combined bending of unsymmetrical beams …e_m.424/unsymm bending.pdfcombined bending of...
TRANSCRIPT
Combined bending of unsymmetrical beams
Internal forces and moments are defined positive as shown
y
z
x
My
Mz
Vy
Vz
Mx =T
Looking down the negative y -axis
O
dx
zV zz
dVV dxdx
+
yM yy
dMM dx
dx+
0
0
O
yy z y
M
dMM V dx M dx
dx
=
⎛ ⎞+ − + =⎜ ⎟
⎝ ⎠
∑
yz
dMV
dx=
+
Looking down the z -axis
O
dx
yV yy
dVV dx
dx+
zM zz
dMM dxdx
+
0
0
O
zz y z
M
dMM V dx M dxdx
=
⎛ ⎞− + + + =⎜ ⎟⎝ ⎠
∑
zy
dM Vdx
= −
+
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
xdw dvu z ydx dx
= − −
w is the displacement of the neutral axis in the z-directionv is the displacement of the neutral axis in the y-direction
Thus, the axial strain developed is2 2
2 2x
xxu d w d ve z yx dx dx
∂= = − −∂
curvatures of the neutral axis
If σxx is the only major stress 2 2
2 2xx xxd w d vEe E z ydx dx
σ⎛ ⎞
= = − −⎜ ⎟⎝ ⎠
Note that there are also strains
2 2
2 2yy zz xxd w d ve e e z ydx dx
ν ν⎛ ⎞
= = − = +⎜ ⎟⎝ ⎠
Relationship between the flexural stress and the internal moments
y
z
x
xxσy
z My
Mz
0xx
xx z
xx y
dA
y dA M
z dA M
σ
σ
σ
=
= −
=
∫∫∫
(1)
(2)
(3)
From (1) 2 2
2 2 0d w d vE zdA E ydAdx dx
− − =∫ ∫
0zdA ydA= =∫ ∫ i.e. neutral axis is at the centroidof the cross section
From (2) 2 22
2 2
2 2
2 2
z
yz zz z
d w d vE yzdA E y dA Mdx dxd w d vE I E I Mdx dx
+ =
+ =
∫ ∫
From (3)2 2
22 2
2 2
2 2
y
yy yz y
d w d vE z dA E yzdA Mdx dxd w d vE I E I Mdx dx
− − =
− − =
∫ ∫
2 2
2 2
2 2
2 2
z yz zz
y yy yz
d w d vM EI EIdx dx
d w d vM EI EIdx dx
= +
= − −
solving these two equations for the curvatures
( )( )( )( )
2
2 2
2
2 2
z yy y yz
yy zz yz
y zz z yz
yy zz yz
M I M Id vdx E I I I
M I M Id wdx E I I I
+=
−
− +=
−
and placing these curvature into the flexural stress relation gives
( ) ( )( )2
y zz z yz z yy y yzxx
yy zz yz
M I M I z M I M I y
I I Iσ
+ − +=
−
( ) ( )( )2
y zz z yz z yy y yzxx
yy zz yz
M I M I z M I M I y
I I Iσ
+ − +=
−
Note: If either the y or z axis is a plane of symmetry then Iyz =0 and we find
y zxx
yy zz
M z M yI I
σ = −
The neutral surface is located where σxx = 0
y
z
neutral surface
λ
tanz yλ=
where
( )( )/
tan/
z y yy yzz yy y yz
y zz z yz zz z y yz
M M I IM I M IM I M I I M M I
λ++
= =+ +
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.
θ
θ
θ
single plane of loading
general loading
/ 2π
P
P
q
q
Flexure Stress Expressions
( ) ( )( )2
y zz z yz z yy y yzxx
yy zz yz
M I M I z M I M I y
I I Iσ
+ − +=
−(1)
(2) can use ( )
( )/
tan/
z y yy yz
zz z y yz
M M I I
I M M Iλ
+=
+to eliminate explicit dependency
on Mz in equation (1) and obtain
( )tantan
yxx
yy yz
M z yI I
λσ
λ−
=− Note: Mz is still in here
(3) can use principal axes of the cross section
y
z
YZ
pθ
sin coscos sin
Y z yZ z y
θ θθ θ
= += −
Y Zxx
YY ZZ
M Z M YI I
σ = −
yθ
z
y’z’
( ) ( )
( ) ( )
( ) ( )
cos 2 sin 22 2
cos 2 sin 22 2
sin 2 cos 22
yy zz yy zzy y yz
yy zz yy zzz z yz
yy zzy z yz
I I I II I
I I I II I
I II I
θ θ
θ θ
θ θ
′ ′
′ ′
′ ′
+ −= + −
+ −= − +
−= +
transformation relations for area moments
parallel axis theorem
y
z
G
yG
zG
y’
z’
( ) ( )22 2zz G z z GG
I y dA y y dA I Ay′ ′′= = + = +∫ ∫similarly
( )( )
2yy y y GG
yz y z G GG
I I Az
I I Ay z
′ ′
′ ′
= +
= +
G is the centroid of A
principal area moments and principal directions
( )2
21, 2
2tan 2
2 2
yzp
yy zz
yy zz yy zzp p yz
II I
I I I II I
θ−
=−
+ −⎛ ⎞= ± +⎜ ⎟
⎝ ⎠
(4) can use neutral surface coordinates
y
z
λ
η
ζ
(eta)
(zeta)
can write ( )2 tany zz z yzxx
yy zz yz
M I M Iz y
I I Iσ λ
+= −
−
but cos sintan
cos cosz yz y λ λ ζλ
λ λ−
− = =
so
2 cosy zz z yz
xxyy zz yz
M I M II I I
ζσλ
+=
−
displacements of the neutral axis
( )( )( )( )
2
2 2
2
2 2
z yy y yz
yy zz yz
y zz z yz
yy zz yz
M I M Id vdx E I I I
M I M Id wdx E I I I
+=
−
− +=
−
we must integrate
twice to obtain the displacements, using boundary conditions on both v and w
( )( )
2
2
2
2
2 2
2 2
tan
tan
z yy y yz
y zz z yz
d vM I M Idx
d w M I M Idxd v d wdx dx
λ
λ
− += = −
+
= −
If λ is a constant and we let ( )2
2
d w F xdx
= − we find, on integration
( )
( )
1 2
3 4tan
w F x dxdx C x C
v F x dxdx C x Cλ
= − + +
−= − + +
∫∫
∫∫
If the boundary conditions on v, w are the same so that
1 3 2 4,C C C C= = then
1tan
wv λ
−= or tan v
wλ −=
y
z
λw
-v
λ
δ = total displacement
Thus, in this case the total displacement
2 2w uδ = + is perpendicular to the
neutral surface.
Example 1
3 m
60o P = 12 kN (through the shear center)
140 mm
200 mm
20 mm all around
Determine:
1. The maximum tensile and compressive stresses and their location
2. The total deflection at the load P
E = 72 GPa
140 mm
200 mm
20 mm all around
1
2
3 y
z
zc
( )( ) ( ) ( )
1 1 2 2 3 3
1 2 3
2 100 4000 10 200010000
82
cz A z A z Az
A A A
mm
+ +=
+ +
× +=
=
( )( ) ( )( )( ) ( )( )
( )( ) ( )( )( )
( )( ) ( )( )( )
3 2 3
6 4
3 2
3 2
6 4
0
1 12 200 20 200 20 70 10 20 10012 12
30.73 1012 20 200 200 20 100 82
121 100 20 100 20 82 10
1239.69 10
yz
zz
yy
I
I
mm
I
mm
=
⎡ ⎤ ⎡ ⎤= × + − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= ×
⎡ ⎤= × + −⎢ ⎥⎣ ⎦⎡ ⎤+ + −⎢ ⎥⎣ ⎦
= ×
Maximum bending moments are at the wall
yz
sin 60P o
cos60P o
6
6
300 sin 60 31.17 10
3000 cos60 18 10y
z
M P N mm
M P N mm
= − = − × −
= = × −
o
o
cot 60 0.577z
y
MM
= − = −o
( )( )
( )
/tan
/
39.690.577 0.74630.73
z y yy yz
zz z y yz
M M I I
I M M Iλ
+=
+
⎛ ⎞= − = −⎜ ⎟⎝ ⎠
0.64036.7
radλ = −
= − o
36.7o
( )
( )
tantan
31.17 0.74639.690.785 0.585
yxx
yy yz
M z yI I
z y
z y
λσ
λ−
=−
−= +
= − −
N-mm mm
mm4
N/mm2 = MPa
36.7o
A
B
At point A: 200 82 11870
z mmy mm= − ==
( ) ( )( ) ( )( )0.785 118 0.585 70
133.6xx A
MPa
σ = − −
= −
At point B: 8270
z mmy mm= −= −
( ) ( )( ) ( )( )0.785 82 0.585 70
105.3xx B
MPa
σ = − − − −
=
alternately, with principal values
31.17 1839.69 30.730.785 0.585
y zxx
yy zz
M z M yI I
z y
z y
σ = −
= − −
= − − which is same as before
Now, compute deflection at the load
yz
sin 60P o
cos60P o
2
2y
yy
Md wdx EI
−=
P
3
3PLwEI
=
L( )
( )( )( )( )
3
3
3 6
sin 603
12,000sin 60 3000
3 72 10 39.69 10
32.72
yy
P Lw
EI
mm
=
=× ×
=
o
o
tan 0.745 24.38v w w mmλ= − = =
2 2 40.8w v mmδ = + =
Example 2
1.5 m
P = 500 N1.5 m
q = 1000 N/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.
y
z
First, we need the reactions at the wall
3.0 m
P = 500 N
0.75 m
500 N 1500 N1125 N-m
1500 N-m
1500 N
500 N1500 N
1125 N-m 1500 n-m
My
Mz
x
1000 N/m
For 0 < x < 1.5
o
2
2
0
1500 1000 / 2 1125 0
500 1500 1125
zo
z
z
M
M x x
M x x N m
=
+ − − =
= − + −
∑ 0
550 1500 0
1500 500
yo
y
y
M
M x
M x N m
=
− + =
= − + −
∑
For 1.5 < x < 3
x
0.75 m
500 N 1500 N1125 N-m
1500 n-m
1500 N
My
Mz
o
01500 1500( 0.75) 1125 00
zo
z
z
MM x xM
=
+ − − − ==
∑ 0
1500 500 0
500 1500
yo
y
y
M
M x
M x N m
=
+ − =
= − −
∑
39.69tan 1.2930.73
yy z z z
zz y y y
I M M MI M M M
λ = = =
0 < x < 1.5
2500 1500 1125tan 1.29500 1500
x xx
λ⎛ ⎞− +
= ⎜ ⎟−⎝ ⎠
1.5 < x < 3
0tan 1.29 0500 1500
0x
λ
λ
⎛ ⎞= =⎜ ⎟−⎝ ⎠=
% 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 degreesx_plot=linspace(0,3,400);ang_plot=[ang zeros(size(x))];plot(x_plot, ang_plot)xlabel('x-distance')ylabel('angle, degrees')
0 0.5 1 1.5 2 2.5 3-45
-40
-35
-30
-25
-20
-15
-10
-5
0
x-distance
angl
e, d
egre
es