aersp 301 bending of open and closed section beams dr. jose palacios
TRANSCRIPT
AERSP 301Bending of open and closed section beams
Dr. Jose Palacios
Today’s ItemsToday’s Items
• HW 2 due today!• HW 3: Extra Credit – 50% of 1 HW worth towards HWs grade• HW 4: will be assigned on Monday. HW helper session coming up. • Bending of Beams -- Megson Chapter 16,
Reference: Donaldson – Chapters Lambda and Mu– Direct stress calculation
– Bending deflections
TodayToday
• HW 3 (EXTRA CREDIT, 50% OF 1 HW): DUE WEDNESDAY• HW 4: ASSIGNED TODAY• EXAM: OCTOBER 20 – 26 HOSLER – 8:15 – 10:15 PM• REVIEW SESSION: OCTOBER 19 – 220 HAMMOND – 6 – 9 PM• WEDNESDAY AFTER CLASS: VOTING REGISTRATION
• Direct stress at a point in the c/s depends on:– Its location in the c/s– The loading– The geometry of the c/s
• Assumption – plane sections remain plane after deformation (No Warping), or cross-section does not deform in plane (i.e. σxx, σyy = 0)
• Sign Conventions!
Megson pp 461
Direct stress calculation due to bending
M – bending moment
S – shear force
P – axial load
T – torque
W – distributed load
Direct stress calculation due to bending (cont’d)
• Beam subject to bending moments Mx and My and bends about its neutral axis (N.A.)
• N.A. – stresses are zero at N.A.
• C – centroid of c/s (origin of axes assumed to be at C).
Neutral Surface Definition
In the process of bending there is an axial line that do not extend or contract. The surface described by the set of lines that do not extend or contract is called the neutral surface. Lines on one side of the neutral surface extend and on the other contract since the arc length is smaller on one side and larger on the other side of the neutral surface. The figure shows the neutral surface in both the initial and the bent configuration.
The axial strain in a line element a distance y above the neutral surface is given by:
• Consider element A at a distance ξ from the N.A.
• Direct Stress:
• Because ρ (bending radius of curvature) relates the strain to the distance to the neutral surface:
Direct stress calculation due to bending (cont’d)
0
0
l
llz
EE zzz
First Moment of Inertia Definition
• Given an area of any shape, and division of that area into very small, equal-sized, elemental areas (dA)
• and given an Cx-Cy axis, from where each elemental area is located (yi and xi)
• The first moment of area in the "X" and "Y" directions are respectively:
dAxAxI
ydAAyI
y
x
• IF the beam is in pure bending, axial load resultant on the c/s is zero:
• 1st moment of inertia of the c/s about the N.A. is zero N.A. passes through the centroid, C
• Assume the inclination of the N.A. to Cx is α
Direct stress calculation due to bending (cont’d)
• Then
The direct stress becomes:
AA
z dAdA 00
cossin yx
cossin yxEE
z
Direct stress calculation due to bending (cont’d)
• Moment Resultants:
• Substituting for σz in the above expressions for Mx and My, and using definitions for Ixx, Iyy, Ixy
dAxM
ydAM
zy
zx
dAxyI
dAxI
dAyI
xy
yy
xx
2
2
cos
sin
cossin
cossin
xyyy
xxxy
y
x
xyyyy
xxxyx
II
IIEM
M
IE
IE
M
IE
IE
M
Direct stress calculation due to bending (cont’d)
• Using the above equation in:
• Gives:
From Matrix Form
y
x
xyyy
xxxy
M
M
II
IIE1
cos
sin
y
x
xyyy
xxxy
xyyyxxM
M
II
II
III
E2
1
cos
sin
cossin yxE
z
yIII
IMIMx
III
IMIM
xyyyxx
xyyyyx
xyyyxx
xyxxxyz
22
Direct stress calculation due to bending (cont’d)
• Or, rearranging terms:
• If My= 0, Mx produces a stress that varies with both x and y. Similarly for My, if Mx=0.
• If the beam c/s has either Cx or Cy (or both) as an axis of symmetry, then Ixy = 0.
• Then:
22xyyyxx
xyxxy
xyyyxx
xyyyxz III
yIxIM
III
xIyIM
yy
y
xx
xz I
xM
I
yM
• Further, if either My or Mx is zero, then:
• We saw that the N.A. passes through the centroid of the c/s. But what about its orientation α?
At any point on the N.A. σz = 0
Direct stress calculation due to bending (cont’d)
xx
xz I
yM
yy
yz I
xMor
tan
022
xyyyyx
xyxxxy
xyyyxx
xyyyyx
xyyyxx
xyxxxyz
IMIM
IMIM
x
y
yIII
IMIMx
III
IMIM
Example Problem
The beam shown is subjected to a 1500 Nm bending moment in the vertical plane.
Calculate the magnitude and location of max σz.
0
8 mm
40 mm 80 mm
80 mm
8 mm
1st: Calculate location of Centroid
mm 528808120
)880(40)8120(60
xx
xxxx
A
Axxc
mm 4.668808120
)880(40)8120(84
xx
xxxx
A
Ayyc
x
y
Example Problem (cont’d)• Calculate Ixx, Iyy, Ixy, with respect to Cxy:
4623
23
23
mm 1009.1)404.66(80812
808
)4.6684(812012
8120
12
cxx dAbt
I
4623
23
23
mm 1031.1)4052(80812
808
)5260(812012
8120
12
cyy dAtb
I
Example Problem (cont’d)
46 mm 1034.0)4.6640()5240(808
)4.6684()5260(8120
A
xy xydAI
Mx = 1500 Nm, My = 0
22xyyyxx
xyxxy
xyyyxx
xyyyxz III
yIxIM
III
xIyIM
mm]in y x,N/mmin [
39.05.12
z
xyz By inspection, MAX at y = -66.4 mm and x = -8 mm(Max stress always further awayFrom centroid)
0x
y
Deflections due to bendingDeflections due to bending
• From strength of materials, recall that [Megson Ch. 16.2.5]:
• Beam bends about its N.A. under moments Mx, My.
– Deflection normal to N.A. is ζ Centroid C moves from CI (initial) to CF (final).
– With R as the center of curvature and ρ as the radius of curvature
Mx
My
wx
w
wy
Distributed Load
Deflections due to bending (cont’d)
• Further,
• Because
Deflections due to bending (cont’d)
• Inverse relation:
• Clearly Mx produces curvatures (deflections) in xz and yz planes even when My = 0 (and vice-versa)
• So an unsymmetrical beam will deflect vertically and horizontally even when loading is entirely in vertical (or horizontal) plane.
What if I have something symmetric?? Like NACA 0012 airfoil?
Deflections due to bending (cont’d)
• If Cx or Cy (or both) are axes of symmetry then Ixy = 0. Then the expressions simplify to:
• Starting with the general expression:
and integrating twice you can calculate the disp. u in the x-direction
Deflections due to bending (cont’d)
• Consider the case where a downward vertical force, W, is applied to the tip of a beam. What is the tip deflection of the beam?
• Integrating,
• Integrating again
Deflections due to bending (cont’d)
• Using b.c.’s: @ z=0 u = 0, u’ = 0– Gives: A = B = 0
• Thus,
– Tip deflection:
• If the c/s has an axis of symmetry, Ixy = 0
You should do this on your own
(z = L)
Simplifications for thin-walled sections
• Thin-walled t << c/s dimensions. – Stresses constant through thickness
– Terms in t2, t3, etc… neglected
• In that case Ixx reduces to:
• What about Ixy for this c/s
• What about Iyy for this c/s
horizontal members
vertical members
Example:
You should do this on your own
Doubly Symmetrical Cross-Section Beam Bending
Beam has a flexural rigidity: EI
)()('''' zpzEIv EI v’’’’ = distributed load
ApzzvEI )(
BpLzpz
zvEI 2
)(2
EI w’’’ = Shear Force
EI w’’ = Moment
CzpLpLzpz
zvEI 226
)(223
EI w’ = Slope
yp(z)
z
pLA
LvEI
0)(
Bc’s
2
0)(2pL
B
LvEI
0
0)0(
C
vEI
DzpLpLzpz
zEIv 4624
)(2234
EI w’ = Displacement
0
0)0(
D
EIv
Doubly Symmetrical Cross-Section Beam Bending
Beam has a flexural rigidity: EI
0)()('''' zpzEIvEI w’’’’ = distributed load
AzvEI )(
BFzzvEI )(
EI w’’’ = Shear Force
EI w’’ = Moment
CFLzFz
zvEI 2
)(2
EI w’ = Slope
y
z
FA
FLvEI
)(
Bc’s
FLB
LvEI
0)(
0
0)0(
C
vEI
DFLzFz
zEIv 26
)(23
EI w’ = Displacement
0
0)0(
D
EIv
F
Concentrated and Partial Span Loads
• Diract delta function:
Example vertical force of magnitude F0 locater at L/2
• Heaviside step function:
Example vertical distributed force of magnitude f0(z) over the second part of the beam only
)/()()( Lzzstpyfzf oy
)2/( LzF
z0
f
Concentrated and Partial Span Loads Example
z
y
L/3L/3 L/3
f0
Fo
z
x
L/3L/3 L/3
f0
Mo
3
2
33)(''''
Lzstp
Lzstpf
LzFzvEI ooxx
3
2
3
23
3)(''''
Lzstp
Lz
L
fLzMxuEI o
oyy
Bc’s
w(0) = w’(0) = v(0) = v’(0) = w(L)= w’(L) = v(L) = v’(L) = 0
Integrating Diract Delta and Heaviside Function
3
2
33)(''''
Lzstp
Lzstpf
LzFzvEI ooxx
3
2
3
2
3324
33662)(
3
2
3
2
336
3322)(
3
2
3
2
33233)(
3
2
3
2
333)(
44
331
22
34
33
221
23
22
12
1
Lzstp
Lz
Lzstp
Lz
f
Lzstp
Lz
FzCzCzCCzvEI
Lzstp
Lz
Lzstp
Lz
f
Lzstp
Lz
FzCzCCzvEI
Lzstp
Lz
Lzstp
Lz
fLzstp
LzFzCCzvEI
Lzstp
Lz
Lzstp
Lzf
LzstpFCzvEI
o
oxx
o
oxx
ooxx
ooxx
Integrating Diract Delta and Heaviside Function
3
2
3
23
3)(''''
Lzstp
Lz
L
fLzMzuEI o
oyy
3
2
3
2
40
33262)(
3
2
3
2
8332)(
3
2
3
2
23)(
3
2
3
2
2
3
3)(
5
231
22
34
421
23
3
12
2
1
Lzstp
Lz
L
f
Lzstp
Lz
MzCzCzCCzuEI
Lzstp
Lz
L
fLzstp
LzM
zCzCCzuEI
Lzstp
Lz
L
fLzstpMzCCzuEI
Lzstp
Lz
L
fLzMCzuEI
o
oyy
ooyy
ooyy
ooyy
OPTIONAL: Macauley’s Method Read Megson Chp 16.)
y
za a a a
W W 2W
AB C D F
Ra Rf
Determine the position and magnitude of the maximum upward and downwarddeflection of the beam:
(upward) 4
3WRA (downward)
4
3WRF
]3[2]2[][ azWazWazWzRM A
The bending moments around the left hand side at any section Z between D and F is: