Beams Session 15-22

Subject : S1014 / MECHANICS of MATERIALSYear : 2008

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Beams

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What is Bending Stresses ?

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What is Bending Stresses ?

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Normal Stress

A normal stress is a stress that occurs when a member is loaded by an axial force. 

The value of the normal force for any prismatic section is simply the force divided by the cross sectional area.

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Normal Stress

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What is Bending Stresses ?

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What is Bending Stresses ?

When a member is being loaded similar to that in figure 1 bending stress (or flexure stress) will result.  Bending stress is a more specific type of normal stress. 

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What is Bending Stresses ?

When a beam experiences load like that shown in figure 1 the top fibers of the beam undergo a normal compressive stress. 

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What is Bending Stresses ?

The stress at the horizontal plane of the neutral is zero.  The bottom fibers of the beam undergo a normal tensile stress. 

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What is Bending Stresses ?It can be concluded therefore that the value of the bending stress will vary linearly with distance from the neutral axis.

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What is Bending Stresses ?

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Shear Stress

Normal stress is a result of load applied perpendicular to a member.  Shear stress however results when

a load is applied parallel to an area. 

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Shear StressLike in bending stress, shear stress will vary across the cross sectional area.

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Shear Stress

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ELASTIC CURVE

The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called the elastic curve, which is characterized by the deflection and slope along the curve. E.g.

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ELASTIC CURVE

Moment-curvature relationship:Sign convention:

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ELASTIC CURVE

Moment-curvature relationship:Sign convention:

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ELASTIC CURVE

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ELASTIC CURVE

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ELASTIC CURVE

Consider a segment of width dx, the strain in are ds, located at a position y from the neutral axis is ε = (ds’ – ds)/ds. However, ds = dx = ρdθ and ds’ = (ρ-y) dθ, and so ε = [(ρ – y) dθ – ρdθ ] / (ρdθ), or

1 ρ

= – ε y

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ELASTIC CURVE

Comparing with the Hooke’s Law ε = σ / E and the flexure formula σ = -My/IWe have

1 ρ = M

EI or

1 ρ

= – σ Ey

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SLOPE AND DISPLACEMENT BY INTEGATION (CONT.)

Boundary Conditions:

• The integration constants can be determined by imposing the boundary conditions, or

• Continuity condition at specific locations

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SLOPE AND DISPLACEMENT BY INTEGATION (CONT.)

Boundary Conditions:

• The integration constants can be determined by imposing the boundary conditions, or

• Continuity condition at specific locations

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SLOPE AND DISPLACEMENT BY INTEGATION (CONT.)

Boundary Conditions:

• The integration constants can be determined by imposing the boundary conditions, or

• Continuity condition at specific locations

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SLOPE AND DISPLACEMENT BY INTEGATION (CONT.)

Boundary Conditions:

• The integration constants can be determined by imposing the boundary conditions, or

• Continuity condition at specific locations

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This assumes that the system is linear-elastic, and therefore the deflection is a linear function of F.

, Deflection at B

Loa

d, F

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The total strain energy stored in the system is the sum of the individual strain energies in each of the truss members numbered i=1 to 7.

ii

ii

i AE

LPU

2

27

1

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Beam Elements:

y

x

y

dxdA

z

y

F(x)

A beam that is symmetrical in x-section about the z-axis, is subjected to bending. Consider a infinitesimal volume element of length dx and area dA as shown. This element is subjected to a normal stress: sx=My/I

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Beam Elements:

y

x

y

dxdA

z

y

F(x)

The Strain Energy Density on this element is:

2

2

12

1

x

xx

E

u

For linear elastic material

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Substituting,

I

Myx dxdA

EI

yMudxdA

2

22

2

and multiplying by the Volume of the element

Hence, the Strain Energy for a slice of the beam, of width dx, is

A

A

dAydxEI

M

udxdAdU

22

2

2

y

x

dx

xxAIdAy 2

dxEI

MdU

2

2

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Strain Energy in Entire Beam

Consider the cantilever beam as shown

I

FL

x

y

M=F(x-L)

dxEI

LxFU

L2

0 2

EI

LF

6

32

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Deflection

I

L

x

y

F

FW2

1 =

EI

LF

6

32

EI

FL

3

3

External Work, Strain Energy

Linear-elastic,F

Classical Solution

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y

x

P

L/2 L/2

She

ar F

orce

PL/4

P/2

-P/2M

omen

t

Determine Elastic Strain Energy due to bending for simply supported 3-point bending member of constant X-section.

For 0 xL/2: M=Px/2

Note by symmetry we can find the total strain energy by doubling the strain energy of the LHS.

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EI

LP

EI

xP

dxEI

xP

dxEI

Mdx

EI

MU

L

L

LL

96

12

82

22

2

32

2/

0

32

2/

0

22

2/

0

2

0

2

y

L/2 L/2

P

B

Determine B…….

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Elastic Strain Energy due to Transverse Shear Stress

y

x

3

2

1aU yxy

xya

xy

G

Gu

xy

xyxyxyxy

2

;2

1

2

xy

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Shear Strain Energy

y

dA

z

y

x

dx

F(x)

dxAG

fT

dAdxGA

TU

forceshearTWheredAT

dVG

udVU

xy

xy

2

2

2

2

2

1

2

1

;/2

1

f is called a form factor:Circle f=1.11Rectangle f=1.2Tube f=2.00I section f=A/Aweb

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Applications:

• Castigliano’s 2nd theorem can be used to determine the deflections in structures (eg, trusses, beams, frames, shells) and we are not limited to applications in which only 1 external force or moment acts.

• Furthermore, we can determine the deflection or rotation at any point, even where no force or moment is applied externally.

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Design Considerations• Stress – Yield Failure or Code Compliance

• Deflection• Strain• Stiffness• Stability – Important in compressive members

• Stress and strain relationships can be studied with Mohr’s circle

Often the controlling factor for functionality

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Deflection [Everything’s a Spring]• When loads are applied, we have

deflection• Depends on

– Type of loading• Tension• Compression• Bending • Torsion

– Cross-section of member– Comparable to pushing on a spring

• We can calculate the amount of beam deflection by various methods

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Superposition• Determine effects of individual loads

separately and add the results • May be applied if

– Each effect is linearly related to the load that produces it

– A load does not create a condition that affects the result of another load

– Deformations resulting from any specific load are not large enough to appreciably alter the geometric relations of the parts of the structural system

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Deflection --- Energy Method

There are situations where the tables are insufficientWe can use energy-methods in these circumstancesDefine strain energy

1

0

x

FdxU

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Deflection --- Energy Method

• Define strain energy density**V – volume

dV

dU

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Deflection --- Energy Method

Put in terms of ,

dVE

U

dUdVdV

dUE

E

x

xxx

xx

2

2

2

1

2

1

2

1

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Example – beam in bending

)(2

2

2

2

2

2

22

2

xfEI

M

dAdxdV

dVEI

yMU

dVE

U

I

My

x

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Example – beam in bending

dxEI

MU

dxEI

dAyMdAdx

EI

yMdV

EI

yMU

dAyI

2

2)(

222

2

22

2

22

2

22

2

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Castigliano’s Theorem

• Deflection at any point along a beam subjected to n loads may be expressed as the partial derivative of the strain energy of the structure WRT the load at that point

ii F

U

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Castigliano’s Theorem

• We can derive the strain energy equations as we did for bending

• Then we take the partial derivative to determine the deflection equation

• AND if we don’t have a force at the desired point:– If there is no load acting at the point of interest, add a

dummy load Q, work out equations, then set Q = 0

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Stability

• Up until now, 2 primary concerns– Strength of a structure

• It’s ability to support a specified load without experiencing excessive stress

– Ability of a structure to support a specified load without undergoing unacceptable deformations

• Now, look at STABILITY of the structure– It’s ability to support a load without undergoing a

sudden change in configuration

Materialfailure

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Unit Load Method

• Deflection at C ???

A BC

L

q

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Unit Load Method

• Procedure 1 : Determine Mo

A BC

L

q

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Unit Load Method

Procedure 1 : Determine Mo

q

VA

HA

VB

MA=0w. ½ L- VB.L = 0qL ½ L- VB.L= 0½ qL2 – VB.L = 0

VB = ½ qL

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Unit Load Method

Procedure 1 : Determine Mo

q

VA

HA

VB

VA = ½ qL

MB=0- w. ½ L+ VA.L = 0- qL ½ L+ VA.L= 0- ½ qL2 + VA.L = 0

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Unit Load Method

Procedure 1 : Determine Mo

q

VA

HA

VB

H=0HA = 0

HA = 0

V=0

VA + VB = w½ qL + ½ qL = qL

OK!!!

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Unit Load MethodProcedure 1 : Determine Mo ( cont’ ) for A-C 0<x1< 1/2 L

HA

VA

q

A

w = qx1

x1

Mx=0w.( ½ .x1) –VA.x1 + Mx1 = 0( qx1.x1 ) .( ½ .x1) – qL.x1 + Mx1 = 0Mx1 = qLx1 – ½ x1

2

Mx1 = qLx1 – ½ x12

Mx1

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Unit Load MethodProcedure 1 : Determine Mo ( cont’ ) for B-C 0<x2< 1/2 L

Mx=0w.( ½ .x2) –VB.x2 + Mx2 = 0( qx1.x2 ) .( ½ .x2) – qL.x2 + Mx2 = 0Mx2 = qLx2 – ½ x2

2

Mx2 = qLx2 – ½ x22

VB

B

x2 q

w = qx2

Mx2

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Unit Load MethodProcedure 2 : Put P = 1 unit on C ( without external loads )

VA

HA

VB

MA=0P. ½ L- VB.L = 01 ½ L- VB.L= 0½ L – VB.L = 0

VB = ½

P = 1 unit

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Unit Load MethodProcedure 2 : Put P = 1 unit on C ( without external loads )

VA

HA

VB

VA = ½

MB=0- P. ½ L+ VA.L = 0- 1. ½ L+ VA.L= 0- ½ L + VA.L = 0

P = 1 unit

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Unit Load MethodProcedure 2 : Put P = 1 unit on C ( without external loads )

VA

HA

VB

H=0HA = 0

HA = 0

V=0

VA + VB = w½ + ½ = 1 OK!!!

P = 1 unit

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Unit Load Method

Procedure 3 : Determine m for A-C 0<x1< 1/2 L

HA

VA

Ax1

Mx=0–VA.x1 + mx1 = 0– ½ .x1 + mx1 = 0mx1 = ½ x1

mx1 = ½ x1

mx1

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Unit Load MethodProcedure 3 : Determine m ( cont’ ) for B-C 0<x2< 1/2 L

Mx=0–VB.x2 + Mx2 = 0– ½ .x2 + Mx2 = 0Mx2 = ½ x2

Mx2 = ½ x2

VB

B

x2 q

Mx2

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Unit Load MethodProcedure 4 : Determine deflection at C

Mx1 = qLx1 – ½ x12 Mx2 = qLx2 – ½ x2

2

mx1 = ½ x1 Mx2 = ½ x2

A-C0<x1< ½ L

B-C0<x2< ½ L

odxEI

mxMx

.

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Unit Load Method

Procedure 4 : Determine deflection at C

2

2/1

0

2221

2/1

0

111 )(1/2x*)2^ x½– (qLx)(1/2x*)2^ x½– (qLxdx

EIdx

EI

LL

dxEI

mxMx

.

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Unit Load Method

Procedure 4 : Determine deflection at C

384

5

EI

qL

2

2/1

0

2221

2/1

0

111 )(1/2x*)2^ x½– (qLx)(1/2x*)2^ x½– (qLxdx

EIdx

EI

LL

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Unit Load Method

Deflection at C =

A BC

L

q 384

5

EI

qL

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Unit Load Method

• Slope Deflection C ???

A BC

L

q

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Unit Load Method

• Procedure 1 : Determine Mo

A BC

L

q

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Unit Load Method

Procedure 1 : Determine Mo

q

VA

HA

VB

MA=0w. ½ L- VB.L = 0qL ½ L- VB.L= 0½ qL2 – VB.L = 0

VB = ½ qL

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Unit Load Method

Procedure 1 : Determine Mo

q

VA

HA

VB

VA = ½ qL

MB=0- w. ½ L+ VA.L = 0- qL ½ L+ VA.L= 0- ½ qL2 + VA.L = 0

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Unit Load Method

Procedure 1 : Determine Mo

q

VA

HA

VB

H=0HA = 0

HA = 0

V=0

VA + VB = w½ qL + ½ qL = qL

OK!!!

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Unit Load MethodProcedure 1 : Determine Mo ( cont’ ) for A-C 0<x1< 1/2 L

HA

VA

q

A

w = qx1

x1

Mx=0w.( ½ .x1) –VA.x1 + Mx1 = 0( qx1.x1 ) .( ½ .x1) – qL.x1 + Mx1 = 0Mx1 = qLx1 – ½ x1

2

Mx1 = qLx1 – ½ x12

Mx1

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Unit Load MethodProcedure 1 : Determine Mo ( cont’ ) for B-C 0<x2< 1/2 L

Mx=0w.( ½ .x2) –VB.x2 + Mx2 = 0( qx1.x2 ) .( ½ .x2) – qL.x2 + Mx2 = 0Mx2 = qLx2 – ½ x2

2

Mx2 = qLx2 – ½ x22

VB

B

x2 q

w = qx2

Mx2

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Unit Load MethodProcedure 2 : Put M = 1 unit on C ( without external loads )

MA=0M - VB.L = 01 - VB.L= 0

VB = 1/L

VA

HA

VB

M = 1 unit

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Unit Load MethodProcedure 2 : Put M = 1 unit on C ( without external loads )

MB=0M + VA.L = 01 + VA.L= 0

VA = -1/L

VA

HA

VB

M = 1 unit

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Unit Load MethodProcedure 2 : Put M = 1 unit on C ( without external loads )

VA

HA

VB

M = 1 unit

H=0HA = 0

HA = 0

V=0

-VA + VB = 0- ½L + ½L = 0 OK!!!

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Unit Load Method

Procedure 3 : Determine m for A-C 0<x1< 1/2 L

HA

VA

Ax1

Mx=0–VA.x1 + mx1 = 0– -1/L .x1 + mx1 = 0mx1 = -1/L x1

mx1 = -1/L x1

mx1

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Unit Load MethodProcedure 3 : Determine m ( cont’ ) for B-C 0<x2< 1/2 L

Mx=0–VB.x2 + Mx2 = 0– 1/L.x2 + Mx2 = 0Mx2 = 1/Lx2

Mx2 = 1/L x2

VB

B

x2 q

Mx2

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Unit Load MethodProcedure 4 : Determine deflection at C

Mx1 = qLx1 – ½ x12 Mx2 = qLx2 – ½ x2

2

mx1 = -1/Lx1 Mx2 = 1/Lx2

A-C0<x1< ½ L

B-C0<x2< ½ L

o

dxEI

mxMx

.

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Unit Load Method

Procedure 4 : Determine deflection at C

2

2/1

0

2221

2/1

0

111 )(1/Lx*)2^ x½– (qLx)(-1/Lx*)2^ x½– (qLxdx

EIdx

EI

L

o

L

o

dxEI

mxMx

.

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Unit Load Method

Procedure 4 : Determine deflection at C

0

2

2/1

0

2221

2/1

0

111 )(1/Lx*)2^ x½– (qLx)(-1/Lx*)2^ x½– (qLxdx

EIdx

EI

L

o

L

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Unit Load Method

Slope Deflection C = 0

A BC

L

q