Q one is Two parallel shafts carrying pulleys 0.6 m diameter and 0.5 m diameter are 3 m apart and are connected by a belt so arranged taht the shafts rotate in opposite directions. If the drive is to be altered so that the shafts will rotate in the same direction by how much should the belts be shortened??

Now got that its a crossed belt, but need a formula to be able to figure this out havent got one.

Q2 Find the diameter and length of a mild steel end journal on which there is a load of 35 kN shock loading. Assume a safe working stress of 4o MPa and a length diameter ration of 1:5

NOTATION KEY

W = TOTAL LOAD ON JOURNAL
F = SAFE WORKING STRESS
E = LENGTH
D = DIAMETER

So its diameter of journal d = 2.26 from a this equation

d = 16WE DIVIDIED BY PIFD SQUARED = 2.26 X WE DIVDIDED BY FD SQUARE

So from the question have this

Diameter of journal D = 2.26

Therefore D = 2.26 35 TIMES 1.5 DIVIDIED BY 4O SQUARE = 1.15

D = 1.15

FROM E : D = 1.5

E = 1.5D 1.5 x D = 1.73 e = 1.73 So therefore a journal of 1.15 mm diameter x 1.73 length would be used

Is this correct according to the question

Q 3 A load of 6400 kg is carried by a journal rotating in a 270 mm bearing diameter and 400 mm long Find the specific load on the bearing and the heat generated per minute in the bearing assuming the shaft speed is 4/r/s and the co efficient of friction is 0.018

NOTATION KEY

W = TOTAL LOAD
U = COEFFICIENT
R = RADIUS OF BEARING
N = SHAFT SPEED

SPECIFIC LOAD ON BEARING = TOTAL LOAD DIVIDED BY PROJECTED AREA

= 6400 X 9.81 X 10 - 3 = 627837 0.135 X 0.400 = 0.027
627837 DIVIDED BY 0.027 = 233kPa

WORK LOST PER SECOND = 2PINWUR
= 2PIX4X6.400X9.81X0.018X 0.135 = 3.83 KNm

Mechanical equivalent of heat is 1 kNm = 1KJ
Hence heat generated per second = 3.83 KJ

Is this correct according to the question thanks

Want 2 formulas so can complete next 2 questions with the information given thanks

Q4 a v belt pulley drive is required to transmit 1.5 kw of power assumning a co efficient of 0.3 between pulleys and belt and a maximum belt tension of 15 kPa of width find the necessarry width of belt. The angle of vee is 30 degrees

Q5 Leather belting 75 mm wide is to be used to transmit power between 2 parallel shafts which run at the same speed the pulleys are 0.6 m diameter as slow speeds the tension in the slack side is half the tension in the driving side and the angle of lap on the pulley is 180 degrees. If the maximum load on the belting is not to exceed 140 N/cm of width neglecting the effect of centrifugal force estimate the max power which can be transmitted at 9.2 r/s