BEAMS: DEFORMATION BY SUPERPOSITION Deflection by Superposition
Lecture 12 deflection in beams
description
Transcript of Lecture 12 deflection in beams
Unit 2- Stresses in Beams
Lecture -1 – Review of shear force and bending moment diagram
Lecture -2 – Bending stresses in beams
Lecture -3 – Shear stresses in beams
Lecture -4- Deflection in beams
Lecture -5 – Torsion in solid and hollow shafts.
Topics Covered
Beam Deflection
Moment-Curvature Equation
Recall: THE ENGINEERING BEAM THEORY
€
σy
=MI
=ER
y
x NA A B
A’ B’
If deformation is small (i.e. slope is “flat”):
v (Deflection)
A’
B’
Alternatively: from Newton’s Curvature Equation
€
⇒1R≈d2ydx 2
€
∴1R≈dθdx
€
δθ ≈δyδx
and (slope is “flat”)
y
x
R
€
y = f (x)
€
⇒1R≈d2ydx 2€
dydx⎛
⎝ ⎜
⎞
⎠ ⎟ 2
<<<<1if
€
1R
=
d2ydx 2⎛
⎝ ⎜
⎞
⎠ ⎟
1+dydx⎛
⎝ ⎜
⎞
⎠ ⎟ 2⎛
⎝ ⎜
⎞
⎠ ⎟
32
R
€
δy
From the Engineering Beam Theory:
€
MI
=ER
€
1R
=MEI
€
=d2ydx 2
€
⇒ EI( ) d2ydx 2
= M
Flexural Stiffness
Bending Moment
Curvature
Relationship A B C
B A C
y
€
Deflection = y
Slope = dydx
Bending moment = EI d2ydx2
Shearing force = EI d3ydx3
Rate of loading = EI d4ydx4
Methods to find slope and deflection
Double integration method
Moment area method
Macaulay’s method
Curvature Slope Deflection
Since,
€
d2ydx 2
=1EI⎛
⎝ ⎜
⎞
⎠ ⎟ M Curvature
€
⇒dydx
=1EI⎛
⎝ ⎜
⎞
⎠ ⎟ M∫ ⋅ dx +C1 Slope
€
⇒ y =1EI⎛
⎝ ⎜
⎞
⎠ ⎟ M∫ ⋅ dx⋅ dx∫ + C1⋅ dx∫ +C2 Deflection
Where C1 and C2 are found using the boundary conditions.
R
€
dydx
€
y
Double integration method
Double integration method Slope Deflection
A B C
yc
L
L/2 L/2
W
€
Slope = dydx
= θA = θB = −WL2
16EI
€
Deflection = yc
= −WL3
48EI
Slope Deflection A B C
yc
L
x w/Unit length
€
Slope = dydx
= θA = θB = −WL2
24EI
€
Deflection = yc
= −5
384WL3
EI
Simple supported
Uniform distributed load
Macaulay’s method The procedure of finding slope and deflection for
simply supported beam with an eccentric load is very laborious.
Macaulay’s method helps to simplify the calculations to find the deflection of beams subjected to point loads.
9 - 11
Moment-Area Theorems
€
dθdx
=d2ydx 2
=MEI
dθθC
θ D
∫ =MEIdx
xC
xD
∫
θD −θC =MEIdx
xC
xD
∫
• Consider a beam subjected to arbitrary loading,
• First Moment-Area Theorem:
area under BM diagram between C and D.
€
dx
€
dθ R
€
CD = Rdθ = dx
9 - 12
Moment-Area Theorems
• Second Moment-Area Theorem: The tangential deviation of C with respect to D is equal to the first moment with respect to a vertical axis through C of the area under the BM diagram between C and D.
• Tangents to the elastic curve at P and P’ intercept a segment of length dt on the vertical through C.
€
dt = xdθ = x MEIdx
tC D = x MEIdx
xC
xD
∫ =1EI
xMdxxC
xD
∫ =1EI
A x−
A= total area of BM diagram between C & D = Distance of CG of BM diagram from C
€
x−
Moment Area Method
Moment Area Method
Ken Youssefi Engineering 10, SJSU 15
An Exercise- Moment of Inertia – Comparison
Load
2 x 8 beam
Maximum distance of 1 inch to the centroid
I1
I2 > I1 , orientation 2 deflects less
1
Maximum distance of 4 inch to the centroid I2
Load 2
2 x 8 beam