A maximum torque of 6.75 kNm can be supplied to the constant diameter steel (G= 80 GPa) line shaft by a motor as shown in Figure Q2. At the current normal operation condition, three machines are driven by gear B, C and D on the shaft and they require torques of 3.x kNm, 1.5y kNm and 1.0z kNm, respectively. Parameter x, y, and z are given by:
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- Suppose an automobile engine can produce 195 N⋅m of torque, and assume this car is suspended so that the wheels can turn freely. Each wheel acts like a 14 kg disk that has a 0.195 m radius. The tires act like 2.15-kg rings that have inside radii of 0.18 m and outside radii of 0.34 m. The tread of each tire acts like a 8.5-kg hoop of radius 0.335 m. The 16-kg axle acts like a solid cylinder that has a 1.75-cm radius. The 29.5-kg drive shaft acts like a solid cylinder that has a 3.25-cm radius. Calculate the angular acceleration, in radians per squared second, produced by the motor if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car.From the following diagram find with k1=20N/m, k2=30 N/m, k3=25 N/m, k4=35 N/m, k5=50N/m and k6=60N/m, m=10kg A) the equivalent spring B) the displacement C) the work performedFor the double slider mechanism shown in the following figure, the crank OA rotates at a uniform speed of 24 rad/s ccw. we need to find the required torque for the crank, if two forces act at sliders B and C as shown in the figure. (P = 4 kN, Q = 2 kN). OA = 10 cm, AB = AC = 70 cm. mg = mc = 5 Kg. Neglect other links weights. (5) (2) (3) B (4) (6) C 45° X. The velocity of slip of slider B in m/s² = Choose.. + The velocity of slip of slider C in m/s? = Choose... + The acceleration of slip of slider B in m/s² = Choose.. + The acceleration of slip of slider C in m/s² = Choose.. + The magnitude of required torque for the crank in N.m = Choose..
- Find the equivalent torsional spring constant (up to two decimal points) of the system shown in the figure. Given : R = 0.2 m k1=1 kN.m/rad, k2 = 2 kN.m/rad , k3 = 3 kN.m/rad , k4 = 4 kN.m/rad , k5 = 5kN/m and k6 = 3.0 kN/mAs seen in the figure, a construction machine is rotated by a drive mechanism consisting of belt-pulley and spur gear mechanisms. Engine power P = 5.5 kW ; engine shaft speednm = 1500 rpm ; drive pulley diameter Dt = 16 cm ; diameter of the opposite pulley Dk = 55 cm ; number of pinion teeth z1 = 21 ; number of teeth of the counter gearz2 = 60 ; pulley mechanism efficiency ηk = 0.95 ; Since the efficiency of the gear mechanism is ηd = 0.97: a) Find the output shaft speed nç and the torque Mç on this shaft.b) The force on the taut arm of the belt S1 = 450 N;coefficient of friction μ = 0.3 ; If the winding angle α = 160⁰, is the frictionally transmitted moment Ms sufficient to compensate for the Mg moment on the input shaft? Calculate.power (N)=9.9kw Input speed (n1)=2350rpm Distance between centers (a)=168mm Total distance between Ave B bearings=440mm module(m)=3mm question 3 Calculate the power on the output shaft since there is 3% loss in gears in gearboxes, rolling bearings in support bearings (2% loss in a pair of bearings) and sealing element spring seal (4% loss in a pair of sealing elements).
- a) A line shaft as shown in Figure Q is driven using a motor placed vertically below it. The pulley on the line shaft is 1.6 m in diameter and has belt tensions 7.5 kN and 2.4 kN on the tight side and slack side of the belt respectively. Both tensions may be assumed to be vertical and the weight of the pulley is negligible. If the pulley is overhang from the shaft, the distance of the centre line of the pulley from the centre line of the bearing being 500 mm.6.Figure Q(i) Predict using distortion energy theory, the appropriate diameter of the shaft that failure will not occur if the yield strength, Sy = 370 Mpa and factor of safety is 2.5. (ii) Assuming the maximum allowable shear stress of 42 MPa, find its diameter using maximum shear stress theory. (iii) Comparing the diameters in (i) and (iii) above, which of them would you have used to design your shaft and why?For the double slider mechanism shown in the following figure, the crank OA rotates at a uniform speed of 100 rad/s CW. we need to find the required torque for the crank, if two forces act at sliders B and C as shown in the figure. (P = 2KN, Q = 1KN). OA = 30 cm, AB = AC = 100 cm. mB = mC = 1 Kg. Neglect other links weights. The velocity of slip of slider B in m/s2 = Answer 1 Choose... The velocity of slip of slider C in m/s2 = Answer 2 Choose... The acceleration of slip of slider B in m/s2 = Answer 3 Choose... The acceleration of slip of slider C in m/s2 = Answer 4 Choose... The magnitude of required torque for the crank in N.m = Answer 5 Choose...A pulley 610 mm diameter is driven at 300 revs/min by a belt, the tensions in the tight and slack sides of the belt being 670 N and 270 N. Find the power transmitted.
- The shaft in the figure, carrying three unbalanced masses, is rotating at a constant speed hızla = 10 r / s. Find the balancing values (kg.mm) and their positions in case of balancing in DI and DII planes.A spring-mass system is shown in the following figure. Use m=3.5 kg and k=100 N/m. a. Determine the static deflection δst . Insert your answer in meters correct up to at least a third decimal place. b. Determine the system period τ. Insert your answer in seconds correct up to at least a third decimal place. c. Determine the maximum velocity vmax which results if the cylinder is displaced 140mm downward from its equilibrium position and released from rest. Insert your answer in m/s correct up to at least a third decimal place.A wagon weighing 5100 Kg moving at a speed of 8 km/h has to be brought to rest. Springs made of wire diameter 25mm with a Dm of 250mm and with 24 turns are available. Find the number of springs required in buffer to stop the wagon at a compression of 180mm. b)A close coiled helical spring is made of diameter wire 25mm. The spring index is 8. Findthe number of turns required and maximum allowable load if the allowable shear stress is100 MPa and elongation is limited to 40mm. [G=80 GPa