Lecture - 03

Internally Unstable Structures and Stability Conditions

ü Some structures are internally unstable due to the presence of internal hinges or flexible joints.

ü To make such structures stable, a suitable arrangement of supports is required.

o The structure is made of two rigid members, AB and BC.

o These members are connected by an internal hinge at point B.

Case 1: Original Support Condition (Fig a)

Support at A: Roller (provides 1 vertical reaction)
Support at C: Hinge (provides 1 vertical and 1 horizontal reaction)

Total external reactions = 3

These 3 reactions would be sufficient to keep a rigid or internally stable structure in equilibrium.

However:

Because of the internal hinge at B, the structure is internally unstable.
It can still move or change shape (i.e. it's a mechanism) even though external equilibrium is satisfied.

Problem:

The internal hinge allows relative rotation between AB and BC.
So, the structure cannot resist horizontal displacement at A.
This causes the whole structure to be unstable.

Case 2: Modified Support Condition (Fig. b)

ü Support at A is changed from roller to hinge, so it now resists:

o Vertical movement

o Horizontal movement

ü Support at C remains a hinge.

Total external reactions = 4

Now the structure is externally stable, even though it has more than 3 reactions.
The extra reaction is justified because the internal hinge introduced internal instability.

Use the Equation of Condition from the Internal Hinge

At the internal hinge B, the bending moment is zero:

ü An internal hinge cannot transmit moment;

→ This means the bending moments at the ends of the connected structural members are zero.

ü When an internal hinge is used to connect two parts of a structure:

→ Each part behaves independently with respect to moment equilibrium at the hinge.

ü Therefore, the algebraic sum of moments about the hinge (caused by loads and reactions) for each portion of the structure must be zero.

ü This condition applies on either side of the hinge (i.e., for both connected members).

ü For example, in the structure of Fig. (b):

→ The presence of the internal hinge at point B requires that the sum of moments about B from:

· the loads and reactions on member AB, and

· the loads and reactions on member BC
must each be individually zero.

This gives an additional equation:

∑MB, AB=0
OR
∑MB, BC=0

This fourth equation makes the system solvable.

ü The two equations:
∑MB, AB=0 and ∑MB, BC=0
are not independent.

ü If one is satisfied, and if you’ve used ∑M=0 for the whole structure, the other is automatically satisfied.

· Therefore, an internal hinge provides only one independent equation of condition.

· This allows you to solve for one extra reaction component.

Equations of Condition (or Construction)

Equations of condition—also called equations of construction—are additional equations used in structural analysis that arise due to specific internal features or constraints within a structure. These equations are not part of the standard three static equilibrium equations but are required to analyze statically determinate structures with internal connections.

1. Internal Hinge:

An internal hinge cannot resist moment.
So, for either side of the hinge:

∑M about hinge=0

Provides 1 equation of condition.

2. Internal Roller:

An internal roller cannot resist moment and cannot transmit force in the sliding direction.
So, you get:

∑M=0 about the roller
∑F parallel to roller surface=0

Provides 2 equations of condition.

Internal Roller Joints in Structures

Special connections are sometimes used in structures:

Allow relative rotation between member ends.
Also allow relative translation in certain directions.

These are modeled as internal roller joints during analysis.

Structure with two rigid members AB and BC connected by an internal roller at B.

The structure is internally unstable.
To resist all coplanar loads, it requires to determine 5 external support reactions to be fully constrained against all possible movements.
Two equations of condition can be used in conjunction with the three equilibrium equations to determine the five unknown external reactions. Thus, the structure is statically determinate externally.

Internal Roller Characteristics:

Cannot transmit moment.
Cannot resist force in the direction parallel to the surface it rolls on.

Equations of Condition Provided by the Internal Roller:

1. ∑Fx, AB=0 → No horizontal force transfer at joint B.

2. ∑MB, AB=0 → No moment transfer at joint B.

· Total of 5 independent equations available.

· Structure now becomes externally statically determinate as there two equations of condition with the three equilibrium equations to determine the five unknown external reactions.

Stability Condition: For Plane internally Unstable Structure

ü r < 3+e_c the structure is statically unstable externally

ü r = 3+e_c the structure is statically determinate externally

ü r > 3+e_c the structure is statically indeterminate externally

· e_c = equations of conditions

· r = external reactions

For an externally indeterminate structure, the degree of external indeterminacy is expressed as

i_c = r – (3+e_c)

Alternative Approach for Static Determinacy of Internally Unstable Structures

Step-by-Step Procedure:

1. Count external reactions: r

2. Count internal force components, f_i:

o Internal hinge: 2 force components

o Internal roller: 1 force component

3. Total number of unknowns:

r + f_i

4. Count number of rigid members/portions: (hinge number +1)

n_r

5. Available equilibrium equations:

3n_r (ΣFx=0, ΣFy=0, ΣM=0 per rigid body)

6. Compare unknowns with equations:

o If r + f_i < 3n_r , the structure is statically unstable externally

o If r + f_i = 3n_r , the structure is statically determinate externally

o If r + f_i > 3n_r , the structure is statically indeterminate externally

7. Degree of Indeterminacy:

=( r + f_i) − 3n_r

Examples:

This beam is internally unstable with r = 5 and e_c = 2. As, r = 3 + e_c, the beam is statically determinate externally

Alternative Method.

f_i = 4, n_r = 3, r + f_i = 5+ 4 = 9 and 3n_r = 3 (3) =9. As, r+f_i = 3n_r, the beam is statically determinate externally

Internally unstable because if the supports are removed, the structure can change its shape by rotating with respect to the hinge located at point C.

The structure possesses four unknown reactions and there is one equation of condition available because of the presence of the hinge at C.

r = 4 and e_c = 1, r = 3 + e_c, Based on this, the structure is classified as statically determinate.

Computation of Reactions

1. Draw the Free-Body Diagram (FBD):

Isolate the structure from its supports and connected bodies.
Show all known forces/couples with arrows (include magnitudes and directions).
Define coordinate system (usually: x = right, y = up).
Indicate all unknown reactions:

Use arrows for forces, curved arrows for moments.
Assume directions (typically positive x/y or counterclockwise).
Use letter symbols for unknowns.

Label dimensions and force locations clearly on the FBD.

2. Check for Static Determinacy:

Use the conditions
If statically or geometrically unstable/indeterminate → Stop analysis here.

3. Apply Equilibrium Equations:

Use the three equilibrium equations:

∑Fx=0
∑Fy=0
∑M=0

Prefer writing equations with only one unknown when possible.
For internally unstable structures:

May need to solve simultaneous equations.
Consider splitting into rigid portions and applying equilibrium to each.
Use FBDs of individual portions, including internal forces (equal & opposite by Newton’s third law).

4. Verify with a Check Equation:

Use a different, previously unused equilibrium equation to check your solution.
Prefer a moment equation about a non-trivial point (i.e., a point not lying on the line of action of any reaction force).
This ensures all reaction forces contribute to the moment equation.
If the equation is satisfied → Reactions are correctly computed.

Example

1. Determine the reactions at the supports for the frame shown in Fig. below

Static Determinacy:

The frame is internally stable with r = 3. Thus, it is statically determinate.

Support Reactions:

2. Determine the reactions at the supports for the beam shown in Fig. below

Static Determinacy:

The beam is internally unstable. It is composed of three rigid members, AB, BE, and EF, connected by two internal hinges at B and E.

The structure has r =5 and e_c = 2; because r = 3 +e_c, the structure is statically determinate.

Support Reactions:

Next, we apply the equation of condition, which involves the summation of moments about B of all the forces acting on the portion AB.

Similarly, by applying the equation of condition on EF portion, we determine the reaction Fy as follows:

The remaining two equilibrium equations can now be applied to determine the remaining two unknowns, Cy and Dy:

It is important to realize that the moment equilibrium equations involve the moments of all the forces acting on the entire structure, whereas the moment equations of condition involve only the moments of those forces that act on the portion of the structure on one side of the internal hinge.

Finally, we compute Dy by using the equilibrium equation,

3. A gable frame is subjected to a wind loading, as shown in Fig. below. Determine the reactions at its supports due to the loading.

Static Determinacy:

The frame is internally unstable, with r = 4 and e_c =1. Since r=3+e_c, it is statically determinate.

Support Reactions:

4. Determine the reactions at the supports for the frame shown in Fig. below

Static Determinacy:

The frame has r = 4 and e_c =1; since r = e_c + 3, it is statically determinate.

Support Reactions:

Solving Eqs. simultaneously, we obtain Ax =14 k and Ay =51 k

5. Determine the reactions at the supports for the three-hinged arch shown in Fig. below

Static Determinacy:

The arch is internally unstable; it is composed of two rigid portions, AB and BC, connected by an internal hinge at B. The arch has r = 4 and e_c =1; since r = 3+e_c , it is statically determinate.

Support Reactions:

Checking Computations. To check our computations, we apply the equilibrium equation ∑ M_B = 0 for the entire structure:

6. The bent beam ABCD shown in Figure is supported by a pin at A and a roller at D. Determine the reactions at A and D.

Determine the statically determinacy:

The bent beam is internally stable as it does not change its shape if the supports are removed.

As there are three reaction components that do not form a system of concurrent or parallel forces and there are three equilibrium equations, the structure is statically determinate.

Determine the reactions:

To obtain an equation involving only V_D, we take moments about A:

To obtain an equation involving only V_A, we can take moments about point F which is the point of intersection of the lines of actions of H_A and V_D, as shown in Figure

[Moment of a force about a point (lies on the line of action of the force, the perpendicular distance is zero) on its line of action is zero]

Summing the forces vertically confirms that the above values are correct (or have compensating errors which is unlikely):

By equating the forces in the x direction to zero, we obtain an equilibrium equation with H_A as the only unknown:

The reaction H_A is therefore acting in the opposite direction to that shown in Figure. The calculated reactions are summarized

Problems:

A. Classify each of the structures shown as externally unstable, statically determinate, or statically indeterminate. If the structure is statically indeterminate externally, then determine the degree of external indeterminacy.

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