7. semester Continuum mechanics Solution Exercise 9

1
7. semester Continuum mechanics
Solution Exercise 9
Question 1
In Figure 1 the deformation for a = 1 and b = 0 is sketched.
Fig. 1: u and v for a = 1 and b = 0
In Figure 2 the deformation for a = 0 and b = 1 is sketched.
Fig. 2: u and v for a = 0 and b = 1
Comments to choice of u:
In Timoshenko's "Theory of Elasticity"the similar problem is treated in Section 21. Ti/-
moshenko nds a dierent solution for u, and indirectly concludes that the shear stiness of
the rectangular prole is 4A. The shear stiness should be 5 A, and Timoshenko's boundary
6 6
conditions were wrong. Timoshenko's solution do not secure orthogonality between the
curvature and the shear. In the present solution this condition is fullled which is shown
in equation (1).
∫ h
2
− h 2
u y dy = b [ 1 4
1
3 y3 − 5
12
1 1
5 y5 (h/2) ] h 2
2 − h 2
= 0 (1)

2
Question 2
The strains are dened as:
ε xx = u ,x = −a y (2l − 2x) (2)
ε yy = v ,y = 0
ε xy = 1 2 (u ,y + v ,x ) = 1 2 [−a(2lx − x2 ) + b( 1 4 − 5 1
4 y2 (h/2) ) + a(2lx − 2 x2 ) + b]
= 1 2 b5 4 (1 − 1
y2
(h/2) ) = 1 2 2 γ xy
The normal strain ε xx has been plotted in Figure 3
Fig. 3: Plot of ε xx
The shear strain ε xy has been plotted in Figure 4
Fig. 4: Plot of ε xy
Question 3
The bending moment around the z−axis is given as:
M =
∫ h
2
− h 2
∫ h
2
−σ xx y t dy = − −E a y (2l − 2x) y t dy (3)
− h 2
= E t(2l − 2x)a[ 1 3 y3 ] h 2
− h 2
= a(2l − 2x)EI

3
For x = l the bending is 0, and for x = 0 the bending moment is:
M x=0 = 2alEI (4)
and therefore
a = 1 M x=0
2 lEI
Question 4
The resulting shear force on line 2-3 is given as:
(5)
P =
∫ h
2
− h 2
Gγ xy t dy = G t
∫ h
2
− h 2
= G t b 5 4 [y − 1 3 y3 1
(h/2) 2 ] h 2
− h 2
b 5 4 (1 − 1
y2 )dy (6)
(h/2)
2
= G t b 5 4 (h − 1 3 h) = b G 5 6 h t
Question 5
The shear force P and the bending moment have to be in equilibrium which gives:
P = M x=0
l
which gives:
= 2 a EI = b G 5 ht (7)
6
b = 12 5
E I
G
A a (8)
Furthermore we have b =
P
GA k
where the so-called shear area A k = 5A.
6
The vertical displacement in the tip is given by:
v x=l = 1 P l 3
3 EI + P l
(9)
GA k
The last part in the expression is the deformation due to shear, and this part is often very
small and therefore neglected. The shear deformation are most pronounced for relative
short beams or beams with low shear stiness G as in timber.