pm sqrtleft(fracsigma_x-sigma_y2 ight)^2+ au_x y^2 (3–14)

au_max =sqrtleft(fracsigma_x2 ight)^2+ au_z x^2=sqrtleft(frac142.6 F2 ight)^2+(76.4 F)^2=104.5 F

Setting this equal to S_y / 2, from Eq. (5–3) via n = 1, and solving for F, we get

au_max =fracS_y2 n quad message or quad sigma_1-sigma_3=fracS_yn (5–3)

F=frac81000 / 2104.5=388 lbf

which is around 7 percent much less than discovered for the DE theory. As declared previously, the MSS theory is more conservative than the DE concept.