234
7 Torsion of Prismatic Bars
∴ τ xz = −Gθ y
and
τ yz = Gθ x
or
τ xz = −
M t x
I P
and
τ yz = Gθ x
= G
M t
G I P x
hence, τ yz =
M t x
I P
.
Therefore, the direction of the resultant shear stress τ is such that, from Fig. 7.4
tan α =
τ yz
τ xz
=
M t x/I P
−M t y/I P
= −x/y
Hence, the resultant shear stress is perpendicular to the radius.
Further,
τ
2
= τ
2
yz + τ
2
xz
τ
2
= M
2
t
x
2
+ y
2
/I
2
p
Fig. 7.4 Circular bar under torsion
7 Torsion of Prismatic Bars
∴ τ xz = −Gθ y
and
τ yz = Gθ x
or
τ xz = −
M t x
I P
and
τ yz = Gθ x
= G
M t
G I P x
hence, τ yz =
M t x
I P
.
Therefore, the direction of the resultant shear stress τ is such that, from Fig. 7.4
tan α =
τ yz
τ xz
=
M t x/I P
−M t y/I P
= −x/y
Hence, the resultant shear stress is perpendicular to the radius.
Further,
τ
2
= τ
2
yz + τ
2
xz
τ
2
= M
2
t
x
2
+ y
2
/I
2
p
Fig. 7.4 Circular bar under torsion