p0
y
width = b
h
x
h
L
p0bL
L
p0bL
Try
σ xx = c1 y + c2 x 2 y + c3 y 3
σ xy = c4 x + c5 xy 2
σ yy = c6 + c7 y + c8 y 3
To satisfy equilibrium:
∂σ xx ∂σ xy
+
=0
∂x
∂y
∂σ xy ∂σ yy
+
=0
∂x
∂y
c5 = −c2
c4 = −c7
c5 = −3c8
σ xx = c1 y + 3c8 x 2 y + c3 y 3
σ xy = c4 x − 3c8 xy 2
σ yy = c6 − c4 y + c8 y 3
To satisfy compatibility
∇ 2 (σ xx + σ yy ) = 0
12c8 + 6c3 = 0
σ xx = c1 y + 3c8 x 2 y − 2c8 y 3
σ xy = c4 x − 3c8 xy
2
σ yy = c6 − c4 y + c8 y 3
To make them simpler, write the stresses as
σ xx = ay + 3bx 2 y − 2by 3
σ xy = cx − 3bxy 2
σ yy = d − cy + by 3
Now, we need to apply the boundary conditions to
solve for a, b, c, d
Boundary Conditions
σ yy ( x, h ) = − p0
(I)
σ xy ( x, h ) = 0
(II)
y
h
h
x
σ yy ( x, −h ) = 0
(III)
σ xy ( x, −h ) = 0
(IV)
(II, IV):
σ xy = 0
At y = ± h
c = 3bh2
(I, III):
At y = h
y = −h
so
σ yy = − p0
σ yy = 0
d = − p0 / 2
b = p0 / 4h
3
− p0 = d − 2bh3
0 = d + 2bh3
and
c = 3 p0 / 4h
which gives
σ xx
σ xy
σ yy
3 p0 2
p0 3
= ay + 3 x y − 3 y
4h
2h
3 p0
3 p0 2
=
x − 3 xy
4h
4h
− p0 3 p0
p
y + 03 y 3
=
−
2
4h
4h
All the constants except a are now known.
We still have to address the boundary conditions at
the ends of the beam
σxx
σxy
h
x
h
L
σxy
σxx
L
p0bL
p0bL
P
M
Saint-Venant boundary conditions
x = -L
x=L
+h
+h
(d)
∫ σ xybdy = − p0bL
(a)
∫ σ xxbdy = 0
(b)
+h
+h
−h
∫σ
xx
bdy = 0
xx
ybdy = 0
(P = 0)
−h
−h
∫σ
bdy = p0bL
+h
+h
(f)
xy
−h
−h
(e)
∫σ
(c)
∫ σ xx ybdy = 0
−h
(M = 0)
+h
(a) :
+h
3 p0 L ⎛ y 2 ⎞
∫− h σ xybdy = p0bL ⇒ 4h −∫h ⎜⎝1 − h2 ⎟⎠dy = p0bL
satisfied identically
+h
(b) :
(c) :
∫σ
−h
xx
bdy = 0
satisfied automatically
(σxx is odd in y)
+h
⎛ 3 3L2 ⎞
∫− h σ xx ybdy = 0 ⇒ a = p0 ⎜⎝ 10h − 4h3 ⎟⎠
(d), (e), (f) are all also satisfied with these same constants
Final solution:
σ xx
σ xy
σ yy
⎛ 3 y 3L2 y ⎞ 3 p0 2
p0 3
= p0 ⎜
−
+ 3 x y− 3 y
3 ⎟
2h
⎝ 10h 4h ⎠ 4h
3 p0
3 p0 2
x − 3 xy
=
4h
4h
p0 3
− p0 3 p0
y+ 3 y
=
−
2
4h
4h
Strength of materials solution in terms of x' = x + L
p0
M(x')
V(x')
p0bL
x'
M(x ′)y
σ xx = −
I
V (x ′)Q(y ) +
σ xy = −
Ib
σ yy = 0
2
⎡
(
x ′) ⎤
M(x ′ ) = p0 b⎢ Lx ′ −
⎥
2
⎢⎣
⎦⎥
⎛L
x ⎞
= p0 b⎜ − ⎟
⎝2
2⎠
2
2
V ( x ′ ) = p0 b( L − x ′ )
= − p0 bx
b(h 2 − y2 )
Q(y ) =
2
2
I = bh 3
3
Our present solution:
σ xx
σ xy
σ yy
σ xx max
M ( x′ ) y
⎛ 3y
y3 ⎞
=−
+ p0 ⎜
− 3⎟
I
⎝ 10h 2h ⎠
V ( x′ ) Q ( y )
=−
+
Ib
⎛ 1 3 y y3 ⎞
= − p0 ⎜ +
− 3⎟
⎝ 2 4h 4h ⎠
3 p0 L2
=
At x = L (center of beam)
2
4h
add
p0
=
σ yy max = p0
5
strength
σ xx max
add
σ xx max
add
⇒
σ xx max
strength
2
4 h
=
15 L2
σ yy
add
max
strength
σ xx max
4 h2
=
3 L2
σ xx
⎛ 6y
y3 ⎞
=⎜
− 3⎟
p0 ⎝ 20h 2h ⎠
The additional flexure stress
(self equilibrated: P = M =0)
_
1
0.8
0.6
+
0.4
y/h
0.2
0
-0.2
_
-0.4
-0.6
-0.8
-1
-0.2
+
-0.15
-0.1
-0.05
0
0.05
additional stress
0.1
0.15
0.2
σ yy
⎛ 1 3 y y3 ⎞
= −⎜ +
− 3⎟
p0
⎝ 2 4h 4h ⎠
The additional stress
1
0.8
0.6
0.4
y/h
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
0.1
0.2
0.3
0.4
0.5
|σ yy/p0 |
0.6
0.7
0.8
0.9
1