CONTINUUM MECHANICS for ENGINEERS

Solution
It is easily verified, by direct substitution, that 4 φ * = 0. The stress components
are directly computed from Eq 6.7-1
σ 11 = 6 D 4 x 1 x 2
σ = 22 0
σ 12 = –B 2 – 3 D 4 x 2 2
These stress components are consistent with an end-loaded cantilever beam,
and the constants B 2 and D 4 can be determined by considering the boundary
conditions. In order for the top and bottom surfaces of the beam to be stressfree,
σ 12 must be zero at x 2 = ±c. Using this condition B 2 is determined in
terms of D 4 as B 2 = –3D 4 c 2 . The shear stress is thus given in terms of single
constant B 2
Bx
=− +
c
σ 12 B2
The concentrated load is modeled as the totality of the shear stress σ on 12
the free end of the beam. Thus, the result of integrating this stress over the
free end of the beam at x1 = 0 yields the applied force P. In equation form
where the minus sign is required due to the sign convention on shear stress.
Carrying out the integration we have B 2 = 3P/4c so that stress components
may now be written as
But for this beam the plane moment of inertia of the cross section is I = 2c 3 /3
so that now
2
2 2
2
P B B x
+ c
2 ⎡ ⎤ 2
=− ⎢−
+ dx ∫ 2 2 ⎥
−c
⎣ c ⎦
σ
σ
σ
11 3 1 2
22
33
3P
=−
2c
= 0
xx
2
3P
⎛ x ⎞ 2
=− ⎜1
− 2 ⎟
4c
⎝ c ⎠
2 2
2 ( 2 )
P
2
σ11 =− 1 2 σ22 = 0 σ33
=− −
I 2
xx
P
; ;
c x
I
in agreement with the results of elementary beam bending theory.