Objective:

To determine the internal torque and the associated shear stress for statically indeterminate problems.

T

A

C

B

LAC LCB

L

1) Reaction at A and B.

2) Internal torque at segment AC and CB.

3) Stress at AC and CB.

Find,

Dr. H. Ali Al-Gadhib

T

A BTA TB

TATAC

TAC = TA

TCBTB

TCB = TB

Equation of equilibrium.

TB – T + TA = 0 ---------------- A

Where TA and TB are reactions at supports.

TCB – T + TAC = 0 ---------------- B

Equation of equilibrium.

Equation A is of better form…

Dr. H. Ali Al-Gadhib

0GJ

LT

GJ

LT

0

0

CBCBACAC

B/CC/A

B/A

TACLAC + TCBLCB = 0 ---------------- C

Solving equations B & C for the internal torque (TAC and TCB)

and then find reactions and stress.

Dr. H. Ali Al-Gadhib

Example:

The solid steel shaft as shown in figure has a diameter of 20 mm. If it is subject to the two torques, determine the reactions at the fixed support A and B.

Dr. H. Ali Al-Gadhib

Solution:

1 0TmN500mN800T ;0M ABx

Equation of equilibrium.

Dr. H. Ali Al-Gadhib

Compatibility:

0B/A

....Ans mN645T mN345T

2 750T2.0T8.1

or

0JG

m3.0T

JG

m5.1mN500T

JG

m2.0T

BA

BA

AAB

Dr. H. Ali Al-Gadhib

Example:

The shaft as shown in figure is made from a steel tube, which is bonded to a brass core. If a torque of T = 250 N·m is applied at its end, plot the shear-stress distribution along a radial line of its cross-sectional area.Take Gst = 80 GPa, Gbr = 36 GPa.

Dr. H. Ali Al-Gadhib

Solution:

Equilibrium.Equilibrium.

0mN250TT brst

Compatibility.Compatibility.

brst

2 mm/N1036mm102/

LT

mm/N1080mm10mm202/

LT

34br

2344st

Dr. H. Ali Al-Gadhib

Solving Eqs. 1 and 2, we get

MPa63.4mm/N63.4mm102/

mm10m/mm10mmN28.7

mN28.7T

mN72.242T

24

3

maxbr

br

st

MPa60.20mm/N60.20]mm10mm202/

mm20m/mm10mN72.242

MPa30.10mm/N30.10]mm10mm202/

mm10m/mm10mN72.242

244

3

maxst

244

3

minst

Dr. H. Ali Al-Gadhib

Dr. H. Ali Al-Gadhib