Stress Due to Torque 1

Find τmax caused by T

1 Write the fundamental equations 2 Write the differential equations for dA

3 Substitute to get dT as f (τmax) 4 Integrate 5 Define J = ∫ r2 dA Solve for τmax

1 Convert hp to ¿ lbs

2 Convert rpm to rads

3 Write power, torque equation, substitute, solve for T4 Write stress, torque equation, substitute, solve

d

T = (F)(r) τ=dFdA

dT = (dF)(r)

dT = (dF)(r) = [(τ)(dA)] (r)

dT=τmaxc

(r ) (dA )(r )∫ dT=

τmaxc ∫ (r )2 (dA )

T=τmaxc ∫ (r )2 (dA )

τ max=T cJ

7.5hp( 550 ft lbs

1hp )¿ (240 revmin )( 2π rad

1 rev )( 1min60 sec )=25.13 rad

s

Power = (T)(ω)

T=49500 ¿lb

s

25.13 rads

=1970∈lb

τ max=TZ

τ max=1970∈lbπ16

¿¿¿

τ max=15770 psi

Conversion Factors

τ= FA

τmaxc

= τr

Proportionalτmaxc

= τr

Stress Due to Torque 2

Statics Required for Torque Calculaltion

1 Draw free-body for section AB

2 Draw free-body for section BC

1 Find Sy and Sys

2 Write stress, torque equation,substitute and solve

Page 714

Belts

MotorApp#1

App#2

ABBC

AB∑T=0T1 + TAB = 0TAB = -T1

∑T=0T1 -T2 + TBC = 0TBC = -T1 + T2

Sy = 441 MPa

Sys=441MPa

2

τ max=TZ

441 Nmm2

2= Tπ

16(15mm)3

T=146100Nmm

Stress Due to Torque with Stress Concentration 3

1 Find τd guideline

2 Find k left side

3 Find T left side

4 Find k right side 5 Find T right side

6 Give maximum torque value, units

Page 714 Page 721τ d=

S y8S y=669 N

mm2

Dd

=24mm12mm

=2rd= 2mm

12mm=0.16 6

Page 730 k = 1.28

τ max=k TZ

T=( 669

8Nmm2 ) [ π16

(12mm)3]1.28

=22170Nmm

Dd

=24mm18mm

=1.5rd= 1mm

16mm=0.625

Page 730 k = 1.54

τ max=k TZ

T=( 669

8Nmm2 ) [ π16

(16mm)3]1.54

=43670Nmm

Torque limited to 22170Nmm

Angle of Twist due to Torque 4

Page 218

Derive Equation for Angle of Twist

1 Write the fundamental equations

2 Substitute to get θ as f (T)

1 Find diameter using angle of twist

2 Find stress

τ max=T cJ

G=τ maxγ

θ= sr

γ= sL

c=r

Common arc length (s)θ= sr

γ= sL

s=γ L

θ= γLr

γ=τ maxG

θ=τmax LGr

τ max=T cJ

θ= TcLJG r

c=r

θ=TLJG

θ=TLJG

(2.2o) πrad180o

=(1360×103Nmm )(820mm)

[ π32(d )4](80×103 N

mm2 )d = 43.85mm

τ max=TZ

τ max=1360×103Nmmπ16

(43.85mm)3

τ max=82.1 Nmm2