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. Extensive glass breakage in tall buildings can be attributed to excessive deflections Large deflections in buildings are unsightly (and unnerving) and can cause cracks in ceilings and walls. Deflections are limited to prevent undesirable vibrations
Beam Deflection Bending changes the initially straight longitudinal axis of the beam into a curve that is called the Deflection Curve or Elastic Curve
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 the moment is negative, the dfl deflection curve is concave downward. Where the moment is positive the deflection curve is concave upward Where the two curve meet is the Inflection Point
Deflected Shape
Example 1 Draw the deflected shape for each of the beams shown
Example Draw the deflected shape for each of the frames shown
Double Integration Method
Elastic Beam Theory Consider a differential element of a beam subjected to pure bending. The radius of curvature ρ is measured from the center of curvature to the neutral axis Since the NA is unstretched, the dx=ρdθ
Elastic Beam Theory Applying Hooke s law and the Flexure formula, we obtain: 1 M = ρ EI
Elastic Beam Theory Theproduct EI is referred to as the flexuralrigidity rigidity. Since dx = ρdθ, then M d θ = dx (Slope) EI In mostcalculus books 1 ρ M EI d v dx d v / dx = 3 = [ 1+ ( dv / dx) ] = d v / dx [ + ( dv dx) ] 1 / M EI 3 ( exact solution lti )
The Double Integration Method Relate Moments to Deflections d v dx = M EI θ( x) ( ) dv = = M x dx dx EI ( x ) ( x ) M v( x) = dx EI ( x) Do Not Integration Constants Use Boundary Conditions to Evaluate Integration Constants
Assumptions and Limitations Deflections caused by shearing action negligibly small compared to bending Deflections are small compared to the cross sectional dimensions of the beam All portions of the beam are acting in the elastic range Beam is straight prior to the application of loads
y L x PL P @ x Examples x P M = PL + d y EI = M d y dx EI = PL + Px dx dy x EI = PLx + P + c dx dy ( 0) = 0 EI 0 = PL 0 + P + c1 c1 = 0 dx 3 PLx x EIy = + P + c 6 3 PL ( 0) y = 0 EI( 0) = ( 0) + P + c c = 0 6 Integrating once 1 @ x = 0 ( ) ( ) Integrating twice PLx EIy = + P y 3 @ x = 0 ( ) ( ) x 6 3 EIy max @ x = L y = y max PL L = 3 L + P = 6 PL 6 3 y max Px 3 PL = 3EI = PL max 3EI
WL y WL W x x L d y dx W = @ x EI ( L x) W M = d y EI = M dx ( L x ) Integrating once ( L x) dy W EI = + c dx 3 3 1 dy dx @ x = 0 = 0 EI( 0) = W ( L 0) 3 3 + c 1 c 1 WL = 6 3 EI dy dx = W 6 ( L x) 3 3 WL 6
Integrating twice 4 3 ( L x) WL W EIy = x + c 6 4 6 4 3 W @ x = 0 ( L 0) WL y = 0 EI( 0) = ( 0 ) y + c c = 6 4 6 WL 4 4 W EIy = ( L x ) x + 4 6 3 4 4 WL WL 4 Max. occurs @ x = L EIy max W L = 6 4 WL + 4 4 WL = 8 4 y max = 4 WL 8EI = max 4 WL 8 EI
Example y x x @ WL WL WL x M = x Wx d y WL x EI = x W dx 3 dy WL x W x EI = + c dx 3 @ L dy x = dx = 3 L L L WL ( ) = W EI 0 + c c1 = 3 Integrating g 1 x Since the beam is symmetric 0 = 1 EI dy dx L WL = 4 W 3 WL x x 6 4 3 WL 4 3
Integrating g 3 4 3 WL x W x WL EIy = x + c 4 3 6 4 4 3 4 3 WL @ x = 0y = 0 ( 0) W ( 0) WL EI ( 0) = ( 0 ) + c c = 0 4 3 6 4 4 EIy = WL 1 x 3 W 4 x 4 WL 4 3 x Max. occurs @ x = L / EIy = max 5WL 384 4 max = 4 5WL 384EI
Example y x P x L/ L/ P P L P for 0 < x < M = x d y P L EI = x for 0 < x < dx dy P x EI = + c dx @ L dy x = dx = L L P = ( ) = @ x EI 0 + c1 c1 = Integrating g 1 Since the beam is symmetric 0 EI dy dx = P 4 PL x 16 PL 16
Integrating g 3 P x PL EIy = x + c 4 3 16 3 P @ x = 0y = 0 ( 0) PL EI ( 0) = ( 0 ) + c c = 0 4 3 16 EIy = P 1 x 3 PL 16 x Max. occurs @ x = L / EIy = max PL 48 3 max = PL 3 48EI
Example
Example 5
Moment Area Theorems
Moment Area Theorems 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. dv M dv = θ = dx EI dx dθ M M = dθ = dx dx EI EI θ B A B M = dx EI A
dt = xdθ M dθ = dx EI B M M tb A= x dx = x dx EI EI A B A
Moment Area Theorems Theorem : 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. B M ta B = x dx EI A B M ta B = x dx EI A
Example 1
Example
Example 3
Example 4
Example 5
Example 6
Example 7
M/EI 30/EI 0/EI t C / B 0 1 10 = ( 1 ) + EI EI 3 53.33 3 = kn m = 0.00741 rad EI
Another Solution
Conjugate-Beam Method
Conjugate-Beam Method dv d M w w dx = dx = dθ M d v M = = dx EI dx EI Integrating V = wdx M = wdx dx M M θ = dx v = dx dx EI EI
Conju ugate-beam Supports
Example 1 Find the Max. deflection Take E=00Gpa, I=60(10 6 )
θ 56. = EI 56.5 ' = (5) EI 5 B = V B ' B = M B = 1406.5 EI
Find the deflection at Point C Example C
7 63 16 C = M C ' = (1) (3) = EI EI EI
Find the deflection at Point D Example 3
3600 D = M D ' = 70 360 EI EI EI
Find the Rotation at A Example 4 10 ft
θ A = 33.3 EI
Example 5
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Example 6
Moment Diagrams and Equations for Maximum Deflection
Example 4 Find the Maximum deflection for the following structure based on The previous diagrams