Required information A countershaft carrying two V-belt pulleys is shown in the

Question

Required information A countershaft carrying two V-belt pulleys is shown in the figure. Pulley A receives power from a motor through a belt with the belt tensions shown. The power is transmitted through the shaft and delivered to the belt on pulley B. Assume the belt tension on the loose side at B is 15 percent of the tension on the tight side. Given: 6(dB) = 210 mm, TA?1720 N, and TA2) = 258 N 230 mm T. 280 mm 30-mm dia 300 mm B(dB) 400-mm dia T(A2) T(A1) Find the magnitudes of the bearing reaction forces, assuming the bearings act as simple supports The magnitude of the bearing reaction force Rc is? N The magnitude of the bearing reaction force Ro isN Draw shear force and bending moment diagrams. At what position along the shaft is the max. bending moment? the bending stress at the point of maximum bending moment is.... the tensional shear stress at the point of maximum bending moment is.

Explanation / Answer

Answer a)

The relation between the tensions in the belt on pulley is given by

T1 = 0.15T2

T(A1) = 1720N, T(A2) = 258N

Sum of torques

(1720-258)(0.2) -(T1-T2)(0.105) = 0

Substituting 0.15T2 = T1

292.4 - (T2-0.15T2)(0.105) =0

292.4 - 0.089T2 =0

T2 = 3285.3 N

T1 = 492.8 N

Magnitude of bearing reaction force Ro

At wheel A, we have (1720-258)(0.2) = 292.4 N

At wheel B, we have (3285.3-492.8)(0.105) = 293.2 N

Therefore, at Ro = (293.2*0.23)-(292.4*0.81) = -169.41N i.e. in downward direction

Ro+Rc = 293.2-292.4

Rc = 0.8+169.4 = 170.2N

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.