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Give lots of explanation for every step or else it won't be marked complete. Make sure to write out equations first before subbing in. Be consistent and carry units properly.

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Q8. A solid cylindrical shaft BCD, with a rigid bar BH has attached at point B, firmly retained by a bearing in C using a contact pressure due to the assembly. The static friction coefficient between the two parts is Luf 0.35. The arm BH and shaft BCD originally being situated in the horizontal plane and a vertical force F is applied at point H. Gradually increasing the intensity of force F, the shaft begins to rotate relative to the housing when F 1.35kN. a) Determine the contact pressure (po), assuming that the pressure is uniform over the entire width e of the bearing. b) For just before the rotation of the bar provided, determine the state of stress (according to xyz axes) acting at the point K (located at the interface between the upper surface of the shaft and just the lower face of the bearing, see Figure 'b'). Then, determine the principal stresses and their directions and also the maximum shear stress at that point. 0.33m bar (d 48mm) e 8mm 0.2m Rigid arm (a) (b)

Explanation / Answer

solution:-

If D descends by 7 mm,C must descend byu C =7×1.54= 2.625mm

since the bar BCD is rigid and pivots aboutB.This displacement is vertical, but we can resolve it into components parallel to and perpen-dicular to the bar AC, as in Figure

From the original figure, we note that AC is the hypotenuse of a triangle with adjacent sidesin the ratio 2:1,

so that cos=25; sin=15

If the perpendicular componen tu C sin were the only displacement, the barACwould notneed to change in length, since it can pivot abou tA. However, the parallel componentu Ccos= 2.625×25= 2.348mmcauses AC to decrease in length by that amount. We therefore conclude that the force in AC is

FAC = EAACACLAC=-210×103××202/4×2.34810001.52+ 32=-46184N=-46.2kN.

Figure shows a free-body diagram of the linkBCD, from which, taking momentsaboutB, we conclude that W=-FAC cos×1.54=46.2×2×1.545= 15.5kN.

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