If T1 and T2 are the tensions on the tight and slack sides of the belt respectively and TC is the centrifugal tension, then initial tension in the belt is equal to

T1 - T2 + TC

T1 + T2 + TC

$$\frac{{{{\text{T}}_1} - {{\text{T}}_2} + {{\text{T}}_{\text{C}}}}}{2}$$

$$\frac{{{{\text{T}}_1} + {{\text{T}}_2} + {{\text{T}}_{\text{C}}}}}{2}$$

Correct Answer: $$\frac{{{{\text{T}}_1} + {{\text{T}}_2} + {{\text{T}}_{\text{C}}}}}{2}$$

The ratio of driving tensions for flat belts, neglecting centrifugal tension, is (where T₁ and T₂ = Tensions on the tight and slack sides of belt respectively, $$\mu $$ = Coefficient of friction, between the belt and pulley and $$\theta $$ = Angle of contact)

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = \mu \theta $$

$$\log \frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = \mu \theta $$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\text{e}}\mu \theta $$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = \log \mu \theta $$

If P₁ and P₂ are the tight and slack side tensions in the belt, then the initial tension P_i (neglecting centrifugal tension) will be equal to (Where, P_c is centrifugal tension)

$${{\text{P}}_1} - {{\text{P}}_2}$$

$${{\text{P}}_1} + {{\text{P}}_2}$$

$$\frac{1}{2}\left( {{{\text{P}}_1} + {{\text{P}}_2}} \right)$$

$$\frac{1}{2}\left( {{{\text{P}}_1} + {{\text{P}}_2}} \right) + {{\text{P}}_{\text{c}}}$$

If P₁ and P₂ are the tight and slack side tensions in the belt, then the initial tension P_i (according to Barth) will be equal to (Where, P_c is centrifugal tension)

$${\left( {\frac{{\sqrt {{{\text{P}}_1}} + \sqrt {{{\text{P}}_2}} }}{2}} \right)^2}$$

$${{\text{P}}_1} + {{\text{P}}_2}$$

$$\frac{1}{2}\left( {{{\text{P}}_1} + {{\text{P}}_2}} \right)$$

$$\frac{1}{2}\left( {{{\text{P}}_1} + {{\text{P}}_2}} \right) + {{\text{P}}_{\text{c}}}$$

Power is transmitted by an electric motor to a machine by using a belt drive. The tensions on the tight and slack side of the belt are 2200 N and 1000 N respectively and diameter of the pulley is 600 mm. If speed of the motor is 1500 r.p.m, find the power transmitted.

46.548 kW

56.548 kW

66.548 kW

76.548 kW

The tension on the slack side of the belt is __________ the tension on the tight side of the belt.

Equal to

Less than

Greater than

None of these

The ratio of belt tensions $$\frac{{{{\text{p}}_1}}}{{{{\text{p}}_2}}}$$ considering centrifugal force in flat belt is given by where,
m = mass of belt per meter (kg/m)
v = belt velocity (m/s)
f = coefficient of friction
$$\alpha $$ = angle of wrap (radians)

$$\frac{{{{\text{p}}_1} - {\text{m}}{{\text{v}}^2}}}{{{{\text{p}}_2} - {\text{m}}{{\text{v}}^2}}} = {{\text{e}}^{{\text{f}}\alpha }}$$

$$\frac{{{{\text{p}}_1}}}{{{{\text{p}}_2}}} = {{\text{e}}^{{\text{f}}\alpha }}$$

$$\frac{{{{\text{p}}_1}}}{{{{\text{p}}_2}}} = {{\text{e}}^{ - {\text{f}}\alpha }}$$

$$\frac{{{{\text{p}}_1} - {\text{m}}{{\text{v}}^2}}}{{{{\text{p}}_2} - {\text{m}}{{\text{v}}^2}}} = {{\text{e}}^{ - {\text{f}}\alpha }}$$

In a band and block brake, the ratio of tensions on the tight and slack sides of band is given by (where $$\mu $$ = Coefficient of friction between the blocks and the drum, $$\theta $$ = Semi-angle of each block subtending at the centre of drum and n = Number of blocks)

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_{\text{2}}}}} = \mu \theta \times {\text{n}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\mu \theta } \right)^{\text{n}}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\frac{{1 - \mu \tan \theta }}{{1 + \mu \tan \theta }}} \right)^{\text{n}}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\frac{{1 + \mu \tan \theta }}{{1 - \mu \tan \theta }}} \right)^{\text{n}}}$$

In a band and block brake, the ratio of tensions on tight side and slack side of the band is (where $$\mu $$ = Coefficient of friction between the blocks and the drum, $$\theta $$ = Semi-angle of each block subtending at the center of drum and n = Number of blocks)

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = \mu \theta {\text{n}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\frac{{1 - \mu \tan \theta }}{{1 + \mu \tan \theta }}} \right)^{\text{n}}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\mu \theta } \right)^{\text{n}}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\frac{{1 + \mu \tan \theta }}{{1 - \mu \tan \theta }}} \right)^{\text{n}}}$$

From an alloy, two specimens are machined and tested separately under tension and compression. The engineering stress-strain curves as well as the true stress-true strain curves for tension and compression are plotted in the same diagram. Identify the correct statements:
P. The engineering stress-strain curves in tension and compression are identical.
Q. The true stress-true strain curve in compression is lower than the true stresstrue strain curve in tension due to Bauschinger effect.
R. The true stress-true strain curve in tension is above the corresponding engineering stress-strain, curve.
S. The true stress-true strain curve in tension and compression are identical.

P, Q

Q, R

Q, S

R, S

The condition for maximum power transmission is that the maximum tension in the flat belt should be equal to
(Where, $${{\text{P}}_{\text{c}}}$$ = tension in belt due to centrifugal force)

$$3{{\text{P}}_{\text{c}}}$$

$${{\text{P}}_{\text{c}}}$$

$$\frac{{{{\text{P}}_{\text{c}}}}}{3}$$

$$2{{\text{P}}_{\text{c}}}$$