Indeterminate By Deflection

Statically Indeterminate Beams We can use the same method that we used for deflection to analyze statically indeterminate beams Deflection – Part 2 Blessed are they who can laugh at themselves for they shall never cease to be amused.

Wednesday, November 20, 2002

Statically Indeterminate Beams

n

If we remove the supports and look at the reactions we have

The left end is supported as a fixed end support The right end is supported on a roller

MA

w0 A

2

Statically Indeterminate Beams

If we start with a beam loaded as shown n

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Ax

B

w0 A

Ay

B L

By

L

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3

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1

Statically Indeterminate Beams

Statically Indeterminate Beams

We have four unknowns and three equilibrium conditions MA Ax

We use up one of those conditions solving for Ax = 0 MA

w0 A

B L

Ay

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A By

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Statically Indeterminate Beams

Ay

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6

MA

B

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We can reduce the distributed load to a point load but that probably won’t help much

w0

L

By

Statically Indeterminate Beams

We still have three unknown reactions and only two equilibrium conditions to solve for them

A

B L

Ay

5

MA

w0

w0 A

By

Ay

7

Wednesday, November 20, 2002

B L

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By

8

2

Statically Indeterminate Beams

Statically Indeterminate Beams

We do know the loading rate on the beam so we can start from that

From that we can look at the change in the shear across the beam

w ( x ) = w0

w ( x ) = w0 MA

MA

w0 A

B L

Ay Wednesday, November 20, 2002

A By

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10

at x = 0,V = A y and x = L, V = −B y MA

w0 B

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V ( x ) = −w0 x + C1

MA

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By

We have two possible boundary conditions that we can use to solve for C1

V ( x ) = −w0 x + C1

L

L

Statically Indeterminate Beams

Integrating to find the shear at any distance into the beam we have

Ay

w0 B

Ay

Statically Indeterminate Beams

A

dV = −w ( x ) = − w0 dx

w0 A

By

Ay 11

Wednesday, November 20, 2002

B L Meeting Thirty Five

By 12

3

Statically Indeterminate Beams

Statically Indeterminate Beams

Using the left hand boundary condition we have

So our expression for the shear is

V ( x ) = − w0 x + C1

V ( x ) = −w0 x + Ay

V (0) = Ay = −w0 ( 0) + C1 C = Ay

MA 1

A

B L

Ay Wednesday, November 20, 2002

MA

w0

A By 13

Statically Indeterminate Beams

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L

Wednesday, November 20, 2002

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MA

A

w0 x 2 + Ay x + C2 2 at x = 0,M = −M A

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B L

w0

A

w0

Ay

14

We can use our boundary condition again M to solve for C2 M ( x) = −

A

By

Statically Indeterminate Beams

If we now utilize the expression for the shear to develop the expression for the moment at any x we have

dM = V ( x ) = − w0 x + Ay dx w x2 M ( x ) = − 0 + Ay x + C2 2

B

Ay

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V ( x ) = − w0 x + Ay

w0

M ( 0) = − M A = −

By

− M A = C2 15

Wednesday, November 20, 2002

w0 ( 0 ) 2

2

Ay

B L

Notice that the sign of MA is negative. This is due to the fact that we are + Ay 0 + C2directing our y -axis downward so a positive moment will be into the page or CW. (i cross j) = Meeting Thirty Five positive k

By

( )

16

4

Statically Indeterminate Beams

Statically Indeterminate Beams

So our expression for the moment at any M point in the beam is

w0

A

A

B L

Ay

By

w x2 M ( x ) = − 0 + Ay x − M A 2

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MA

w x2 M ( x ) = − 0 + Ay x − M A 2 M x ( ) dθ =− dx EI

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Statically Indeterminate Beams

MA

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Ay

L

Ay

By

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MA

B L

B

Integrating for the slope we have

w0 A

dθ w x2 EI = + 0 − Ay x + M A dx 2

Wednesday, November 20, 2002

w0 A

Statically Indeterminate Beams

An easier to use form can be generated by multiplying both sides by EI and taking the negative sign inside of the expression for the moment

Wednesday, November 20, 2002

We can continue with the process utilizing our slope equation

w0 A

By

Ay

EIθ ( x ) = +

19

B L

By

3 A x2 w0 x − y + M A x + C3 6 2

Wednesday, November 20, 2002

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5

Statically Indeterminate Beams

Statically Indeterminate Beams

We have a boundary condition that we can utilize here to solve for C3 At the support at A, the slope is equal to 0 MA

w0 A

EIθ ( x ) = +

B

2

w0 x 3 Ay x − + M A x + C3 6 2

EIθ (0 ) = 0 = +

w0 ( 0 )

3

6

−

Ay (0 ) 2

So the slope equation has the form

L

Ay

By

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MA

B L

Ay

Ay

By

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Utilizing our boundary condition at the support at A to solve for C4 At the support we have a deflection of 0 MA

B L

w0 A

By

3

w0 x 4 Ay x M x2 − + A + C4 24 6 2 Meeting Thirty Five

Wednesday, November 20, 2002

w0 A

dv dx

Wednesday, November 20, 2002

w0 A

Statically Indeterminate Beams

Finally utilizing the deflection relationship we have

EIv ( x ) = +

MA

+ M A ( 0) + C3

Statically Indeterminate Beams

θ ( x) =

3 A x2 w0 x − y + M Ax 6 2

2

0=C

3 November 20, Wednesday, 2002

EIθ ( x ) = +

EIv ( x ) = +

3 w0 x4 Ay x M x2 − + A + C4 24 6 2

EIv ( 0 ) = 0 = +

Ay ( 0) w0 (0 ) − 24 6 4

3

Ay

B L

By

M A ( 0) + +C 4 2 2

0 = C4 23

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6

Statically Indeterminate Beams

Statically Indeterminate Beams

So our final form for the deflection is

There are actually still two boundary conditions that we know but we haven’t used M At x = L, M=0 A B L At x = L, v =0 A A

MA

w0

y

A L

By

3 w0 x 4 Ay x M x2 − + A 24 6 2

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Statically Indeterminate Beams If we use these two conditions in the expression below, we will have two equations with two unknowns M At x = L, M=0 A At x = L, v =0 L

Ay

EIv ( L ) = 0 = + M ( L) = 0 = − Wednesday, November 20, 2002

w0 ( L ) − 24

w0 ( L ) 2

2

Ay ( L )

3

6

+

M A ( L) 2

3 w0 x 4 Ay x M x2 − + A 24 6 2 2 wx M ( x ) = − 0 + Ay x − M A 2

EIv ( x ) = +

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Statically Indeterminate Beams Working on the math

MA

w0 A

Ay

w0

A

4

By

B

Ay

EIv ( x ) = +

w0

B

0=+

By

2

0=−

B L

By

w0 L2 Ay L − + MA 12 3 w0 ( L ) 2

2

+ Ay ( L ) − M A

+ Ay ( L) − M A

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7

Statically Indeterminate Beams Adding the equations together

Statically Indeterminate Beams

MA

w0 A

B L

Ay

By

Substituting the Ay into the second equation we have MA

0=−

5 w L2 2 Ay L 0=− 0 + 12 3 2 5w0 L 3 5w L Ay = = 0 12 2 L 8 Wednesday, November 20, 2002

29

Statically Indeterminate Beams

w0 ( L )

+ Ay ( L ) − M A

MA

w0 A

Ay

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By

2

B L

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+

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By

If you can write the expressions for the loading, shear, moment, slope, and deflection you have very powerful tools for solving indeterminate structures To be able to use these tools you must be able to identify the boundary conditions that can be used MA

2

w L 5w L wL MA = − 0 + 0 = 0 2 8 8 Wednesday, November 20, 2002

B L

Boundary Conditions

You can substitute the values for MA and Ay in to the expressions for slope and deflection You can also solve for By using the equilibrium expression

2

2

w0 A

Ay

5 w0 L ( L) − MA 2 8 w L2 5 w L2 w L2 MA = − 0 + 0 = 0 2 8 8

0=−

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2

w0 ( L )

2

w0 A

Ay

31

Wednesday, November 20, 2002

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B L

By

32

8

Boundary Conditions

Boundary Conditions

If you are at a fixed end support n n

If you are at a point in a beam where two solutions meet

The slope (θ ) is equal to 0 The deflection (v) is equal to 0

n

If you are at a pin or roller n n

n

The deflection (v) is equal to 0 The moment (M) is equal to 0

n MA

w0 A

Ay

Wednesday, November 20, 2002

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B L

By

33

n

The slope (θ ) from one direction is equal to the slope from the other direction The deflection (v) from one direction is equal to the deflection from the other direction The moment (M) from one direction is equal to the moment from the other direction The shear (V) from one direction is equal to the shear from the other direction

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Homework 9-2.1 9-2.4 9-2.7 In Class test next Wed

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