Beams - Mechanics of Structures: Variational and Computational Methods - page 741

With the definitions

N

= + = α

N N T , N T E TdA

A

M y =

M y + M Ty , = E α TzdA (12.37)

Ty

A

M z

= M z + M Tz , =− E α TydA

Tz

A

and y, z centroidal axes so that the expressions A = A zdA ydA = 0 hold, Eq. (12.36)

becomes

N
E =

A du
dx

M y
E =−

I yz d

θ

I zz d

θ

z
dx +

y
dx

(12.38)

M z
E

d θ z

dx −

d θ y

dx

= I yy I yz

/ θ / θ /

The normal stress is found by solving Eqs. (12.38) for du dx, d dx, d dx and placing

z y

the resulting expressions into Eq. (12.35). This leads to

N
A +

M y I yy

+ M z I yz
I yy I zz

M y I yz

+ M z I zz
I yy I zz

σ = z − y − E α T (12.39)

x

− I yz − I yz

Shear Stress Due to Shear Forces
To derive the shear stress corresponding to the transverse shear force, consider the equi-
librium of a slice (cross-sectional area A 0 ) of an element of a bar as shown in Fig. 12.10.
Although in Fig. 12.10, the slice is taken to be below the location at which τ is desired, the

slice could just as well be taken as being above this location. If τ = τ xz = τ zx is assumed to

FIGURE 12.10
Calculation of shear stress.