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Indeterminate beams
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INDETERMINATE BEAMS INTRODUCTION
STRENGTH OF MATERIALS 3B (SLR 3B21)
• Chapter 8 in “Mechanics of Engineering Materials” – Benham, Crawford and Armstrong • Chapter 12 in “Mechanics of Materials” – Hibbeler (12.6-12.9)
INDETERMINATE BEAMS INTRODUCTION
STRENGTH OF MATERIALS 3B (SLR 3B21)
• Statically determinacy depends on the number of support reactions. So far only statically determinate beams have been considered having two unknown reactions which can be solved using conditions of equilibrium ∑ F =0 and ∑M=0.
• Only forces perpendicular to beams are considered for SF, BM and beam deflections making the third condition, ∑ F = 0, of no use for this purpose, only for end thrusts.
• Therefore, with statically determinate beams: Two equations resolve two supports.
• On the other hand, statically indeterminate beams have more than two supports, so other techniques have to be employed to solve all the support moment and forces.
INDETERMINATE BEAMS
INTRODUCTION
STRENGTH OF MATERIALS 3B (SLR 3B21)
STATICALLY DETERMINATE BEAMS
F
F
INDETERMINATE BEAMS INTRODUCTION
STRENGTH OF MATERIALS 3B (SLR 3B21)
• A structural member of any type is said to be statically indeterminate if the number of unknown reactions exceeds the available number of equilibrium equations.
• The additional support reactions on the beam or structure that are not needed to keep it in stable equilibrium are called redundants.
• The number of these redundants is referred to as the degree of indeterminacy.
INDETERMINATE BEAMS INTRODUCTION
STRENGTH OF MATERIALS 3B (SLR 3B21)
STATICALLY INDETERMINATE BEAMS
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INDETERMINATE BEAMS INTRODUCTION
STRENGTH OF MATERIALS 3B (SLR 3B21)
As can be seen, these beams have more than two reactions. Therefore they require more equations to be fully resolved. These additional equations can be provided by considering: 1. The basic differential equation of deflection for the beam, i.e. direct integration; 2. The method of superposition; 3. The area-moment method. These approaches will be considered in turn.
INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
This method, as the name implies, requires two integrations of the differential equation:
𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
= 𝑀𝑀𝐸𝐸𝐸𝐸
• Once the integrations have been done, the resulting equation has two constants of integration and unknown redundancies which can all be determined from the boundary conditions.
• Any outstanding redundancies can be determined from equilibrium and compatibility conditions.
• The starting point therefore is to identify the redundancy that would be incorporated into the internal momentum equation and be determined from the boundary conditions.*
* - Section 12.7 Hibbeler [Example 12.17 p.629]
INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
Example [Example 12.18 p.630]
Consider the example illustrated below.
A B w [kg/m]
L RA RB
MA MB
+ - -
BMD
INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
SOLUTION
From geometrical and loading symmetry, we note that MA = MB and RA = RB. From equilibrium of forces in the vertical direction as well we get:
Substituting this into the differential equation we have:
2wxM
2wLxM
dxydEI
2
A2
2++−=−=
2LwRR BA ==
Integrating for the first time yields:
A6
wxxM4
wLxdxdyEI
3
A
2+++−=
INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
Integrating for the second leads to:
BAx24
wx2
xM12
wLxEIy42
A3
++++−=
The unknown terms in this equation include the integration constants A and B and the redundancy MA. These can be obtained from the boundary conditions:
0yiv.)
0dxdyiii.)
0y ii.)
0dxdyi.)
Lx
Lx
0x
0x
=
=
=
=
=
=
=
=
INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
Applying these boundary conditions gives:
24wLMM 0,B 0,A
2
BA ====
Maximum displacement occurs at x = L/2 and is:
IE384Lwyy
4
2Lxmax == =
INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
Example
Next consider the example illustrated below where an point load is applied in the place of a distributed load.
a
L
RA RB
F
MA MB
INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
SOLUTION You will notice that a single expression for bending moment does not apply over the whole length of the beam (unless the origin is moved to the load point) so, keeping origin at A, Macaulay's method is used..
axFxRMMdx
ydEI AA2
2−+−=−=
Integrating once gives:
Aax2F
2xRxM
dxdyEI 2
2
AA +−+−=
Integrating again yields:
BAxax6F
6xR
2xMyIE 3
3
A
2
A ++−+−=
INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
Boundary conditions for the problem are:
i.) At x = 0, y = 0
ii.) At x = L, y = 0
iii.) At x = 0, dy/dx = 0
iv.) At x = L, dy/dx = 0
Applying condition i.) we have B = 0.
Applying condition iii.) we have A = 0.
Now applying condition ii.) we have: 3
3
A
2
A aL6F
6LR
2LM0 −+−=
INDETERMINATE BEAMS
DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
Applying condition iv.) we get:
22
AA aL2F
2LRLM0 −+−=
Solving these two equations simultaneously for MA and RA we get:
( ) ( )( )3
2
A2
2
A LaLa2LFR
LaLFaM −+
=−
= and
The remaining reactions can be determined from equilibrium conditions. For equilibrium of forces in the vertical direction we have:
0FRR BA =−+
And for equilibrium of moments about A we have:
0LRMM- BBA =−+
INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
Solving these two equations for RB and MB we have:
( ) ( )3
2
B2
2
B L2bLFaRand
LaLFaM +
=−
=
Once the reactions have been determined, the complete beam deflection function can be determined enabling the evaluation of the displacement at any given position along the beam.
INDETERMINATE BEAMS
SUPERPOSITION METHOD*
STRENGTH OF MATERIALS 3B (SLR 3B21)
This method involves breaking a complex system down into simpler 'standard‘ systems, analysing them independently then summing the resulting deflections to get the response of the complex system. As an example consider the system shown below where the system is split into two simple configurations. *-Section 12.5 Hibbeler
= + A B
ω
y’B
y’’B
RB
ω
RB RA
MA
L
INDETERMINATE BEAMS SUPERPOSITION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
By super position, the total deflection at point B is given by: ''B
'BB yyy +=
And we already know that:
EI8Lωy
4'B −=
and:
EI3LRy
3B''
B +=
Therefore the total deflection at point B is given by:
EI3LR
EI8ωLy
3B
4
B +−=
INDETERMINATE BEAMS SUPERPOSITION METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
However we know that from compatibility of displacements, the deflection at B must be zero, hence:
0EI3LR
EI8ωL 3
B4
=+−
Solving for RB we have:
8Lω3RB =
Application of equilibrium of forces in the vertical direction gives:
8Lω5RA =
Taking moments about B gives the moment at A i.e.
8LωM
2
A =
INDETERMINATE BEAMS MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
Consider a general beam subjected to general loading as shown below.
INDETERMINATE BEAMS MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
You will recall that the elastic curve for the beam is given by: d2ydx2
= MEI
This can be expressed as: d
dxdydx
= MEI
where we know that dy/dx = θ, the slope at any point along the elastic curve. Hence:
dθdx
= MEI
and
dθ = MEI
dx
INDETERMINATE BEAMS MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
This means that the change in angle between any two pints in the elastic curve is equal to the area under the M/EI distribution function. This is termed the first moment area theorem and is stated as:
� dθ B
A = θB − θA = ∆θB/A = �
MEI
dx B
A
where ∆θB/A is the change in angle between point B and A. If the slope at point A is known i.e. θA , then the slope at point B is given by:
θ𝐵𝐵 = θA + ∆θB/A
If the small change in slope, ∆θ, is multiplied by an arbitrary distance from a reference state, we get a vertical displacement that we can represent by dv i.e.
dv = ∆θ dx
INDETERMINATE BEAMS
MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
INDETERMINATE BEAMS MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
This leads us to second moment-area theorem which can be expressed as:
� dv B
A = vB − vA = ∆vB/A = �
MEI
x dx B
A
where ∆vB/A is the change in vertical displacement between point B and A with respect to a particular reference point. This can also be expressed in the form:
vA/B = φ x�
where φ is area of the M/EI diagram between point B and A and x� is the horizontal distance to the centroid of the area from A. This can also be computed from point B.
INDETERMINATE BEAMS MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
EXAMPLE Find the maximum downward deflection of the small aluminium beam shown in the figure below due to the applied force P = 100 N. The beam’s constant flexural rigidity is EI = 60 Nm2.
P = 100 N
0.15 m 0.10 m
RA RB
MA EI = 60 Nm2
INDETERMINATE BEAMS
MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
SOLUTION The first part of this solution is to determine the redundant reaction so as to be able to establish the bending moment distribution. The expected elastic curve is as shown in the figure below.
P = 100 N
0.15 m 0.10 m
RA RB
MA
INDETERMINATE BEAMS MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
The tangent at the built-in end remains horizontal after the application of the load. To apply this condition using moment area theorem, we need the bending moment distribution diagram as shown below.
P = 100 N
0.15 m 0.10 m
RA RB
MA
6 Nm
INDETERMINATE BEAMS MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
Now applying the moment area approach we have: 1
6012
0.15 613
0.15 + 0.1 +12
0.1 623
0.1 +12
0.25 𝑀𝑀𝐴𝐴23
0.25 = 0
Solving this equation for MA yields: MA = −4.4 Nm
Having obtained the redundant reaction, we can then proceed to determine the other reaction using equilibrium conditions. This gives:
RA = 57.6 N and
RB = 42.4 N
INDETERMINATE BEAMS MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
The maximum deflection occurs where the tangent is zero. Once all the reactions are found we can determine the exact bending moment diagram as shown below.
P = 100 N
0.15 m 0.10 m
RA RB
MA
INDETERMINATE BEAMS
MOMENT AREA METHOD
STRENGTH OF MATERIALS 3B (SLR 3B21)
The location of zero slope occurs when the negative moment is cancelled by the positive moment and is determined to occur at 0.148 m from A. The maximum deflection is then:
vmax = 160
12
0.074 4.2 0.074 + 23
0.074 + 12
0.074 −4.2 13
0.074
= 0.256 mm (down)