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