Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

y
A
(1)
0.85 m 1.00 m 0.70 m
550 N.m
900 N.m
B
−350 N.m
Shear Stresses
The maximum shear stress magnitude in each segment can be calculated with the use of
the elastic torsion formula:
Tc 1 1
(550 Nm)(36 ⋅ mm/2)(1, 000 mm/m)
τ 1 = =
= 60.0 MPa Ans.
4
J
164,895.9 mm
1
Tc 2 2
(350 Nm)(30 ⋅ mm/2)(1, 000 mm/m)
τ 2 = =
= 66.0 MPa Ans.
4
J
79,521.6 mm
2
Tc 3 3
(250 Nm)(25 ⋅ mm/2)(1,000 mm/m)
τ 3 = =
= 81.5 MPa Ans.
4
J
38,349.5 mm
3
Angles of Twist
Before rotation angles can be determined, the angles of twist in each segment must be
determined. In the preceding calculation, the sign of the internal torque was not considered
because only the magnitude of the shear stress was desired. For the angle-of-twist
calculations here, the sign of the internal torque must be included. We obtain
TL 1 1
(550 Nm)(850 ⋅ mm)(1, 000 mm/m)
φ 1 = =
= 0.035439 rad
4 2
JG 1 1 (164,895.9 mm )(80,000 N/mm )
TL 2 2
( 350 Nm)(1, 000 mm)(1, 000 mm/m)
φ 2 = = − ⋅ =−0.055017 rad
4 2
JG (79,521.6 mm )(80,000 N/mm )
2 2
TL 3 3
(250 Nm)(700 ⋅ mm)(1, 000 mm/m)
φ 3 = =
= 0.057041 rad
4 2
JG (38,349.5 mm )(80,000 N/mm )
3 3
Rotation Angles
The angles of twist can be defined in terms of the rotation angles at the ends of each segment:
φ = φ − φ φ = φ − φ φ = φ −φ
1 B A 2 C B 3 D C
The origin of the coordinate system is located at flange A. We will arbitrarily define the
rotation angle at flange A to be zero (φ A = 0). The rotation angle at gear B can be calculated
from the angle of twist in segment (1):
φ = φ −φ
600 N.m
(2) (3)
C
250 N.m
250 N.m
Internal torque diagram for compound shaft.
0.0354 rad
−0.01958 rad
D
0.0375 rad
Rotation angle diagram for compound shaft.
x
1
∴ φB = φA + φ1
= 0 + 0.035439 rad
B
= 0.035439 rad = 0.0354 rad
Similarly, the rotation angle at C is determined from the angle of
twist in segment (2) and the rotation angle of gear B:
φ2
= φC
−φB
∴ φC = φB + φ2
= 0.035439 rad + ( −0.055017 rad)
=− 0.019578 rad =−0.01958 rad
Finally, the rotation angle at gear D is
3
D
A
φ = φ −φ
∴ φD = φC + φ3
= − 0.019578 rad + 0.057041 rad
= 0.037464 rad = 0.0375 rad
A plot of the rotation angle results can be added to the torque
diagram to give a complete report for the three-segment shaft.
C
150