5-3. The shaft has an outer diameter of 1.25 in. and an inner diameter of 1 in. If it is sUbjected to the applied torques as shown, determine the absolute maximum shear stress developed in the shaft. The smooth bearings at A and B do not resist torque.

T,. .. = 1500 lb· in.

_ Te _ 1500(0.625) = 6.62 ksi Ans f mu

- J - ~ (0.625)' - (0.5)']

*5-4. The shaft has an outer diameter of 1.25 in. and an inner diameter of 1 in. If it is subjected to the applied torques as shown, plot the shear-stress distribution acting along a radial line lying within region EA of the shaft. The smooth bearings at A and B do not resist torque.

T = 1500 lb . in.

Te 1500(0.625) = 6.62 ksi t"m .. = J = K (0.625)' - (0.5)']

t" = Tp = 1500(0.5) = 5.30 ksi 2 J I[ (0.625)' - (0.5)']

5-5. The solid 30-rnrn-diameter shaft is used to transmit the torques applied to the gears. Determine the shear stress developed in the shaft at points C and D. Indicate the shear stress on volume elements located at these points.

Internal Torque: As shown on FBD. Maximum Sh.ar Stress: Applying torsion formula

Te e 200(0.015)

t"e = J = ~(O.015') =37.7MPa

TD c 400(0.015) TD = J = Y(O.OIS')

= 75.5 MPa

Ans

Ans

180

)00 N.. 500:01 ..

5- 6. The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at 8 . The smaller pipe has an outer diameter of 0.75 in. and an inner diameter of 0.68 in., whereas the larger pipe has an outer diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is tigh tly secured to the wail at C, determine the maximum shear stress developed in each section of the pipe when the couple shown is applied to the handles of the wrench.

1.'1lh

e

<1/0 lb · ;"

~TA8·<IV'6.j"

.etO 'H .. ~ , J » T8( 'ZIO"'"

Tc 210(0.375) = 7.82 ksi Ans TAB = J = r(0.375' - 0.34')

Tc 210(0.5) = 2.36 ksi TBe = J = r(0.5' - 0.43')

Ans

5-7. The solid aluminwn shaft has a diameter of 50 mm and an allowable shear stress of 'Tallow = 6 MPa. Determine tbe largest torque T\ tbat can be applied to the shaft if it is also subjected to tbe otber torsional loadings. It is required tbat T\ act in the direction shown. Also, determine tbe maximum shear stress within regions CD and DE.

I"t,,"al Torq"": At; shown on FBD. Mazi","", She'" 5'''$1 : Applying the torsion Fonnula and assume failure at region Be.

T.ec 'fill .. = 1'.110_ = J

( 6) _ (7j - 68)( 0.025)

6 10 - ;(0.025')

7j =215.26N · m=2ISN · m An.

Mtximum torque occurs within region BC as indicated on the torque diagram.

Mazimu", She", Str,url at Other Reg/Oil : From the torque diagram. the inlema! torque at region CD and DE are .TCD = 98.26 N· m and TDE .. 63.26 N · m respectively.

TcDC 98.26(0.025) _ 4 00 MP An. (Tmu)CD"j" r(0.02S') -. a

.. TDlc = 63.26(0.02S) = 2.S8 MPa An. (T .... )DE J f(0.02S')

~.o

181

•

•

•

.. •

-..

5-69. The glass tube is confined within a rubber stopper, so that when the tube is twisted at constant angular velocity the stopper creates a constant distribution of frictional torque along the contacting length AB of the tube. If the tube has an inner diameter of 2 mm and an outer diameter of 4 mm, determine the shear stress developed at a point located at its inner and outer walls at a section through level C. Show the shear-stress distribution acting along a radial line segment at this section. Also, determine the angle of twist at A with respect to B. Gg = 10 GPa.

Equilibrium : FBD (a) . -

~=O; 0_15-1(0_04)=0 1=3.7SN.mfm

InUrnol Torque: As shown on FBD (b) and (e) .

Shear S lress : Applying lorsion fonnula, we have

-r. = 0.0375(0.002) ;(0.002'-0.001') =3.18MPa

0.0315(0.001) -r, = ;(0.002' -0.001') = 1.59 MPa

Angle of Twisl :

ftT(X)dx

.pM. = --o JG

Ans

An.

1 (o .o'm

= ;(0.002' - 0.001') IO( 10') J 0 3.7S.fdx

= 0.01273 rad = 0.730· Ans

5-70. Th.e 60.~-diarneter solid shaft is made ot A-36 stee! and J5 s~bJected to tbe distributed and concentrated torslOnalloadlDgs shown. Determine the aogle of twist at the free end A of t~.e shaft due to these loadings.

Inl.rnaI Toriu.: As shown on FBD. An,l. 01 Till 1" :

TL ;A =LJG

-400(0.6) f. •. lm (200-2000..)dx - + - ;<0.03')75.0(10') • r(0.03')75.0( 1()9)

= -0.007545 rod = I 0.432· I Ans

209

•

•

•

• •

•

( 5-78. The bronze C86100 pipe has an outer diameter of 1.S in. and a thickness of 0.125 in. The coupling on it at Cis being tightened using a wrench. If the applied force is F .. 20lb, determine the maximum shear stress in the pipe.

E'Iulllbrlllm :

1;. +T, -20(12) =0

Compallbllill :

fie" = flc/A T,(8) 1;. (10) Jo=JG

T. = \.251;.

Solving Eqs. [I) and (2) yields :

1;. • 106.67 lb· in. T, .. 133.33 lb · in.

MII%/mum ,II.u ,,,." :

T. c 133.33(0.75) . ~ .... = T = f(0.75' -0.625') so 389 p."

[I)

(2)

An.

5-79. A rod is made from two segments: AB is A-36 steel and has a diameter of 30 mID and BD is C83400 red brass a~d has a diameter of 50 mm. It is fixed at its ends and subjected to a torque of T = 500 N . m. Determine the torsional reactions at the walls A and D.

Equilibrlllm :

[I)

ComplJl/billll :

1;. (0.75)

r(0.OZS')(37.0)( 1()9) i(0.015')(75.0)( 10')

1;. = 0.29807T"

Solving Eqs. [I) and (2) yields :

T" =385 N · m 1;. = 115 N· m

1;.(0.5) + f(o.o25')(37.0)( 10')

Ans ADS

(2)

214