IA: 1P2: Structures Flashcards
1
Q
What is a force?
A
A force is the action of one body on another. The action of a force is characterised by its magnitude, by the direction of its action, and by its point of application. Force is a vector. F = F eғ where the force of magnitude F is acting in direction eғ.
2
Q
What are the 2 types of forces in structures?
A
- External Forces: These may cause acceleration
- Internal Forces: Forces that resist external forces and cause deformation
3
Q
What is a moment?
A
It is the turning effect of a force around a point. It is equal to the magnitude of the force multiplied by the perpendicular distance from the line of action of the force to the point. Therefore it is given by by the cross product between the line of action of the force and the displacement from the point to the point of application of the force: M₀ = r x F
It obey’s the right hand rule
4
Q
What is a couple?
A
A pair of equal and opposite forces whose lines of action do not coincide, creating pure rotation. These forces cannot be combined into a single force. Their combined moment about a point O in the plane is the couple M which has magnitude Fd. The moment of a couple is the same about any point.
5
Q
Why is the moment of a couple the same about any point?
A
The magnitude of the moment produced is independent of the reference point, it is only dependent on the magnitude of the force and their separation (Fd). Therefore the moment of the couple is the same about any point - it is a purely rotational effect.
6
Q
What is the resultant of a system of forces and couples?
A
The simplest force combination that can replace the original system without altering the effect of the system on the body.
7
Q
How can you find the resultant of congruent forces?
A
The resultant F of any number of intersecting forces can be found with a force polygon. Draw the applied forces to scale, connected heel-to-toe. The resultant force is then teh vector that connects the toe of the first force artrow to the heel of the last: Fᵣₑₛ = ΣFₓ
8
Q
What is Varignon’s theorem for congruent forces?
A
For a body acted on by congruent forces, even if the resultant force F is non-zero, the moment about point O of the original system of forces is zero: the moment of the resultant is equal to the sum of the moments of the individual forces about O, and these are all zero because all forces pass through O.
Additionally, the sum of the moments of all forces about any other point P is equal to the moment of the resultant of the forces about P. This can be given by: ΣMᵢ = (O - P) x Fʀ
9
Q
What are congruent forces?
A
Forces that all intersect at one point
10
Q
How can you find the resultant of non-intersecting forces?
A
Consider the case where the line of action of F₃ does not pass through O, the intersection point of F₁ and F₂:
Firstly, two directly opposed new forces are introcued at O equal to ±F₃. As these cancel each other, they do not effect the resultant forces and momemtns on the body. However, the original F₃ and one of the new ones can now be replaced by the couple C = F₃d₃. The three forces acting at O are now congruent and the resultant can be determined using a polygon of forces. Finally, a pair of opposite couples of magnitude C are added, which again cancel each other so do not effect the resultant forces and moments on the body. The distance d is chosen so that the couple arises from a pair of opposite ±F.
11
Q
What is the general formula for the resultant of forces and couples?
A
12
Q
What are Newton’s laws of motion?
A
- First law: A particle remains at rest or moving in a straight line with constant velocity, providing there is no unbalanced (or resultant) force acting on it.
- Second law: A particle acted upon by a resultant force F experiences an acceleration a that has the same direction as the force, and a magnitude directly proportional to that of the force. If the particle has a mass m, the second law may be stated as F = ma
- Third law: The mutual forces of action and reaction between two interacting bodies are equal in magnitude, opposite in direction, and collinear
13
Q
What are the 2 conditions for static equilibrium?
A
- The resultant force on the body must be zero: F = ΣFᵢ = 0
- The resultant moment about any point must be zero: M = ΣMⱼ = 0
This implies that the vector polygon of all forces, and the polygon of all moments acting on the body are closed.
14
Q
What are distributed loads?
A
Real forces are always spread out over an area rather than being a point load, point loads are simplifications that are used when the contact area is small. Examples of distributed loads include: constant fluid pressure, hydrostatic pressure, and gravitational attraction.
15
Q
What is the equation for the centre of mass of a body?
A
16
Q
What is the centroid of area?
A
The point which defines the geometric centre of a body
17
Q
How can you find the centroid of area of an object?
A
- Centroids of area for common shapes (with uniform density) are listed in the mechanics data book
- If the body has any lines or planes of symmetry, the centroid must lie on them
- Use the equation below
18
Q
What are the Kuhn-Tucker conditions for contact forces?
A
At any point on the interface of two contacting bodies one of two conditions must apply:
* Either, the distance between the bodies is zero so the force between them is positive.
* Or, the distance between the bodies is positive so the force between them is zero.
19
Q
For frictionless contact, what are the properties of the contact force?
A
- A distributed force acts equally and oppositely on both bodies
- Frictionless contact forces act normally to the contact interface (in the direction of the contact normal)
20
Q
What is the equation for static friction?
A
For static contact (no sliding):
μₛ = Coefficient of static friction
21
Q
What is the equation for dynamic friction?
A
For dynamic contact (sliding):
μd = Coefficient of dynamic friction
22
Q
How do the static and dynamic coefficients of friction compare?
A
μₛ > μd
coefficient of static friction > coefficient of dynamic friction
Therefore, once an object starts sliding the friction decreases
23
Q
What is the angle of friction, ψ?
A
The angle of friction is the angle between the normal reaction force (N) and the resultant force (R) where the resultant force is the resultant of the normal contact force and friction. R = F + N
24
Q
What is the equation for the angle of friction?
A
25
For static friction, what is the value of the angle of friction?
26
For dynamic friction, what is the value of the angle of friction?
27
How do the static and dynamic angles of friction compare?
**ψₛ > ψd**
**angle of static friction > angle of dynamic friction**
Therefore, once an object starts sliding the friction decreases
28
What are the 3 different scenarios where the value of friction differs?
* **Static contact, without impending motion:** in these problems, F is not proportional to N. F must be found purely from consideration of equilibrium of the body. The constraints on F are that **F < μₛN or ψ < ψₛ**
* **Static contact, on the point of slipping:** where the bodiues are on the verge of slipping, the coefficient or angle of static friction reach their maximum values, so **F = μₛN or ψ = ψₛ**
* **Dynamic contact, surfaces slipping:** When slip is occuring the coefficient or angle of friction falls to the **lower** constant values given by **F = μdN or ψ = ψd**
29
What is the purpose of the angle of friction?
* The angle of friction gives us a geometric way to visualize how strong the frictional force is between two surfaces
* In problems where an object is sliding or about to slide on a slope (like a hill or ramp), the angle of friction helps you relate the inclination angle (θ) with the frictional forces
30
What is distributed friction?
Distributed friction is the idea that frictional forces act continuously over a surface area or length, rather than being concentrated at a single point.
31
For a rough carpet being pulled (and sliding) across a rough surface, how does the tension differ through the carpet? the coefficient of dry friction is constant.
Although the coefficient of dry friction is constant, as friction is distributed rather than being concentrated at a single point the forces within the carpet vary continuously. This is due to the integration of friction along the interface. Therefore tension is greatest at the front of the carpet and is zero at the back of the carpet.
32
Derive the relationship between T₂ and T₁ on this rough pulley if it is on the point of slipping?
## Footnote
* The coefficient of static friction is μ
* T₂ > T₁
T₂/T₁ = e^(μφ)
33
What is the capstan equation
**T₂/T₁ = e^(μφ)**
* Pulley is on the point of slipping
* The coefficient of static friction is μ
* T₂ > T₁
34
What are the two types of external forces that can be persent in a structure?
* Forces can be **applied** to the structure by an exernal agency in which case the force is known and the displacement of the structure under the action of the force must be found.
* Forces can be **reactions** of the structure at a geometric contraint or support, in which case the displacement is zero and the force is found by overall equilibrium.
35
What are pin-joints?
Pin-joints are a type of mechanical connection that have 1 degree of freedom - they constrain 2 degrees of freedom. They apply a single point reaction force to the structure.
* **Allows rotation** (they cannot apply a moment to the structure)
* **Prevents translation**
36
For a body acted upon by 2 forces, when is it in equilibrium?
**The forces must be colinear.** If one force is the reaction at a pin-joint, then the line of action of the applied force must pass through the pin
37
For a body acted upon by 3 forces, when is it in equilibrium?
**The forces must be concurrent (they must all pass through a single point).** If the forces are parallel, equilibrium is still possible as this point is at infinity
38
How many degrees of freedom does a general 3-dimensional body have?
6
39
How many degrees of freedom does a general 2-dimensional body have?
**3**
rotation, vertical translation, and horizontal translation
40
What is the symbol for a pin joint?
41
What is the symbol for a roller joint?
42
What are roller joints?
Roller joints are a type of mechanical connection that have 2 degrees of freedom - it only contrains one degree of freedom.
* **Allows rotation** (they cannot apply a moment to the structure)
* **Allows translation in ONE direction (usually lateral)**
* **Prevents translation in the other direction (usually vertical)**
43
What is over constraining a body?
A general 2-dimensional body has 3 degree's of freedom, if there are 4 constraints on the body then it is not possible to determine the reactions at the supports from just equilibrium alone and hence it is "over constrained"
44
How is a roller joint represented?
The "formal" definition of a roller joint should be 2 rollers (as shown on the right hand side) as it prevents motion upwards and downwards, however in practise it is usually represented by a single roller (as shown on the left).
45
What is a built-in or 'encastre' support?
Built-in/encastre supoorts are a type of mechanical connection that have 0 degrees of freedom - they contrain all 3 degrees of freedom. Acts like the structure is embedded or clamped into something solid.
* **Prevents all rotation** (they **CAN** apply a moment to the structure)
* **Prevents all translation**
46
What are the 4 types of support options for two-dimensional structures?
* Pin-joint
* Roller joint
* Built-in/encastre joint
* Slider/translation support joint
47
What is a slider/translational support joint?
Slider/translational support joints are a type of mechanical connection that have 2 degrees of freedom - they constrain 1 degrees of freedom.
* **Prevents all rotation** (they **CAN** apply a moment to the structure)
* **Prevents translation in one direction**
* **Allows translation in one direction**
48
What is a truss?
A truss is a framework of members joined at their ends
49
What are the assumptions of an ideal truss?
* **Members are connected by idealised frictionless pin joints** (no moments transferred)
* **Loads are applied only at the joints** (not along the length of the members)
* **All of the members are straight**
* **The self weight of all the members is neglected**
50
How do forces behave in an **ideal** truss?
All the forces in the members are **axial** - they are purely compressive or tensile, there are no bending moments or shear forces
51
What is a statically determinate frame?
A truss where:
* All internal forces can be found using equilibrium equations (no need for material properties).
* The structure is neither under-constrained or over-constrained (for a 2-dimensional truss they require three independent external restraints)
52
What is an under-constrained frame?
The structure has fewer than three independent external restraints and is therefore a **mechanism**, it can move and may collapse without the application of any load
53
What is a statically indeterminate frame?
An **over-constrained** frame, it has greater than 3 independent external restraints and therefore the internal forces depend on the interaction of components which in turn depends on their material properties. It makes analysis far more difficult.
54
How can you determine if a truss is a mechanism (under-constrained)?
**Dj > b + r**
* D = Number of dimensions
* j = Number of joints
* b = Number of bars
* r = Number of restraints
55
How can you determine if a truss is statically determinate?
**Dj = b + r**
* D = Number of dimensions
* j = Number of joints
* b = Number of bars
* r = Number of restraints
56
How can you determine if a truss is statically indeterminate (over-constrained)?
**Dj < b + r**
* D = Number of dimensions
* j = Number of joints
* b = Number of bars
* r = Number of restraints
57
Do the equations shown always hold?
## Footnote
D = Number of dimensions
j = Number of joints
b = Number of bars
r = Number of restraints
No, it is indicative but not universal. For example, it may fail if a structure included both a mechanism and a separate indeterminacy (the equation would indicate it is statically indeterminate). Most well-designed statically determinate trusses will be built up of pin-jointed triangles, so any structure that does not have this appearance may be suspect.
58
What is the method of joints?
The method of joints is used to find **all** the bar forces:
1. Draw a free body diagram for each pin-joint
2. Consider the conditions of force equilibrium at each joint: ΣF = 0. You can do this by graphical methods (force polygon) or by resolving forces (ΣFₓ = 0 and ΣFᵧ = 0)
3. Solve for up to two unknowns per joint (since you get two equations)
## Footnote
Note for stage 2: By definition the forces must be concurrent (meet at the join) and so the moment equilibnrium is automatically satisfied, therefore we only consider the equilibrium of forces
59
When are graphical methods more suitable than purely mathematical methods?
When the geometry is particularly unusual or complex
60
What is the method of sections?
The method of sections is used to find the bar forces in only specific members of a truss:
1. Draw a free-body diagram for a section of the structure chosen such that you cut through the member of interest. Ensure that the cut does not go through more than three unknown member forces!!
2. Draw on the free-body diagram any external forces that act on it, and the axial forces in cut members
3. Resolve and/or take moments to impose the conditions of equilibrium on the section as required: ΣM = 0, ΣFₓ = 0 and ΣFᵧ = 0.
## Footnote
Note for step 3: Try to take moments about a point where the lines-of-action of two of the unknown forces intersect, this allows you to directly determine the third unknown force. (this can be a virtual point, does not necessarily have to be a joint)
61
What are the 3 simplifications that can be made for **unloaded** joints in a planar pin-jointed truss?
62
What is the principle of superposition?
**T = ΣλᵢWᵢ**
When there are many applied forces acting on a truss, W₁, W₂, W₃, etc., and the axial force in a particular member due to Wᵢ alone is λᵢWᵢ, then the force T in that member is given by the principle of superposition: T = λ₁W₁ + λ₂W₂ + λ₃W₃ + ... where the coefficients λᵢ depend on the truss geometry. The principle of superposition is not resitrcted to bar forces in trusses, it applied to all linear systems. In particular, it applies to displacements of trusses aswell
63
When is the behaviour of a truss linear?
The behaviour of a truss is linear if:
* The material of the truss remains in the linear elastic range
* The geometry changes (distortion of the structure) caused by the loads are small and hence the equilibrium equations written in the undeformed configurations are also valid after the structure has deformed.
64
How does a truss being symmetrical make calculations easier?
If a truss is symmetrical and is loaded symmetrically, then the bar tensions (and support reactions) must be symmetrical across the same line. This makes the calculations far easier as we only need to analyse one half of the structure and then flip it across the line of symmetry and use the principle of superposition.
## Footnote
Example of "mirror symmetry" where the truss is symmetrical about a vertical centre-line
65
What is a truss that has been loaded "anti-symmetrically"?
66
How can you analyse a truss that has been loaded "anti-symmetrically"?
Similarly to a symmetric truss, you only need to analyse one half of the structure, as the internal forces in the other half will have the same magnitudes but opposite signs. Therefore you only need to analyse one half of the structure and then flip it across the line of symmetry, flip the signs of all the bar forces and then use the principle of superposition.
67
What is true for any symmetric structure?
It is always possible (and sometimes convenient) to represent any **set** of applied loads by superposing a symmetric and an anti-symmetric set of loads
68
When do shear forces and bending moments arise in members?
* When the joints are not perfectly pin-jointed
* When a member is not completely straight
* When the load is not applied at the joint
69
What is a shear force?
A force perpendicular to the axial force
70
What is a bending moment?
A bending moment is an internal moment that develops in a structural member when an external force tries to bend it.
71
What is the data book sign convention for axial forces, shear forces, and bending moments?
* Axial forces: +ve = tension
* Shear forces: +ve = anticlockwise
* Bending moments: +ve = "hogging"
72
What is "hogging"?
This is when the middle of the beam bends upwards, it is concave downwards.
73
What is the **generalised** method of sections?
For any structural member (cables, struts, beams etc) you can apply the method of sections as long as the internal forces at the cut include axial forces, shear forces, and bending moments. You can then apply equilibrium to the cut part:
* ΣFₓ = 0
* ΣFᵧ = 0
* ΣM = 0
74
What are arches?
Arches are curved structures that efficiently carry loads using mainly compression. They rely on their supports, called abutments, to resist lateral movement.
75
What are segmented arches (and the accompanying assumptions)?
Segmented arches (those made of blocks, such as stone arches in bridges and churches) generally can't carry tensile loads at joints. Three assumptions are typically made for analysing them:
* The blocks are "infinitely" strong (no compression failure)
* The block interfaces are unable to carry bending moments (no tensile capacity)
* Block interfaces have infinite friction (no lateral sliding along the interfaces)
76
How is the capacity of a segmented arch evaluated?
The concept of a **thrust line** is used. The line of thrust comprises of two straight lines that connect the point load to any point within the arch where it meets the abutments to either side of the arch.
* If it is possible to draw a line of thrust to each side of the load that remains entirely within the arch, the arch can support the load.
* If the thrust exits the structure: the arch fails (requires bending, which it can't resist)
## Footnote
Note: It is assumed that the point load is so large that the weight of the stones can be ignored
77
What is a continuous arch?
Continuous (thin) arches (e.g. made of steel or reinforced concrete) use bending moment capacity to resist alternate loading conditions, instead of using geometric thickness. (material property rather than pure geometry)
* To analyse a continuous arch you use method from beam theory (such as method of sections)
78
When is the bending moment in a continuous arch 0?
If it is a **parabolic** arch with a contrinuous distributed load
79
For a 3-pin arch with a non-central load, how do you determine the direction of the abutment reactions?
## Footnote
IMPORTANT
Because the structure is a three-pin arch, each section must be in equilibrium. To determine the direction of the abutment reaction at C, consider the segment OC:
* This segment is acted on by: a reaction force at C, and a hinge force at O (which can transmit force but not moment).
* For this segment to be in equilibrium, it must satisfy the condition of zero net moment. Since the hinge at O cannot resist a moment, any force acting at C must not create a moment about O. The only way for that to be true is if the reaction force at C acts directly along the line connecting C to O. If it didn’t, it would create a moment about O, which the hinge couldn’t resist.
* Therefore, the reaction at C must act along the line CO
To determine the direction of the abutment reaction at B, you can then simply use the rule that all 3 forces must be congruent (meet at a common point).
80
For this 3-pin arch with a non-central load, determine the position of the points where the bending moment is at a maximum in each segment.
Maximum bending moment in OC:
* Occurs where the tangent to the arch is parallel to OC, because this is the point where the lever arm to the reaction at C is maximized.
Maximum bending moment in OB:
* At point D, which is where the moment caused by the load and reactions is greatest in that section.
81
What is a 3-pin continuous arch?
A continuous arch that has three pin joints: one at each abutment, and one at the crown (the highest point in the middle of the arch)
82
What are the 3 components of the state of stress in 2-dimensions?
* Normal stresses: σₓ, σᵧ
* Shear stress: τₓᵧ
83
What is normal stress, σ?
Normal stress is the stress component that acts perpendicular to a surface, it tends to stretch or compress the material along an axis.
84
What is shear stress?
Shear stress is a type of stress that occurs when a force is applied parallel or tangential to the surface of a material (shear force). It tends to shear the material from a small square to a rhomboid form.
85
How can the internal forces (stresses) be related to external loads?
Integrating them with respect to area over the boundary influenced by the load.
86
What is a thin-walled pressure vessel?
For a thin-walled pressure vessel, we can assume the stress distribution is uniform throughout the thickness. Additionally, for an axi-symmetric shape (such as a sphere or cylinder) the stress distribution is uniform along any circumferential or longitudinal section.
87
What is circumferential/hoop stress?
**σθ = circumferential/hoop stress**
For a cylindrical thin-walled pressure vessel, the circumferential/hoop stress is the stress that acts around the circumference of the pressure vessel due to internal pressure. It acts perpendicular to the axis and along the circular cross-section.
88
What is axial stress?
**σz = axial stress**
For a cylindrical thin-walled pressure vessel, the axial stress is the stress that acts along the axis of the pressure vessel due to internal pressure.
89
How do you find the stress in a thin walled pressure vessel?
Using the method of sections:
* A cut is made around one section of the loaded pressure vessel, typically a plane-cut through the vessel in some direction.
* Asserting equilibrium across this plane, the external force (pressure times the projected area of the vessel's interior) must be equal to the internal force (found as the stress in the wall acting normal to the plane multiplied by the area of the cut surface of the vessel)
**Projected area x p = σ x area of cut material**
90
What is the equation for the circumferential/hoop stress (σθ) in a cylindrical thin walled pressure vessel?
## Footnote
p = internal pressure
R = inner radius
t = wall thickness
91
What is the equation for the axial stress (σz) in a cylindrical thin walled pressure vessel?
## Footnote
p = internal pressure
R = inner radius
t = wall thickness
92
What is the equation for the stress in a spherical thin walled pressure vessel?
σθ = σφ = circumferential/hoop stress
93
What are cables (and the associated assumptions)?
Cables (or strings, ropes, chains etc) are structural elements that are only able to carry uniaxial tension, they are unable to carry compression or bending moments as they have negligble bending stiffness. We can assume:
* Tension at any point acts in the direction of the tangent to the cable at that point.
* The cable length will remain approximately constant throughout, they can be treated as inextensible.
Therefore the deformed geometry of an inextensible cable can be predicted purely by considering the equilibrium of external and internal forces.
94
What is the shape of a cable subjected to a **concentrated** load?
If a cable is subjected only to concentrated loads, then it will adopt a stright line between loading points. The overall shape of the cable is therefore a broken line, consisting of straight segments each of which carry constant tension.
95
How can you determine the tension in a cable subjected to a concentrated load?
Use the method of sections or the method of joints
96
What is the shape of a cable subjected to a **uniformly distributed** load (per horizontal length)?
They adopt the shape of a parabolic curve (independent of the magnitude of the loading)
## Footnote
Note: the example used is self weight, but we assume the self weight is w per unit *horizontal* length as it makes analysis easier
97
Derive the shape of this cable subjected to a uniformly distributed load (w) per unit horizontal length using the equilibrium of a free body:
## Footnote
Note: horizontal span of cable is **2**L
y = x²d/L²
## Footnote
d = Mid-point displacement
L = **Half** the horizontal length of cable
98
Derive the shape of this cable subjected to a uniformly distributed load (w) per unit horizontal length using integration:
## Footnote
Note: horizontal span of cable is **2**L
y = x²d/L²
## Footnote
d = Mid-point displacement
L = **Half** the horizontal length of cable
99
What is the equation for the shape of a cable subjected to a **uniformly distributed** load (per horizontal length)?
## Footnote
Note: horizontal span of cable is **2**L
## Footnote
w = Uniform load per horizontal unit length
H = Horizontal component of the tension
d = Mid-point displacement
L = **Half** the horizontal length of cable
100
How do you find the length of a parabolic curve (for a cable)?
Given in data book!
## Footnote
d = Mid-point displacement
L = **Half** the horizontal length of cable
101
How do you approach a question about the deflection of cables?
The deflection of cables can be determined solely from equilibrium arguments:
* Generally, you can find the vertical component of the two reaction forces by equilibrium arguments only
* In order to find the horizontal component of the reaction forces, one additional piece of information about th shape of the cable must have been provided: use it!
* Haveing found the full reaction forces, you can now take a free body diagram at any loacation in the cable to find its deflected shape, angle, and its tension.
102
How does the tension in a cable subjected to a uniformly distributed load (per horizontal length) differ across its length?
It is greatest at the supports and at a minimum at the midpoint
103
What is the equation for strain?
Strain = extension / length
104
What is the equation for stress?
Stress = axial force / cross sectional area
105
What is the equation for Young's modulus?
Young's modulus = stress / strain
106
What is the equation for the extension of a truss member?
## Footnote
e = extension
T = axial force
L = length
A = cross sectional area
E = Young's modulus
107
When will thermal strain lead to an internal stress?
* For a statically determinate pin-jointd truss, uneven thermal expansion can be compensated by displacement of the nodes. Therefore no stress is induced.
* However, for statically indeterminate (over-constrained) structures, uneven thermal expansion cannot be compensated by nodal movement, so will lead to an internal state of stress in the structure. Therefore, even without any applied external loads, the structure will experience internal forces.
108
What is the equation for thermal strain?
α = coefficient of thermal expansion
Δθ = temperature change
109
What is the equation for total strain (due to an axial force and a temperature change)?
110
What is the equation for total extension (due to an axial force and a temperature change)?
It is just the sum of the individual extensions
111
What is compatibility?
Each individual element of the whole structure can only deform in a way that is compatible with the overall deformation of the total structure, otherwise it will break.
112
If all the members in a structure change length, how can you determine what the overall structure looks like?
Using one of two methods:
* **Displacement diagrams** (graphical method)
* **Virtual work** (calculation based)
Both of these methods where the nodes move, how much they move, and in what direction they move.
113
How do displacement diagrams work?
For displacement diagrams, you are **not** redrawing the entire truss, you are simply drawing the movements (displacements) of the nodes caused by the bars changing length. When a bar changes length 2 things happen:
* **There is extension along the bar**. If the bar is increasing in length, then the nodes try to move apart along the bar's axis. If the bar is decreasing in length, then the nodes try to move together along the bar's axis.
* **There is a rotation perpendicular to the bar.** Since the structure has multiple bars joined at a node, rotations must also be considered. Due to small angle approximations, these rotations can be represented as a straight line perpendicular to the extenson of the bar.
114
What is the method for constructing a displacement diagram?
1. **Pick a reference point - the "pole" (o):** Choose one or two nodes that are fixed (e.g., pinned to a wall or ground). These points have zero displacement. The displacement diagram is drawn relative to these fixed points.
2. **Draw the extensions/compressions of bars:** For each bar connected to a known node: Draw an arrow along the bar’s original direction if extended (or opposite if compressed!). The length of this arrow is proportional to the extension/compression.
3. **Account for rotation effects:** For each bar: Draw a line perpendicular to the bar’s original direction. This represents the rotation-induced displacement.
4. **Find the new position of free nodes:** The intersection of these "rotation" lines give the new displaced position of the node relative to fixed nodes (o).
5. **Repeat for all nodes**
6. **Rigid body rotation:** If needed. *Separate flashcard on how to apply a rigid body rotation.*
## Footnote
Note: A bar which has not extension can still rotate! You simply draw the rotation arrow perpendicular to its original direction
115
What is important when constructing a displacement diagram?
* Use large diagrams
* Choose a scale so that tiny extensions (which could be microns in real life) are visible as millimeters.
* Use precise angles — errors over 1° or 1mm can cause big misinterpretations later.
* Label everything: nodes (e.g., A, B, C), extension directions, and whether each is due to stretching or compression.
116
Why may you need to apply a rigid body rotation after constructing a displacement diagram?
Sometimes, even after careful drawing, you find that the structure doesn’t meet certain physical constraints (e.g. a node that should stay level with another doesn’t). This happens as the displacement diagram assumes bar movements but does not initially know about overall structure rotation. To fix this you must apply a small rigid-body rotation to the whole diagram:
* Rotate the whole structure slightly.
* This shifts node positions without stretching any bars.
* Maintains compatibility between bars.
* Makes the displaced structure satisfy boundary conditions (e.g., two supports at same height).
After correction, the displacement diagram now satisfies both: bar extensions and structure-level constraints.
117
How do you apply a fixed body rotation after constructing a displacement diagram?
1. **Find which nodes violate a known constraint** (e.g., Node C should be level with Node A).
2. **Calculate the small angle needed:** θ = height difference / horizontal distance (small angle approximation).
3. **Apply rotation:** Move each node perpendicularly to the line connecting it's original position to the pole (fixed node). The length it should move is given by: "Distance moved = (distance from pole) × (small angle)"
## Footnote
Note: The direction of rotation should be perpendicular to the **original** position vector from the pole to the node, NOT its displaced position!!!
118
What information can be extracted from a displacement diagram?
* **New shape of the structure:** Difference between two node displacements gives the relative movement. From this you can construct the new shape of the structure under load.
* **Strain in a bar:** Strain = change in length / original length. To determine this from the diagram measure the extension arrow along the bars original direction and divide thios by the bars original length
* **Angle of rotation of the bar:** From the perpendicular rotation arrow:
Angle of rotation θ ≈ (length of rotation arrow) / (original bar length). This uses small angle approximations.
119
What is the virtual work equation?
## Footnote
Note: For a pin-jointed planar truss with i nodes and j bars
External work done on nodes = change in internal energy stored in truss members
## Footnote
Fᵢ = External force applied to node i
dᵢ = displacement of node i
Tⱼ = Internal force in truss member j
eⱼ = Extension of truss member j
* = virtual
120
How does the method of virtual work work?
Virtual work is based on the principle of conservation of energy, the external work done on nodes is equal to the change in the internal energy stored in the truss members (external force × displacement = internal force × member extension). Virtual Work allows you to compute displacements or internal forces without fully solving the structure. The equation indicates that F and T must be in equilibrium and that d and e must be compatible and consistent with supports.
121
How do you perform the method of virtual work in order to find real displacements/extensions?
To find real displacements we use real extensions (or vice versa) and we must invent a virtual load (typically a unit force at node and direction of interest). The equation is shown below.
1. Calculate real internal forces Tⱼ due to real loads Fᵢ
2. Calculate the real bar extensions using the equation eⱼ = TⱼLⱼ/EⱼAⱼ (hookes law)
3. Apply a virtual unit load F**'** at the location and direction you care about
4. Find virtual internal internal forces Tⱼ**'**
5. Use the virtual work equation to find the real displacement at the point you wanted **in the direction of the virtual force applied**
To find the total displacement of each node, you must apply a horizontal virtual force and then a vertical virtual force. This will give you the real horizontal and virtual displacement and hence you can determine the total real displacement.
122
How do you perform the method of virtual work in order to find real external/internal forces?
To find real internal forces (or support reactions), we use real external loads and invent a virtual displacement (usually a virtual bar extension or a small rigid-body rotation). The equation is shown below:
1. Apply a virtual displacement d**'** at the location and direction you care about:
.- Either a unit virtual extension in the bar you are interested in (e.g., stretch bar DF by 1 unit),
.- Or a small rigid-body rotation of the structure (if you want a support reaction).
2. Find the virtual bar extensions, e**'**, caused by the virtual displacement.
4. Use the virtual work equation to find the real internal force or support reaction that you are interested in.
## Footnote
Note: The virtual displacement must be in the direction of the real force or bar tension you are trying to find
123
Virtual work and displacement diagrams are both methods for finding the displacement of nodes in a loaded truss, when should each method be used?
**Displacement diagrams:**
* Displacement diagrams find the displacement of **all** nodes in a truss
* Displacement diagrams are more suitable when you want the displacement of many nodes or if the structure is simple/symmetrical
* Cannot be used to find the bar forces or external forces
**Virtual work:**
* Virtual work finds the displacement of **one** node in a specific direction in a truss
* The virtual work method is more suitable when you are trying to find the displacement of just a single node or if the structure is very complicated
* Can also be used to find bar forces and external forces individually
124
How can virtual work be used for structural optimisation for the efficient use of mass in a truss?
Example of how virtual work can be used to determine which bar, if strengthened, would most reduce deflection of a particular node.
125
What are beam structures?
Structures that do not only resist axial loads, but also carry transverse loads such as shear forces and bending moments
126
What is a simply supported beam?
A beam is simply supported if the end supports:
1. Prevent vertical movement (often we assume these supports prevent movement up aswell as down, even though it may look as though the beam is able to lift off)
2. Allow rotation
3. But only one of the supports prevents horizontal movement
Equivalently, the end supports:
1. Can provide vertical forces
2. Cannot transmit moments
3. Onlt one of the supports can provide horizontal forces (and hence unless horizontal loads are applied, there will be no horizontal forces in the system at all)
127
What is a cantilever beam?
A beam is a cantilever if the end support:
1. Prevents vertical movement
2. Prevents rotation
3. Prevents horizontal movement
Equivalently, the end support:
1. Can transmit vertical forces
2. Can transmit moments
3. Can transmit horizontal forces (but in most examples, no horizontal loads will be applied)
128
How do you find the internal forces at a particular point on a beam?
You make a cut on the beam and consider the equilibrium of the piece that has been cut off - which is now a free body. The internal forces include a shear force, a bending moment, and sometimes an axial force
129
What are the assumptions we make when dealing when beam problems?
* The moments are small - there is not permanent deformation
* We can assume taht the structure does not deflect significantly under load and hence we can consider equilibrium in the original configuration (unless otherwise stated!!)
130
What is the sign conventions for axial forces, shear forces, and bending moments?
131
How do you draw a bending moment diagram?
1. Determine any support reactions
2. Take cuts at appropriate positions along the beam, such as a length x from a free end or an end with a known reaction force
3. Write a moment equation about the cut section
4. Solve for the moment expression, M(x) (Note: This may be a piecewise function and hence requires multiple cuts to obtain)
5. Sketch the bending moment diagram on the tension side of the beam - you may have to flip it horizontally depending on the side you have measured from (i.e. if the free end is on the right and you have measured going left)
## Footnote
The same process can be applied to plot the axial force and shear force diagrams
132
What is the bending moment at the free end of a beam?
At a free end of a beam, the bending moment is always zero (unless an external couple/moment is applied directly at that free end)
133
If a beam is subject to a more complex distributed loading (such as sinusoidal) how can you determine the internal forces?
Through the use of integration
134
What are the differential equations of equilibrium for a beam subjected to a distributed force of intensity q (with units of force/length)?
135
Derive the differential equations of equilibrium for a beam subjected to a distributed force of intensity q (with units of force/length)?
136
When is the bending moment in a beam at a maximum?
When the shear force is zero
137
For a beam subjected to a distributed load q(x) across its length, how can you find the expression for the bending moment, M(x)?
1. Write down the expression q(x)
2. Use dS/dx = q and then integrate to find S(x)
3. Use dM/dx = S and then integrate to find M(x)
4. Use the boundary conditions at the supports to obtain the values of the constants of integration
138
When may using these equations pose an issue?
* For concentrated loads
* Distributed loads with an abrupt change in intensity
For these cases it would be necessary to have different differential equations for different sections of the beam, with boundary conditions used to ensure continuity between the sections. Alternatively, you could just use the method of taking a cut and considering the equilibrium of a free body.
139
What is the theorem for a beam subjected to a point-force?
When a straight beam is loaded exclusively by point forces and reactions, then between any two consecturive poitns of application of force:
* The shearing force is constant
* The bending moment varies linearly
Therefore, to find the bending moment diagram for a beam under point loads you simply need to calculate M at all points of application of the force (but not anywhere else) and join consecutive poitns up with straight lines. M(x) is piecewise linear
140
What is a propped cantilever?
A propped cantilever is a type of beam structure that is similar to a cantilever, but it has an additional support at some point along its length, usually at the free end. It is a statically indeterminate structure.
141
A propped cantilever is a statically indeterminate structure, how can we determine Q, F, and R?
As it is statically indeterminate, we cannot use equilbrium alone to find the internal forces. Therefore we must also use compatability (matching deflections) to solve fully. Equilibrium alone cannot find a unique solution, however there is only one value of R which will satisfy equilibrium and give zero tip deflection.
142
How can you recognise statically determinate and statically indeterminate structures?
Any "tree structure" i.e. any series of catilever beams connected to one another which does not form any closed loops, is statically determinate
143
What is curvature?
The curvature of a plane curve is the rate of turning of the tangent to the curve, i.e. the angle turned through by the tangent per unit length of the curve
144
What is the equation for curvature?
ψ = rotation
s = length along arc
ρ = local radius of curvature
## Footnote
Assuming ψ is positive clockwise
145
What is deflection, v?
Deflection is the displacement of a point on a structurefrom its original position when the structure is subjected to a load
146
When performing a deflection calculation along a beam, what is a common assumption made?
If the rotation ψ is small, we can assume that the arc length ≈ horizontal distance (s ≈ x). Therefore the curvature equation can be written as:
## Footnote
Assuming ψ is positive clockwise
147
What is the equation for rotation in terms of deflection?
## Footnote
Assuming ψ is positive clockwise and s ≈ x
148
What is the equation for curvature in terms of deflection?
## Footnote
Assuming ψ is positive clockwise and s ≈ x
149
For a **slender beam**, how does internal forces lead to distortion in their shape?
In a **slender beam** the distoritions due to the tensile and shear forces are often negligible in comparison to those due to the bending moment. Therefore for slender beams we can often neglect the distortions due to S and T and only consider those due to M. Bending moment dominates deformation.
150
What is the relationship between the bending moment in a beam and the curvature?
## Footnote
"Moment-curvature" relation
M = bending moment
Δκ = change in curvature
B = Bending stiffness
## Footnote
If the initial curvature is zero, then the equation just becomes: M = Bκ
151
What is the equation for bending stiffness of a beam?
E = Young's modulus
I = second moment of area of the cross-section
## Footnote
Also known as the flexural stiffness or flexural rigidity - it is an elastic constant of the beam
152
How can you find the deflection of a beam from the from a load q(x)?
1. From the distributed load you can determine the shear force S(x), q = dS/dx
2. From the shear force S(x) you can determine the bending moment M(x), S = dM/dx
3. From the bending moment M(x) you can determine the curvature κ(x), M = Bκ
4. From the curvature κ(x) you can determine the deflection v(x) by integrating twice, κ = - d²v/dx²
| Alternatively you could find the bending moment M(x) through equilibrium
## Footnote
Note: You must stick to all the databook sign conventions
153
What is the tip rotation and deflection of a cantilever loaded by a pure couple at the end?
End rotation = QL/B clockwise
End deflection = QL²/2B down
154
How can superposition be used to determine deflection?
You can use linear superposition of "simple" deflection cases to determine more complex deflection scenarios.
155
Determine the deflection of this overhanging beam:
156
Determine the deflection of this overhanging beam:
157
What is the reaction force acting on this propped cantiliever?
Using the principle of linear superposition:
158
If there is a resultant axial force T acting along a beam, how is the stress distributed across the cross-section?
Under uniaxial tension the stress is **uniform** over the cross-section
159
If there is a resultant axial force T acting along a beam, where does the resultant force pass through?
Resultant force acts through the centroid of the cross-sectional area
160
If a beam is subjected to bending moments as shown below, what is the direction of the tensile force acting through the beam?
* The top fibres are in tension
* The bottom fibres are in compression
161
This is the side view of an originally straight beam which has been curved following the application of a bending moment. What are the 2 assumptions we can make about the fibres?
## Footnote
a "fibre" of this beam is a line parallel to the z-axis at a distance y from the zx plane
1. The line which does not change in lengthtraces the position of the neutral axis.
2. The "sleeper" lines remain straight and perpendicular to the "rail" lines - this is an example of the "Euler-Bernoulli" hypothesis.
162
What is the neutral axis?
The neutral axis is the line (or plane) within a beam's cross-section where no longitudinal stress or strain occurs under a pure bending moment. It is perpendicular to the plane of bending.How
163
What is the Euler-Bernoulli hypothesis?
Plane cross-sections of a beam remain plane and perpendicular to the neutral axis after bending. They rotate rather than shearing
164
For a beam subjected to pure bending, what is the expression for strain in the beam?
ε = strain
y = distance from the neutral axis
R = local radius of curvature
κ = curvature
165
Derive this expression
166
For a beam subjected to pure bending, what does the "plane sections remain plane" assumption imply?
The strain in a fibre is proportional to the curvature and is also proportional to the distance from the neutral axis
167
For a beam subjected to pure bending, what is the expression for stress in the beam?
168
What is the second moment of area of a cross-section?
* It measures how far the area is spread from the axis - the further out the area, the higher the value.
* A higher second moment of area means the beam resists bending better.
* It's purely geometric - it doesn’t depend on material
169
What is the general formula for the bending of linear-elastic beams?
170
For a beam subjected to pure bending, what is the slope of the strain diagram (for distance from the neutral axis)?
Curvature, κ
171
How do you calculate the second moment of area (I) for a given cross-section?
We assume that there is one axis of mirror symmetry, along which we will place our y-axis.
1. Find the centroid and put the origin O of the x,y coordinate system there
2. Find Iₓₓ = ∫ y² dA, with y measured from the centroid. Sometimes the integration can be simplified by use of the parallel axis theorem or standard results can be used.
172
When calculating the second moment of area for a given cross section, how do you determine the position of the centroid?
The centroid is the position at which the first moment of area is zero, ∫ y dA = 0. It's the point at which the area of the shape is evenly distributed in all directions. To find it:
1. First set up some temporary axes, X, Y in a convenient location, and define the position of the centroid (the origin, x,y) relative to these
2. We can find the position of the centroid if we know the area, and the first moment of area, about the temporary axes.
3. The formula for the (y in this scenario) coordinate of the centroid is given by:
**∫ y dA / ∫ dA**
**First moment of area / total area**
173
Once you have determined the position of the centroid, how do you determine the second moment of area for a given cross section?
Once the centroid has been located, the second moment of area about a given axis can be found from I = ∫ y² dA.
Please see the 2 examples below:
174
What are shortcuts that can be used when finding the centroid of a cross section?
* For a doubly-symmetric shape (has 2 axes of symmetry) the centroid will be located at the centre of the object
* For a singly-symmetric shape, define the y axis along the axis of mirror symmetry. The centroid will then lie along this line somewhere.
175
Determine the position of the centroid in this isoceles triangle:
176
Once you have determined the position of the centroid, how do you determine the second moment of area for a given cross section **using parallel axis theorem**?
You can break the shape into smaller simpler shapes and then use the parallel axis theorem to find the second moment of area from the centroid.
177
What is the general stress equation for the stress in a beam subjected to pure bending?
178
Once you have determined the position of the centroid, how do you determine the second moment of area for a given cross section **using standard results for I**?
The structures data book includes the second moment of area for some simple shapes (such as solid circular section, a thin walled circular section, and a solid rectangle). These standard results can be used to calculate the second moment of area for more compelx shapes.
179
What is the second moment of area for an I-beam?
You can simply use the databook result for a rectangle twice
180
When can you simply sum second moments of area to obtain the total second moment of area?
**When they are centred on the same axis**. The second moment of area measures how spread out an area is relative to an axis. For complex shapes each part contributes its own "resistance to bending" about that axis. Because integration is linear, the total second moment of area is the sum of each part’s contribution - provided everything is referred to the same axis.
If they are not originally centred on the same axis. You can simply use the parallel axis theorem so that they are centred on the same axis and then sum them.
This may be useful for more complex shapes as you can break them up into smaller pieces, find the individual second moments of area and then simply sum (or use the parallel axis theorem and sum) them to determine the total second moment of area.
181
What is the elastic section modulus?
182
What is the equation for the maximum bending moment in a beam in terms of the elastic section modulus?
183
When there is both a bending moment and an axial force acting upon a beam, what is the equation for the total stress distribution acting in the beam?
It is simply the result of superposing the stresses due to the axial force and the bending moment
184
What is the middle third rule?
If a vertical load is applied within the middle third of a cross section, then the entire section remains in compression.
185
What is a composite beam?
A composite beam is a beam made from two (or more) different materials.
186
For a composite beam how can you determine the stress within each material?
* We first assume compatible strains across materials, the strain is based purely on geometry not on the material itself: **ε = κy**
* The stresses can then simple be found from hookes law: **σₛ = Eₛκy, σ𝓌 = E𝓌κy**
187
How can you find the bending moment acting on a composite beam?
## Footnote
**IMPORTANT**
Using the "transformed section" method, this is based on the fact that a beam of one material has a bending stiffness about the x-axis equal to the bending stiffness of a beam of another material with equal depth but a different width. Therefore it can be replaced by this equivalent beam. The method is as follows:
* Determine the width of the new equivalent beam by using the expression: **β = b E₁/E₂**, where β is the width of the new equivalent beam, b is the width of the original beam, E₁ is the youngs modulus of the original material, and E₂ is the youngs modulus of the new material. This transformed section now has the same young's modulus everywhere and has a second moment of area about the neutral axis. You can either use this to shrink the material with a lower youngs modulus or lengthen the material with a greater youngs modulus (see below)
* The bending stiffness of the transformed sections are equal to the bending stiffness of the original composite section: **B = E₁I₁ = E₂I₂**.
* Under an applied bending moment the curvatures and strains can be correctly calculated from either of the transformed sections. **ε = κy = My/E₁I₁ = My/E₂I₂**
* **However**, the stresses calculated are only correct for the "untransformed" part of the section. **σ₁ = My/I₁, σ₂ = My/I₂**
## Footnote
Note: This method can also be used if you are bending around the y-axis, but rather than chanding the width you would change the depth.
188
In a beam how are shear forces, axial forces, and bending moments carried?
* Bending moments and axial forces are carried by different distributions of longitudinal stresses that are normal to the cut face
* Shear forces are carried by shear stresses, τ, that are parallel with the cut face. These shear stresses are not uniform across the face
189
How do shear stresses arise for an object in equilibrium in a state of pure shear?
A state of simple shear requires shear stress on four faces of an arbitrary small block. The stresses τ' are complementary to τ, and vice versa
190
What is the principle of complementary shear?
**If τ acts on a plane, at a point in a stressed body, then τ acts also on the orthogonal plane**. It ensures moment equilibrium for a small material element
191
How do you find the intensity of shear stress, τ, at the mid-depth of a cross section?
**τ = Q/bc = 3S/2bd**
* Find the longitudinal shear stress on a horizontal cut at mid-depth, by considering the longitudinal equilibrium of the lower part of the beam
* A longitudinal force was needed for equilibirium as M varied along the beam
* Apply the principle of complementary shear stress.
This demonstrates when S = constant, it can also be applied if S is not constant along a beam by replacing the separation c of the 2 cuts by dx and let dx → 0
192
What is shear flow, q?
**q = Q/c**
Shear flow tells you how shear force is distributed along a line (shear force per unit length). It is contant along a beam as long as both the shear force and wall thickness remain constant.
193
How does shear stress vary through a rectangular beam?
Shear stress varies parabolically, it is at a maximum at the neutral axis (where it is 1.5x the average shear stress) and it is zero at the outer fibres
194
How do you find how the intensity of shear flow, q, varies across a cross section?
195
What is the intensity of shear stress, τ, at the mid-depth of a cross section?
**τ = Q/bc = 3S/2bd**
1.5x the average shear stress
196
How do you find how the intensity of shear stress, τ, varies across a cross section?
* Find the shear flow (equation in data book)
* Shear stress = shear flow / width of beam
Once you plug in the expressions for each term you will get a function τ(y) that describes how shear stress varies continuously across the section (not just at a point).
197
How does the shear flow, q, vary across a cross section?
198
Find the shear stress distribution in a rectangular section
199
What is the equation for the shear stress distribution in a beam?
S = Applied shear force acting on beam (can be found from dM/dx)
A = Cross-sectional area of the "cut out" section
y = Distance of centroid of cut to neutral axis
I = second moment of area of whole section
b = Width of the cut
## Footnote
Ay is tge first moment of area of the "cut out" section
200
What are shear stresses like in a thin-walled beam?
The shear stress at a point of a cross-section of a thin-walled beam must be parallel to the surface of the section. This is because shear stresses in a direction perpendicular to the surface would be accompanied by complementary shear stresses acting on the surface of the beam.
201
How do you know the direction of shear stress?
It is aligned with the shear flow, you can use the principle of complementary shear stress to determine the direction of the perependicular complementary stress
202
Determine the shear stresses in a box beam:
203
How would shear stress vary in an I beam?
204
What is buckling?
Buckling is a sudden and highly destructive non-linear mode of failure in which a structure that is supposed to carry certain given loads in compression, and without much deflection, suddenly undergoes large lateral deflections in an unexpected mode. This behavior is not due to material yield or crushing but rather due to instability.
205
What is a bifurcation load?
A critical load at which a structure can suddenly switch from one equilibrium state to another - typically from a straight, unbuckled configuration to a buckled or deformed shape.
206
What does the fact buckling is non-linear mean?
We must consider the equilbrium of the deformed shape - equilibrium of the undeformed shape tells us nothing.
207
What is an Euler column?
A pin-ended bar in compression. The column buckles when a non-straight equilibrium shape becomes possible.
208
What is the formula for Euler buckling?
For a perfect elastic pin-ended compression member:
209
Derive the expression for the Euler load (and the buckling shape)
210
What is the Euler load?
The lowest critical value of P for a pin-jointed column which will result in buckling
211
What does the plot of P against the central deflection of the bar δ look like for an Euler column?
There are 2 distinct lines:
* Either δ = 0 for all P, whether P > 0 (compressive in our analysis) or P < 0 (tensile). In this case the bar does not buckle, it remains straight.
* At P = Pᴇ = constant, δ can have any value: The axial load remains constant while the buckle grows.
These 2 paths in the P, δ space intersect at a point of bifurcation
212
When you are at a point of bifurcation, what determines whether the beam buckles or remains straight?
If there are any **intial imperfections** (such as slight curvature or asymmetry) in the beam or any slight **external disturbances** (such as lateral forces, vibrations, or thermal effects) the beam will buckle
213
What is a fixed end column?
**An encastre-jointed bar in compression.** It is more complex than an Euler column as there are additional unknowns - fixed-end moments at each end (which by symmetry must be the same). However there are also additional boundary conditions - no rotation at the ends)
214
Derive the expression for the critical load for the first buckling mode of this fixed end column (and the buckling shape):
215
How can you use the "effective length" method to determine the critical load for a fixed end column?
On a buckled fixed end column there are points of inflection, these are points where the curvature is zero. The section of the bar between these points of inflection behaves as though it is a pin-ended column. The length of this section is known as the "effective length" and can be used to calculate the buckling load. Rather than calculating the buckling load of the whole fixed-end column, you can simply calculate the buckling load of an Euler column with a length that of the effective length. This is far simpler as you can simply use the standard result.
216
217
What is the effective length of a fixed-end column?
* Theoretically: Lₑ = 0.5L
* Practically: Lₑ = 0.7L
218
What is the expression for the critical compressive stress at which a perfect strut will fail by Euler buckling, σᴇ?
## Footnote
r = radius of gyration
219
What is the equation for the radius of gyration?
220
What is the slenderness of a column?
Slenderness = L / r
## Footnote
L = **effective** length of column
r = radius of gyration
221
What determines the critical compressive stress, σ at which a perfect strut will fail?
It will either fail by general yielding when σ𝒸ᵣ = σᵧ, or it will fail by bucklign when σ𝒸ᵣ = σᴇ. Whichever is lower will be the manner in which a perfect strut will fail.
222
How is the Euler buckling stress related to the slenderness of the beam?
Inversely proportional to slenderness **squared**
223
What is an imperfect column?
In practise bars are never absolutely straight, they may have an initial imperfection such as an intial deflection.
224
How can you find the shape of an imperfect beam under load and the central deflection?
## Footnote
Assume the imperfection can be represented as an initial bow of: v₀ = δ ₀sin(πx/L)
This is the same as a perfect beam, however Δκ ≠ κ as there is an initial curvature.
225
What is the equation for the shape of an imperfect beam under load and the central deflection for first mode buckling?
## Footnote
Assume the imperfection can be represented as an initial bow of: v₀ = δ ₀sin(πx/L)
It is simply an amplficiation of the intial deflection, by a factor of:
**1 / (1 - [P/Pᴇ])**
226
What does the load-deflection graph look like for a an intially imperfect beam compared to an intially perfect beam?
For an imperfect bar there is no longer a bifurcation point at which the beam suddenly buckled, instead it tends asymptotically towards the buckling load of the perfectly straight bar
227
If an intially imperfect beam doesn't "suddenly buckle", how can we determine the point at which the bar fails?
Using Perry's failure criterion: The average stress in the bar reaches a critical value, when the maximum stress in the bar reaches the yield point of the material.
σₘₐₓ = σᵧ
σₐᵥₑᵣₐ𝓰ₑ = σ𝒸ᵣ
228
According to Perry's failure criterion, what is the equation for the maximum stress in the beam?
## Footnote
r = radius of gyration = √(I/A)
229
Derive this equation for the Perry maximum stress
## Footnote
r = radius of gyration = √(I/A)
230
What is the equation for the deflection in an imperfect beam when yield first occurs?
231
What is the Perry Equation?
Where η is a dimensionless measure of the initial imperfection
232
How is the Perry equation derived?
233
What is η and what is the equation for it?
It is a dimensionless measure of the intial imperfection
234
What does the plot of the critical stress (in an imperfect beam) against slenderness ratio look like and what does it tell us?
There are difference curves depending on the value of η:
* When η is 0, the beam behaved like a traditional perfect beam
* The larger the value of η, the lower the critical stress. Therefore intially curved (imperfect) beams are weaker than their straight (perfect) counterparts
235
Explain how you can use the transformed section method on reinforced concrete:
Concrete is a material that is strong in compression but incredibly weak in tension, steel is equally strong in both tension and compression. Therefore for bending in reinforced concrete, the steel bars carry all the tensile forces and concrete carries all the compressive forces, these compressive and tensile forces must be equal to one another, forming a couple and hence an internal bending moment. To use the transformed section method we must make 2 transformations:
* Since the concrete in the tensile section carries no load, it simply disappears, the transformed material has zero width
* The steel is "transformed" into concrete so that the whole beam has the same Young's modulus. Therefore the steel must be widened by the ratio Es/Ec
Then we must find the position of the neutral axis (position where first moment of area is equal to zero), to do this the sum of the first moment of area of the tensile and compressive sections must be zero [see below].
236
When performing the transformed section method, when must you find a new neutral axis?
* When the original neutral axis doesn't align with the centroid (material distribution is not uniform/symmetrical
* When there has been a non-symmetric transformation in the materials (such as in a reinforced concrete beam when you ignore the tensile concrete)
