Torque Solution - UCFileSpace Tools - Homehomepages.uc.edu/~caldwelm/Courses/MMI/Torque_sol.docx ·...
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Stress Due to Torque 1 Find τ max caused 1 Write the fundamental equations 2 Write the differential equations for dA 3 Substitute to get dT as f (τ ) 4 Integrate 5 Define J = ∫ r 2 dA 1 Write stress, torque equation 2 Substitute values and d T = (F) (r) τ= dF dA dT = (dF)(r) dT = (dF)(r) = [(τ) (dA)] (r) dT = τ max ( r )( dA ) ( r ) ∫ dT = τ max c ∫ ( r ) 2 ( dA) T= τ max c ∫ ( r ) 2 ( dA ) τ max = Tc J τ max = Tc J τ max = ( 560 × 10 3 Nmm ) ( 35 mm 2 ) π 32 ¿¿ N τ= F A τ max c = τ r Proportional τ max c = τ r
Transcript of Torque Solution - UCFileSpace Tools - Homehomepages.uc.edu/~caldwelm/Courses/MMI/Torque_sol.docx ·...
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 Write stress, torque equation
2 Substitute values and units and 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
τ max=T cJ
τ max=(560×103 Nmm )( 35mm
2)
π32
¿¿
τ max=89.9 Nmm2
τ= FA
τmaxc
= τr
Proportionalτmaxc
= τr
Stress Due to Torque 2
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
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
HW#14
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
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=441 MPa
2
τ max=TZ
441 Nmm2
2= T
π16
(15mm)3
T=146100 Nmm
Stress Due to Torque with Stress Concentration 3
1 Write τmax equation (include k for stress conc)
2 Find k @ step
3 Substitute & solve for τmax @ step
4. Use τmax for stress @ hole, solve for k5. Use stress conc chart to find diameter of hole (a)
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
HW#15
Stress Conc
Page 714 Page 721
τ max=k τnom τnom=TZ
Dd
=2∈ ¿1.5∈¿=1.33¿
¿rd=0.08∈ ¿
1.5∈¿=0.053 ¿¿
Page 730
Page 728
k = 1.56
τ max=k TZ
τ max=(1.56 )(7500∈lb)
π16 ¿¿¿
τ max=17660 psi
τ max=k TZ
17660 lb¿2 =
(k )(7500∈lb)π
16 ¿¿¿
k= 3.7
For k=3.7aD
=0.1
a = (0.1)(2in) = 0.2 in
τ d=S y
8S 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
=24 mm18mm
=1.5rd= 1mm
16mm=0.625
Page 730 k = 1.54
τ max=k TZ
T=( 669
8N
mm2 ) [ π16(16mm)3]
1.54=43670 Nmm
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=1360×103 Nmmπ16
(43.85mm)3
τ max=82.1 Nmm2
Angle of Twist due to Torque 5
2 Find stress
3 Find guideline & calc required Sy 4 Compare to 6061-T6
1 Find diameter for angle of twist
HW#16
Material PropertiesGuidelines
θ=TLJG
(10o) πrad180o =
(5×103 Nmm) (150mm)
[ π32(d )4](26×103 N
mm2 )d = 6.406mm
τ max=TZ
τ max=5×103Nmmπ16
(6.406mm)3
τ max=98.9 Nmm2
τ d=S y
4S y=(98.9 N
mm2 )(4) = 395.6 MPa
Page 718
6061-T6 Sy = 276 MPa < required 395.6 MPaSelect a different aluminum w/same G but Sy>396MPa
