Chapter-8

3 4
⇒ c2 = 0
WL 3 W 4 WL3 ∴ EIy = x − x − x 12 24 24
Max occurs @ x = L /2 Max.
EIymax ∆ max
5WL4 =− 384 5WL4 = 384 EI
Example
y
x
P x
L P M= x 2 2 2 d y P L EI 2 = x f 0< x< for dx 2 2 dy P x 2 = + c1 EI Integrating g g d dx 2 2 L dy Since the beam is symmetric @ x= =0 2 dx 2 ⎛L⎞ ⎜ PL2 L P ⎝2⎠ c1 = − EI (0 ) = @ x= + c1 ⇒ 16 2 2 2 for 0 < x <
y x PL P L
Examples
x P
M = − PL + Px
d2y EI 2 = − PL + Px @ x dx x2 dy EI = − PLx + P + c1 Integrating once dx 2 2 ( dy 0) = 0 ⇒ EI (0 ) = − PL(0) + P + c1 ⇒ c1 = 0 @ x = 0 dx 2 2 3 PL PLx x Integrating twice EIy = − + P + c2 2 6 3 ( PL 2 0) @ x = 0 y = 0 ⇒ EI (0 ) = − (0) + P + c2 ⇒ c2 = 0 2 6
P 3 PL2 ∴ EIy = x − x 12 16
Max occurs @ x = L /2 Max.
EIymax ∆ max
PL 3 =− 48 PL3 = 48EI
Example
Example 5
Moment‐Area Theorems
Moment‐Area Theorems
Max occurs @ x = L Max.
EIymax
W L4 WL4 WL4 WL4 =− + =− ⇒ ymax = − 6 24 8 8 EI ∆ max WL4 = 8 EI
Example
y
x x
WL x M= x − Wx 2 2 2 d y WL x2 EI 2 = x −W dx 2 2 dy WL x 2 W x 3 = − + c1 EI Integrating g g d dx 2 2 2 3 L dy Since the beam is symmetric @ x= =0 dx 2 3 2 ⎛L⎞ ⎛L⎞ ⎜ ⎜ WL3 L WL ⎝ 2 ⎠ W ⎝ 2 ⎠ EI (0 ) = @ x= − + c1 ⇒ c1 = − 24 2 2 2 2 3
d 2v M = 2 dx EI dθ M = dx EI
dv ⇒ θ= dx ⎛M ⎞ ⇒ dθ = ⎜ ⎟ dx ⎝ EI ⎠
θB
A
M = ∫ dx EI A
B
dt = xd θ ⎛M dθ = ⎜ ⎝ EI
B
⎞ ⎟ dx ⎠
B
tB
A
M M = ∫x dx = x ∫ dx EI EI A A
Deflected Shape
Example 1
Draw the deflected shape for each of the beams shown
Example 2
Draw the deflected shape for each of the frames shown
Double Integration Method
3 3
W 2 M = − (L − x ) 2 2 d y EI 2 = M dx
Integrating once
@ x = 0
dy W (L − 0 ) WL3 = 0 ⇒ EI (0 ) = + c1 ⇒ c1 = − dx 2 3 6
dy W WL3 3 EI = (L − x ) − dx 6 6
Theorem 1: The change in slope between any two points on the elastic curve equal to the area of the bending moment diagram between these two points, divided by the product EI.
Beam Deflection
• To determine the deflection curve:
– Draw shear and moment diagram for the beam – Directly under the moment diagram draw a line for the beam and label all supports – At the supports displacement is zero – Where h the h moment is i negative, i the h deflection d fl i curve i is concave downward. – Where the moment is positive the deflection curve is concave upward – Where the two curve meet is the Inflection Point
Elastic‐Beam Theory
• Applying Hooke Hooke’s s law and the Flexure formula, formula we obtain:
M = ρ EI
1
Elastic‐Beam Theory
• The product EI is referred to as the flexural rigidity. rigidity • Since dx = ρdθ, then M dθ = dx ( Slope) EI
Moment‐Area Theorems
Theorem 2: The vertical distance of point A on a elastic curve from the tangent drawn to the curve at B is equal to the moment of the area under the M/EI diagram between two points (A and B) about point A .
Structure Analysis I
Chapter 8
Deflections
Introduction
• Calculation of deflections is an important part of structural analysis • Excessive beam deflection can be seen as a mode of failure.
WL 2
L
WL 2
∴
dy WL 2 W 3 WL3 EI = x − x − 4 6 24 dx
Integrating g g
WL x 3 W x 4 WL3 EIy = − − x + c2 4 3 6 4 24
@ x = 0 y = 0
WL (0 ) W (0 ) WL3 (0) + c2 ⇒ EI (0 ) = − − 4 3 6 4 24
Beam Deflection
• Bending changes the initially straight longitudinal axis of the beam into a curve that is called the Deflection fl Curve or Elastic Curve
Relate Moments to Deflections
d v M = 2 dx EI
2
dv M ( x) θ ( x) = = ∫ dx d dx EI( x)
M (x ) 2 dx v ( x ) = ∫∫ EI ( x )
Do Not Integration Constants Use Boundary Conditions to Evaluate Integration Constants

In most calculus books
1
ρ
=
[1 + (dv / dx ) ]
d 2 v / dx 2
d 2 v / dx 2
3 2 dx ) ]
3 2 2
(exact t solution l ti )
d 2v M = 2 dx EI
The Double Integration Method
PLx 2 x3 EIy y=− +P 2 6
@ x = L y = ymax
d2y EI 2 = M dx
∆ max
PL3 = 3EI
EIymax
PL L2 L3 PL3 PL3 =− +P =− ⇒ ymax = − 2 6 6 3EI
y
W
x
x L
WL2 2
WL
@ x
d2y W 2 EI 2 = − (L − x ) dx 2 dy W (L − x ) = EI + c1 dx 2 3
Elastic‐Beam Theory
• Consider a differential element of a beam subjected to pure bending. g • The radius of curvature ρ is measured from the center of curvature to the neutral axis • Since the NA is unstretched, the dx=ρdθ