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Unformatted text preview: lellnm 3.44 A solid circular bet ABC cunslsts of [Wu segments.
as shown in the ﬁgure. One segment has diameter dl = 50 mm and
length L. = 1.25 m: the other segment has diameter :12 = 40 mm
and. length l'.2 = LO m. What is the allowable torque 3';th if the shear stress is not to
exceed 30 MPa and the angle of twist between the ends of the bar
is not to exceed 1.5"? (Assume G = 30 GM.) Salutinn 3.44 Bar consisting of two segments d;;50mm d2;40mm '''—"—x_ T A I..=1.25m B Lz=1Dm C
7,le ; 30 MPa ALLOWABLE. TORQUE BASED LI'PON munz OF thsT
gem“: t.s==u.02518rad ¢_ EE_ 111+ TL, ::(ﬂ 2)
G _ 80 Spa 61,, G!“ Gin G [p] I» d. _ 32T(ﬂ + AUDWABLE. TORQUE BASED Upon SHEAR mm “C 0,1,, d;
Segment BC has the smaller diameter and hence the mtdl G
larger stress. a.“ = —L1 = 343 N ' m
l
_ tar _ fm’i‘rmo. 32(d? + d?) =_. _=3I N.
Tm: 7rd] annu ‘6 I” “1 ANGLE 0F TWle GOVFRNS Tﬂm=34ENm 4— Prulrlem 1412 A prismatic bar A3 of length L and solid circular crass section [diameter d) is loaded by a distributed torque of constant intensity
r per unit distance {see ﬁgure]. {a} Determine the maximum shear stress rum in the bar.
(b) Determine the angle Uftwisl tit between the ends of the bar. Snitlllnn 3.412 Bar with distributed torque (a) MAXIMUM SHEAR 5111155? IOT iﬁri.
Tm, '— rL = m" = —
T“ 170" mi“
(b) ANGIJT. OF TWIST ‘ 1rd" ft): = L: .l' = — J P 32 T
W = J% = 32 txdx
_ I Gt", rrGa“
I = Intensityr of distributed torque ( r. I _ ' _ 32:  15:13
d=dtameter ‘h— d¢_néniUXdX=1rﬁd‘ .—
G = shear modulus of elasticity
Problem 3.34 A circular bar A3 with ends ﬁxed against rotation has l‘ 25 in' 25 m' l a hole extending for half of its length (see ﬁgure}.'I‘h: outer diameter
of the bar is d2 = 3.0 in. and the diameter of the hole is d" = 14 in.
The total length of the bar is L L 50 in. At what. distance x from Lhe lefthand end of the but should a torque
To be applied su that the reactive torques at the support: will be equal? Solution 3.8! Bar willI a huh: L ‘ 50 in. Substitute Eq. [1] into Eq. [2) and simplify:
U2"25in. d'_I“D[L+L x L L
d2 = uutcr dimmer s G “P! 4"?“ In 2"”: 2’91 = 3.0 in. Comm11mm den = 0
d‘ = diameter of hole I y L
=2.4in. " E13!» _ﬁ
TD = Torque applicd at distance I SOLVE ma
 1:
Find 1 so [hat TA = T8 1 I
x L 4(3  E)
EQUILIBRIUM 4 I”
y _ _ _ 1' I d' — d“ d ‘
TA+TRZIO ‘I‘TA='[E_—u (1) {TE— 3 '1 l—l—(J)
2 I'm d2 In‘2 L[ (or. )‘
x — ' '2 + — ‘—
4 d2
SUBSTITUTE NUMERICAL VALUES: 50'. I '.‘
x: m [2+(24in)]=30_l2in. 1—
4 3.0m. 9b” = Aug}: aflwisl at B in — Pniar moment of mania at left—hand and
I” = Polar moment of inertia at righthand and
_ TAM) Tatm) Top: —Lf2]
" _ '6!» + THAI _ GI»:
_ Tum'2]
GI” {21 Prﬂhlﬂm 3.104} A torque I is applied to a thinwalled tube having
a cross scdiun in Lhc shape or II regular hexagon with wnmm wall
thickness r and side length is {see figure). Obtain formulas for the shear sites: 7 and Lhe rate of twist 9. Solution 3.108 Regular hexagon SHEAR STRS
T I‘VE
'r = — = . *—
m,‘ 95: ANGLE OF was‘r m1 UNIT LliNlel (RATE or TWIST)
J_ 4.4:: _M3.:_9b3t id: L, 2
T 7 f _ a 7 i 7 27 7 2'." _
I; _ LensIh a side, 6; 5.195%) _ 96%
I = "ﬂinchms {radians per unit lengﬂI) L“ = 6!: FROM APPENDIX D, CASE 25: ,8 = 60" a: = 6
b: A, — "rang: Tm: 3D”
3V3]? ...
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This test prep was uploaded on 04/06/2008 for the course EM 319 taught by Professor Kennethm.liechti during the Fall '08 term at University of Texas at Austin.
 Fall '08
 KennethM.Liechti

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