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