Lecture 12 deflection in beams

transcript

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

x NA A B

A’ B’

If deformation is small (i.e. slope is “flat”):

v (Deflection)

Alternatively: from Newton’s Curvature Equation

⇒1R≈d2ydx 2

∴1R≈dθdx

δθ ≈δyδx

and (slope is “flat”)

y = f (x)

⇒1R≈d2ydx 2€

dydx⎛

⎝ ⎜

⎠ ⎟ 2

<<<<1if

d2ydx 2⎛

⎝ ⎜

⎠ ⎟

1+dydx⎛

⎝ ⎜

⎠ ⎟ 2⎛

⎝ ⎜

⎠ ⎟

From the Engineering Beam Theory:

=d2ydx 2

⇒ EI( ) d2ydx 2

Flexural Stiffness

Bending Moment

Curvature

Relationship A B C

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.

Double integration method

Double integration method Slope Deflection

L/2 L/2

Slope = dydx

= θA = θB = −WL2

Deflection = yc

= −WL3

Slope Deflection A B C

x w/Unit length

Slope = dydx

= θA = θB = −WL2

Deflection = yc

= −5

384WL3

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

=d2ydx 2

dθθC

∫ =MEIdx

θD −θC =MEIdx

• Consider a beam subjected to arbitrary loading,

• First Moment-Area Theorem:

area under BM diagram between C and D.

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

∫ =1EI

xMdxxC

∫ =1EI

A x−

A= total area of BM diagram between C & D = Distance of CG of BM diagram from C

Moment Area Method

Ken Youssefi Engineering 10, SJSU 15

An Exercise- Moment of Inertia – Comparison

2 x 8 beam

Maximum distance of 1 inch to the centroid

I2 > I1 , orientation 2 deflects less

Maximum distance of 4 inch to the centroid I2

Load 2

2 x 8 beam

Lecture 12 deflection in beams

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