Combined bending of unsymmetrical beams
Internal forces and moments are defined positive as shown
z
y
Mz
My
Mx =T
Vy
Vz
x
Looking down the negative y -axis
Vz +
Vz
My
dVz
dx
dx
O
dx
+
∑M
O
=0
dM y ⎞
⎛
M y + Vz dx − ⎜ M y +
dx ⎟ = 0
dx
⎝
⎠
dM y
dx
= Vz
My +
dM y
dx
dx
Looking down the z -axis
Vy
Vy +
Mz
dVy
dx
dx
O
dx
+
∑M
O
=0
dM z ⎞
⎛
− M z + Vy dx + ⎜ M z +
dx ⎟ = 0
dx
⎝
⎠
dM z
= −Vy
dx
Mz +
dM z
dx
dx
Assume that plane sections remain plane but that is bending
induced about both the y and z axes even if the moment might only
be about one axis
ux = − z
dw
dv
−y
dx
dx
w is the displacement of the neutral axis in the z-direction
v is the displacement of the neutral axis in the y-direction
Thus, the axial strain developed is
∂u x
d 2w
d 2v
exx =
= −z 2 − y 2
dx
dx
∂x
curvatures of the neutral axis
If σxx is the only major stress
⎛ d 2w
d 2v ⎞
σ xx = Eexx = E ⎜ − z 2 − y 2 ⎟
dx
dx ⎠
⎝
Note that there are also strains
⎛ d 2w
d 2v ⎞
eyy = ezz = −ν exx = ν ⎜ z 2 + y 2 ⎟
dx ⎠
⎝ dx
Relationship between the flexural stress and the internal moments
z
Mz
y
z
σ xx
My
∫ σ dA = 0
∫ yσ dA = − M
∫ zσ dA = M
(1)
xx
y
xx
xx
x
y
z
(2)
(3)
From (1)
d 2w
d 2v
− E 2 ∫ zdA − E 2 ∫ ydA = 0
dx
dx
∫ zdA = ∫ ydA = 0
From (2)
i.e. neutral axis is at the centroid
of the cross section
d 2w
d 2v 2
E 2 ∫ yzdA + E 2 ∫ y dA = M z
dx
dx
d 2w
d 2v
E 2 I yz + E 2 I zz = M z
dx
dx
From (3)
d 2w 2
d 2v
− E 2 ∫ z dA − E 2 ∫ yzdA = M y
dx
dx
d 2w
d 2v
− E 2 I yy − E 2 I yz = M y
dx
dx
d 2w
d 2v
M z = EI yz 2 + EI zz 2
dx
dx
d 2w
d 2v
M y = − EI yy 2 − EI yz 2
dx
dx
solving these two equations for the curvatures
d 2 v ( M z I yy + M y I yz )
=
2
dx
E ( I yy I zz − I yz2 )
d 2 w − ( M y I zz + M z I yz )
=
dx 2
E ( I yy I zz − I yz2 )
and placing these curvature into the flexural stress relation gives
σ xx
M
(
=
I + M z I yz ) z − ( M z I yy + M y I yz ) y
y zz
(I
I − I yz2 )
yy zz
σ xx
M
(
=
I + M z I yz ) z − ( M z I yy + M y I yz ) y
y zz
(I
I − I yz2 )
yy zz
Note: If either the y or z axis is a plane of symmetry then Iyz =0 and
we find
σ xx =
Myz
I yy
Mz y
−
I zz
The neutral surface is located where σxx = 0
z
z = tan λ y
where
neutral surface
λ
tan λ =
y
M z I yy + M y I yz
M y I zz + M z I yz
M
(
=
z
/ M y ) I yy + I yz
I zz + ( M z / M y ) I yz
Although the bending moments may individually be functions of x, if
the ratio Mz / My is independent of x (called single plane of loading),
then the angle of the neutral axis will be a constant in the beam.
Otherwise, the angle will vary with x.
single plane of loading
q
θ
θ
P
general loading
q
θ
P
π /2
Flexure Stress Expressions
(1)
σ xx
(2) can use
M
(
=
I + M z I yz ) z − ( M z I yy + M y I yz ) y
y zz
M
(
tan λ =
(I
z
I − I yz2 )
yy zz
/ M y ) I yy + I yz
I zz + ( M z / M y ) I yz
to eliminate explicit dependency
on Mz in equation (1) and obtain
σ xx =
M y ( z − tan λ y )
I yy − I yz tan λ
Note: Mz is still in here
(3) can use principal axes of the cross section
z
Y
Z
θp
MY Z M ZY
σ xx =
−
IYY
I ZZ
Y = z sin θ + y cos θ
Z = z cos θ − y sin θ
y
z
z’
y’
θ
y
transformation relations for area moments
I y′y′ =
I z′z′ =
I y′z′ =
I yy + I zz
2
I yy + I zz
2
I yy − I zz
2
+
−
I yy − I zz
2
I yy − I zz
2
cos ( 2θ ) − I yz sin ( 2θ )
cos ( 2θ ) + I yz sin ( 2θ )
sin ( 2θ ) + I yz cos ( 2θ )
z’
parallel axis theorem
z
G is the centroid of A
yG
y’
G
zG
y
I zz = ∫ y 2 dA = ∫ ( yG + y′ ) dA = ( I z′z′ )G + AyG2
2
similarly
I yy = ( I y′y′ ) + AzG2
G
I yz = ( I y′z′ ) + AyG zG
G
principal area moments and principal directions
tan ( 2θ p ) =
I p1, p 2 =
−2 I yz
I yy − I zz
I yy + I zz
2
⎛ I yy − I zz ⎞
2
± ⎜
+
I
⎟
yz
2
⎝
⎠
2
(4) can use neutral surface coordinates
(zeta)
ζ
z
η
λ
can write
but
so
σ xx =
z − tan λ y =
σ xx =
(eta)
y
M y I zz + M z I yz
I yy I zz − I
2
yz
( z − tan λ y )
z cos λ − y sin λ
ζ
=
cos λ
cos λ
M y I zz + M z I yz
ζ
I yy I zz − I yz2
cos λ
displacements of the neutral axis
we must integrate
d 2 v ( M z I yy + M y I yz )
=
dx 2
E ( I yy I zz − I yz2 )
d 2 w − ( M y I zz + M z I yz )
=
2
dx
E ( I yy I zz − I yz2 )
twice to obtain the displacements, using
boundary conditions on both v and w
d 2v
dx 2 = − ( M z I yy + M y I yz ) = − tan λ
d 2w
M y I zz + M z I yz )
(
dx 2
d 2v
d 2w
= − tan λ 2
2
dx
dx
If λ is a constant and we let
d 2w
= −F ( x)
dx 2
we find, on integration
w = − ∫∫ F ( x ) dxdx + C1 x + C2
−v
= − ∫∫ F ( x ) dxdx + C3 x + C4
tan λ
If the boundary conditions on v, w are the same so that
C1 = C3 , C2 = C4
w
−1
=
v tan λ
w
tan λ =
or
-v
then
z
λ
δ
−v
w
= total displacement
Thus, in this case the total displacement
λ
y
δ = w2 + u 2
is perpendicular to the
neutral surface.
Example 1
3m
140 mm
60o P = 12 kN (through the shear center)
E = 72 GPa
200 mm
Determine:
20 mm all around
1. The maximum tensile and
compressive stresses and their location
2. The total deflection at the load P
z
140 mm
zc =
z1 A1 + z2 A2 + z3 A3
A1 + A2 + A3
2 × (100 )( 4000 ) + (10 )( 2000 )
=
10000
= 82 mm
1
zc
3
200 mm
2
20 mm all around
I yz = 0
3
2⎤
3⎤
⎡1
⎡1
I zz = 2 × ⎢ ( 200 )( 20 ) + ( 200 )( 20 )( 70 − 10 ) ⎥ + ⎢ ( 20 )(100 ) ⎥
⎣12
⎦ ⎣12
⎦
= 30.73 × 106 mm 4
3
2⎤
⎡1
I yy = 2 × ⎢ ( 20 )( 200 ) + ( 200 )( 20 )(100 − 82 ) ⎥
⎣12
⎦
3
2⎤
⎡1
+ ⎢ (100 )( 20 ) + (100 )( 20 )( 82 − 10 ) ⎥
⎣12
⎦
= 39.69 × 106 mm 4
y
Maximum bending moments are at the wall
z
M y = −300 P sin 60o = −31.17 × 106 N − mm
y
M z = 3000 P cos 60o = 18 × 106 N − mm
P cos 60o
P sin 60o
Mz
= − cot 60o = −0.577
My
M
(
tan λ =
z
/ M y ) I yy + I yz
I zz + ( M z / M y ) I yz
⎛ 39.69 ⎞
= ( −0.577 ) ⎜
⎟ = −0.746
⎝ 30.73 ⎠
λ = −0.640 rad
= −36.7o
36.7o
N-mm
σ xx =
mm
M y ( z − tan λ y )
I yy − I yz tan λ
A
mm4
−31.17
( z + 0.746 y )
39.69
= −0.785 z − 0.585 y
36.7o
=
At point A:
B
N/mm2 = MPa
z = 200 − 82 = 118 mm
y = 70 mm
(σ xx ) A = − ( 0.785)(118) − ( 0.585)( 70 )
= −133.6 MPa
At point B:
z = −82 mm
y = −70 mm
(σ xx ) B = − ( 0.785)( −82 ) − ( 0.585)( −70 )
= 105.3 MPa
alternately, with principal values
σ xx =
Myz
I yy
Mz y
−
I zz
31.17
18
z−
y
=−
39.69
30.73
= −0.785 z − 0.585 y
which is same as before
Now, compute deflection at the load
z
y
d 2 w −M y
=
2
dx
EI yy
P cos 60o
PL3
w=
3EI
P sin 60o
L
P sin 60 ) L
(
w=
o
3
3EI yy
12, 000sin 60 ) ( 3000 )
(
=
3 ( 72 ×10 )( 39.69 × 10 )
3
o
3
6
= 32.72 mm
v = − tan λ w = 0.745w = 24.38 mm
δ = w2 + v 2 = 40.8 mm
P
Example 2
z
y
q = 1000 N/m
1.5 m
P = 500 N
1.5 m
Determine the angle that the neutral surface makes with
respect to the y-axis as a function of x. Use the cross section
properties of example 1.
First, we need the reactions at the wall
500 N
1125 N-m
1500 N
1500 N-m
0.75 m
1500 N
3.0 m
P = 500 N
For 0 < x < 1.5
500 N
1500 N
1500 n-m
1125 N-m
Mz
My
1000 N/m
o
x
∑M
zo
=0
∑M
yo
=0
M z + 1500 x − 1000 x 2 / 2 − 1125 = 0
M y − 550 x + 1500 = 0
M z = 500 x 2 − 1500 x + 1125 N − m
M y = −1500 + 500 x N − m
For 1.5 < x < 3
500 N
1125 N-m
1500 N
1500 n-m
Mz
My
o
0.75 m
1500 N
x
∑M
zo
=0
M z + 1500 x − 1500( x − 0.75) − 1125 = 0
Mz = 0
∑M
yo
=0
M y + 1500 − 500 x = 0
M y = 500 x − 1500 N − m
I yy M z 39.69 M z
M
tan λ =
=
= 1.29 z
I zz M y 30.73 M y
My
0 < x < 1.5
⎛ 500 x 2 − 1500 x + 1125 ⎞
tan λ = 1.29 ⎜
⎟
500
x
1500
−
⎝
⎠
1.5 < x < 3
0
⎛
⎞
tan λ = 1.29 ⎜
⎟=0
⎝ 500 x − 1500 ⎠
λ =0
% script neutral_axis
x=linspace(0,1.5,200);
num=1.29.*(500*x.^2-1500.*x+1125);
denom = 500.*x-1500;
ang=atan(num./denom);
ang = ang*180/pi; % change angle to degrees
x_plot=linspace(0,3,400);
ang_plot=[ang zeros(size(x))];
plot(x_plot, ang_plot)
xlabel('x-distance')
0
ylabel('angle, degrees')
-5
-10
angle, degrees
-15
-20
-25
-30
-35
-40
-45
0
0.5
1
1.5
x-distance
2
2.5
3