Beams

BeamsBeams:

Comparison with trusses, plates

Examples:

1. simply supported beams 2. cantilever beams

L, W, t: L >> W and L >> t

L W

t

Beams - loads and internal loadsLoads: concentrated loads, distributed loads, couples (moments)

Internal loads: shear force and bending moments

q

q > 0 q < 0

Shear Forces, Bending Moments - Sign Conventions

Shear forces:

Bending moments:

left section right section

positive shear:

negative shear:

positive moment

negative moment

Shear Forces, Bending Moments - Static Equilibrium Approach

Procedure: 1. find reactions; 2. cut the beam at a certain cross section, draw F.B.D. of one piece of the beam; 3. set up equations; 4. solve for shear force and bending moment at that cross section; 5. draw shear and bending moment diagrams.

Example 1: Find the shear force and bending diagram at any cross section of the beam shown below.

Relationship between Loads, Shear Forces, and Bending Diagram

dV dMq Vdx dx

Beam - Normal Strain

Pure bending problem

no transverse load

no axial load

no torque

Observations of the deformed beam under pure bending

Length of the longitudinal elements

Vertical plane remains plane after deformation

Beam deforms like an arc

MM

Normal Strain - Analysis

x

neutral axis (N.A.):

radius of curvature:

Coordinate system:

longitudinal strain:

y

q

r

'x

L x xL xy y

r q rqrq r

N.A.

Beam - Normal Stressx x

EyE r

Hooke’s Law:

Maximum stresses:

MMM x

y

Neutral axis:0 0 0

0 0

x xA A

cA

EyF dA dA

ydA y

r

Flexure Formula

M M

22 1 x

A A

y E MdM dA y M E dA y dAEI

r r r

2 : second moment of inertial (with respect to the neutral axis)A

I y dA

M x

y

Moment balance:

x xEyE r

xM yI

Axially loaded members Torsional shafts:

Comparison:

Moment of Inertia - IdAyI

A2

Example 2:

Example 3:

w

h

4hww

w

h

Design of Beams for Bending Stresses

Design Criteria:

nuY

allowableallowable or , 1.

2. cost as low as possible

Design Question: Given the loading and material, how to choose the shape and the size of the beam so that the two design criteria are satisfied?

Design of Beams for Bending Stresses

Procedure:

• Find Mmax

• Calculate the required section modulus

• Pick a beam with the least cross-sectional area or weight

• Check your answer

; : section modulusxM y M ISI S y

Design of Beams for Bending Stresses

Example 4: A beam needs to support a uniform loading with density of 200 lb /ft. The allowable stress is 16,000 psi. Select the shape and the sizeof the beam if the height of the beam has to be 2 in and only rectangular and circular shapes are allowed.

6 ft

Shear Stresses inside Beamsshear force: V

Horizontal shear stresses:

V

21

tH

1

1

, : first momenth

Hy

VQ Q ydAIw

t

x

yh1

h2

y1

Shear Stresses inside Beams

HVQIw

t t

Relationship between the horizontal shear stresses and the vertical shear stresses:

Shear stresses - force balance

IwVQ

tV: shear force at the transverse cross sectionQ: first moment of the cross sectional area above the level at which the shear stress is being evaluated

w: width of the beam at the point at which the shear stress is being evaluatedI: second moment of inertial of the cross section

xh1

h2

y1

y

1

1

h

y

Q ydA

Shear Stresses inside Beams

2L

4L

Example 5: Find shear stresses at points A, O and B located at cross sectiona-a. P

a

a 4h

4L

4h4h

4h

w

A

BO

Shear Stress Formula - Limitations

IwVQ

t - elementary shear stress theory

Assumptions: 1. Linearly elastic material, small deformation 2. The edge of the cross section must be parallel to y axis, not applicable for triangular or semi-circular shape 3. Shear stress must be uniform across the width 4. For rectangular shape, w should not be too large

Shear Stresses inside BeamsExample 6: The transverse shear V is 6000 N. Determine the vertical shear stressat the web.

Beams - ExamplesExample 7: For the beam and loading shown, determine (1) the largest normal stress (2) the largest shearing stress (3) the shearing stress at point a

Deflections of Beam

1 MEIr

2

2

3 22

1

1

d ydxdydx

r

Deflection curve of the beam: deflection of the neutral axis of the beam.

x

y

x

y

Derivation:

2 2

2 2 d y M d yEI Mdx EI dx

2

2 dM d d yV V EIdx dx dx

Moment-curvature relationship:

Curvature of the deflection curve:

Small deflection:

(1)

(2)

(3)

Equations (1), (2) and (3) are totally equivalent.

P

Deflections by Integration of the Moment Differential Equation

Example 8 (approach 1):

Deflections by Integration of the Load Differential Equation

Example 8 (approach 2):

Method of SuperpositionP

q

P

Deflection: y

Deflection: y1Deflection: y2

1 2y y y

Method of Superposition

Example 9

P

F

Fq

A B B C

+ dB

Method of Superposition

Example 9 P

F

A B

+ dB

𝜃𝐴=𝜃bending+𝜃𝛿𝐵

𝜃bending 𝜃𝛿𝐵

𝜃bending=4 𝑃𝑎2

81𝐸𝐼 𝜃𝛿𝐵≈ tan 𝜃𝛿𝐵

=𝛿𝐵

𝑎

Method of Superposition

Example 9 Fq

B C

CB C

F q

𝛿𝐵=𝛿𝐹+𝛿𝑞

𝛿𝐹=𝐹𝑏3

3𝐸𝐼 𝛿𝑞=𝑞𝑏4

8𝐸𝐼

𝜃A=4 𝑃𝑎2

81𝐸𝐼 +2𝑃 𝑏3

9𝐸𝐼 𝑎 +𝑞𝑏4

8𝐸𝐼 𝑎

Statically Indeterminate Beam

Number of unknown reactions is larger than the number of independentEquilibrium equations.

Propped cantilever beam

Clamped-clamped beam

Continuous beam

Statically Indeterminate Beam

Example 10. Find the reactions of the propped beam shown below.