Basic Equations: Differential Form - Mechanics of Structures: Variational and Computational Methods - page 50

y

f

-dy

a

q z

ds

P

y

q y dz

f

q z

q y

r

a

a

z

z

(c) Circular cross section

(d) Arbitrary cross section

FIGURE 1.17
The cross-sectional rotation of a bar due to torque M t . Point P moves to point P .

( )

is rotated to P =

( + + w). φ φ ≈

The point P y, z P y dv, z d Since d is small, cos d 1 and

φ ≈ φ.

sin d d It follows from Fig. 1.17b that

dv = r cos (φ + d φ) − r cos φ = r ( cos φ cos d φ − sin φ sin d φ) − r cos φ

≈ r cos φ − d φ r sin φ − r cos φ =− d φ r sin φ

=− φ dx z

=− d φ z

d w = r sin (φ + d φ) − r sin φ = r ( cos φ sin d φ + sin φ cos d φ) − r sin φ

≈ d φ r cos φ + r sin φ − r sin φ = d φ r cos φ

= φ dx y

= d φ y

or

v =− φ xz

(1.139)

w = φ xy

φ = The term d φ( x )/ dx is the angle of twist per unit length. Here, it is assumed to be constant

along the bar axis.