example 13contents.kocw.net/kocw/document/2015/hanyang/hanseog...statically indeterminate...

EXAMPLE 13.1

5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS

• Procedure for analysis:– use both equilibrium and compatibility equations

Equilibrium• Draw a free-body diagram of the shaft in order to

identify all the torques that act on it. Then write the equations of moment equilibrium about the axis of the shaft.

Compatibility• To write the compatibility equation, investigate the way

the shaft will twist when subjected to the external loads, and give consideration as to how the supports constrain the shaft when it is twisted.

Copyright ©2014 Pearson Education, All Rights Reserved

STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS (cont)

• Express the compatibility condition in terms of the rotational displacements caused by the reactive torques, and then use a torque-displacement relation, such as Φ= TL/GJ, to relate the unknown torques to the unknown displacements.

• Solve the equilibrium and compatibility equations for the unknown reactive torques. If any of the magnitudes have a negative numerical value, it indicates that this torque acts in the opposite sense of direction to that indicated on the free-body diagram.


(1) Equilibrium eq.

A BT T T(2) Compatibility eq.

/ / C A C B

(3) Torsion –deformation relationship

/ /,

A AC B BCC A C B

T L T LGJ GJ

, BC ACA B

L LT T T TL L

EXAMPLE 5.8

The solid steel shaft shown in Fig. 5–23a has a diameter of20 mm. If it is subjected to the two torques, determine thereactions at the fixed supports A and B.


EXAMPLE 5.8 (cont)

• It is seen that the problem is statically indeterminate since there is only one available equation of equilibrium and there are 2 unknowns

• Since the ends of the shaft are fixed, the angle of twist of one end of the shaft with respect to the other must be zero.

Solutions

0800 500 0 (1)

x

B A

MT T

0/ BA


EXAMPLE 5.8 (cont)

• Using the sign convention established,

• Using Eq. 1,

• The negative sign indicates that acts in the opposite direction of that shown in Fig. 5–23b.

Solutions

(Ans) mN 645

03.03005.18002.0

B

BBB

TJGT

JGT

JGT

mN 345 AT


EXAMPLE 13.1

5.6 SOLID NON-CIRCULAR SHAFTS


SOLID NON-CIRCULAR SHAFTS (cont)


EXAMPLE 5.10 The 6061-T6 aluminum shaft shown in Fig. 5–27 has a cross-sectional area in the shape of an equilateral triangle.Determine the largest torque T that can be applied to the endof the shaft if the allowable shear stress is allow = 56 MPa.and the angle of twist at its end is restricted to Φallow = 0.02rad. How much torque can be applied to a shaft of circularcross section made from the same amount of material?


EXAMPLE 5.10 (cont)

• By inspection, the resultant internal torque at any cross section along the shaft’s axis is also T.

• By comparison, the torque is limited due to the angle of twist.

Solutions

3 3

3

4 4 3

20 20; 56 1779.2 Nm40

46 1.2 1046 ; 0.02 24.12 Nm (Ans)40 26 10

allow

allowal

T T Ta

TT Ta G


EXAMPLE 5.10 (cont)

• For circular cross section, we have

• The limitations of stress and angle of twist then require

• Again, the angle of twist limits the applied torque.

Solutions

(Ans) Nm 10.33

102685.142/102.102.0 ;

Nm 06.28885.142/

85.1456 ;

34

3

4

TTJGTL

TTJ

Tc

alallow

allow

mm 85.1460sin404021c ; 2 cAA trianglecircle


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