Task 6. Torsion with bending

The shaft of belt transmission (see fig.9) is turned round with n frequency and passed K capacity. On the shaft are fixed two pulleys. Diameter of the first pulley D ₁, diameter of the second pulley D ₂. Conducting branch of the belt transmission is stretched twice more than the force conducted. Using the third theory of strength determine the necessary diameter of the shaft.

The allowable stress =100 MPa.

Take the data which are necessary for computation from table 1 and table 2.

Order of work

1. Determine the torsional moments put to the shaft.

2. Construct the twisting moment diagram M _t.

3. Determine the environing forces t ₁ and t ₂ acting on pulleys.

4. Construct the bending moment diagrams M _b^hor due to the horizontal forces and M _b^vert due to the vertical ones.

5. Construct the total bending moment diagram as the geometric sum of the two previous diagrams M _b^tot = .

Fig.9

Table 7

Variant number
D 1 , m	0.15	0.25	0.35	0.45	0.55	0.1	0.2	0.3	0.4	0.5
K,kVt	0.2	0.4	0.5	0.6	0.7	0.25	0.35	0.45	0.55	0.65

6. Find the dangerous cross - section of the shaft and determine themaximum conventional moment value M _con using the third theory of strength

M _con = .

7. Using the condition of strength determine the necessary diameter

Table 8

Variant number	a 1, m	a 2 ,m	a 3 , m	a 4 , m	D 2 , m	n, turn/min
	0.2	0.5		0.8	0.50
	0.4	0.2		0.7	0.55
		0.5	0.2	0.7	0.60
		0,8	0,3	0,6	0,55
	0.6		0.2	0.8	0.60
	0.8		0.3	0.5	0.55
	0.2	0.4		0.7	0.55
	0.6	0.3		0.8	0.55
	0.5	0.3		0.9	0.50
	0.2	0.4		0.8	0.60
	0.3	0.8		0.5	0.45
	0.9	0.4		0.6	0.45
	0.5	0.3		0.7	0.55
	0.8		0.4	0.6	0.55
	0.6		0.3	0.8	0.40
	0.5		0.2	0.7	0.40
		0.8	0.3	0.5	0.35
		0.4	0.2	0.7	0.35
		0.9	0.2	0.5	0.40
		0.5	0.3	0.9	0.45
	0.3	0.9		0.5	0.45
	0.4	0.8		0.6	0.40
	0.7	0.3		0.5	0.35
	0.6	0.2		0.8	0.35
	0.6		0.2	0.9	0.35
	0.9		0.2	0.7	0.40
	0.7		0.3	1.0	0.45
		0.9	0.3	0.7	0.30
		0.6	0.2	0.8	0.35
		0.7	0.2	0.4	0.30

Дата добавления: 2015-11-14