Structural Mechanics Flashcards

(55 cards)

1
Q

Structural redundancy/ Indeterminancy number (α) ?

A

The number of extra unknowns in relation to a statically determinate system

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2
Q

Redundancy formula?

A

External (de) + internal (ai) redundancy

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3
Q

How do hinges affect redundancy?

A

-1 redundancy for each internal hinge because it adds an equation

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4
Q

What redundancy do local mechanisms have?

A

Local mechanisms can appear >_ 0 total redundancy

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5
Q

How do cells affect redundancy?

A

+3* number of cells to redundancy

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6
Q

How do bars in trusses that connect to existing nodes effect redundancy?

A

In a truss add redundancy for each extra bar connecting two existing nodes in a truss minus the number of internal hinges

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7
Q

d1(U)=Δd(W)

A

Virtual internal forces* real deformations = virtual external forces* real displacements

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8
Q

How to work out displacements for trusses?

A

Sum of all virtual axis forces (N) x Real strains (ε) = 1* real displacement

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9
Q

How to work out displacements for beams and frames?

A

Sum of all virtual bending moments (M) * Real curvatures(k) = 1 x real displacement

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10
Q

Structure (SIS) = +*_

A

Structure (SIS) = Case 0 (SDS) + X1 (unknown) * Case 1 (SDS)

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11
Q

Flexibility method of statically indeterminate structures ( removing supports)?

A
  • Obtain the redundancy
  • Identify α supports that can be removed maintaining equilibrium
  • Replace the supports by the unknown reactions
  • Decompound the initial structure in α+1 load cases. Apply the external loads in Case 0, and the reactions equal to 1 times the unknowns for the other α cases
  • Obtain the vertical displacements at the supports that have been removed (using the unit load method, with unit loads equal to the load cases 1 to α, and applying superposition)
  • Impose the compatibility equation displacement = 0 at each removed support and obtain the unkowns
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12
Q

Flexibility method for statically indeterminate structures (adding hinges)?

A
  • Obtain the redundancy
  • Identify α hinges that can be added maintaining equilibrium
  • Replace each continuous section by a hinge and the corresponding unknown bending moment (couple of moments)
  • Decompound the initial structure in α+1 load cases. Apply the external loads in case 0, an a couple of unityoments times the unknowns for the other α cases
  • Obtain the angular dislocations at the hinges (using the unit load method, with unit loads equal to the load cases 1 to α, and applying superposition)
  • Impose the compatibility equation dislocation=0 at each added hinge , and obtain unkowns
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13
Q

F1= Σ ((N0 L)/(EA) + ΔL0) N1

A

N0= axial forces in case 0 (real)
EA= youngs modulus*area
L= length of beam
ΔL = change in length
N1= axial force in case 1
F1= unknown force

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14
Q

F2= Σ ((N1)^2* L)/EA

A

E= Young’s modulus
A= area
N1= Axial force of case 1
L= length of beam

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15
Q

F1 + F2 * X1 = 0

A

X1= unknown force

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16
Q

What is the product integral of two rectangles?

A

Lab

Length* height of one rectangle * height of second rectangle

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17
Q

What is the product integral of a triangle and rectangle?

A

1/2 * L * a * b

L= length
a= height of triangle
b= height of rectangle

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18
Q

What is the product integral of two triangles?

A

1/3 * L * a * b

L= length
a= height of triangle one
b= height of triangle two

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19
Q

What is the product integral of two triangles in opposite directions?

A

1/6 L a b

Length * height of one triangle * height of two triangle

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20
Q

What is the product integral of a not right angled triangle and a right angle triangle ?

A

1/4 Lab

L= length
a= height of one triangle
b= height of second triangle

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21
Q

What is the product integral of a rectangle and semi circle?

A

2/3 Lab

Length* height of rectangle * radius

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22
Q

Number of independent displacements for a system of axial loaded bars?

A

β = 2j - R

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23
Q

Number of independent displacements for rigid jointed frames?

24
Q

How do pins effect the number of independent displacements β

25
What do you do for inextensible directions?
Us acting in an inextensibke direction = 0 which means it results to one U. Vertical direction Us = 0 for u frames
26
Stiffness?
The resistance to deformation p/u Variation of the force/ the increment in displacement
27
σn = Fx / A
Normal stress uniform across the cross sectional area = the axial force along the x axis / cross sectional area
28
σn = My*z / Iyy
σn = normal stress at depth z in cross sectional area My= moment acting about the y axis Z= distance from the y axis Iyy = the second moment of area of the cross section about the y axis (Same can be done for z instead of y)
29
σxx = σn = Fx/A + Myz/Iyy + Mzy/Izz
The distribution of normal stress in a beam cross section due to axial force and bending moments about the x and y axis
30
I = db^3/ 12
I = second order of momentum d = depth parallel to direction b = depth perpendicular to direction
31
32
τ(s) = Vz Qy(s) /Iyy t(s)
τ (s) = Vz= shear force acting the vertical direction along the z axis Qy = first moment area about the y axis t = thickness of the section where the shear stress is being calculated (y axis) τ = shear stress at point s below the top of the section
33
τ = Tx r / J For thin walled closed circular section
τ = shear stress Tx = moment acting about the x axis J= torsion constant r = radius
34
J = 2π r^3 t
J = torsion constant r = radius t= constant wall thickness
35
What is normal stress due to?
Axial force Bending moments
36
What are shear stresses due to
Shear forces Torsion moments
37
DF = Mx / M
Distribution factor = element moment / total moment at joint
38
M = K θ Κ = 4E/L1 + 4E/L2
Moment = K * angle of rotation at joint K= value for element before and element after
39
Mx = 4EI/L * θ
Moment in one element = 4* stiffness * angle of rotation
40
Σ DFs
Sum of distribution factors always equals 0
41
Restraint for moment distribution method for pinned end?
Mx=0
42
Restraint for moment distribution method for fixed end?
Θx = 0
43
K for a pinned joint element?
3*EI/ L
44
What’s the carry over coefficient?
Mxy/ Myx
45
How to do the method distribution?
- find distribution factors - find fixed end moments - find the balance at pin joints - distribute the balance as balance* DF to each moment - then distribute the carry over which is balances diagonal on the table x carry over (0.5) - then sum over the carry overs and balances to = member end moment
46
47
Minor bending axis?
Associated with the lowest value of second moment of area and is the acid which it is easiest to bend the section
48
Major bending axis?
Associated with highest value of second moment of area and is the axis about which it is most difficult to bend the section
49
Biaxial bending?
Simultaneous bending about the major and minor axes of a cross section
50
Asymmetric bending?
Bending about axes that are not the principal bending axes
51
Mel,y = Wel,y Fy = Iyy* fy/z
Mel,y = the moment My at which yielding will begin to take place in the top and bottom fibres of the cross section Wel,y = the elastic section modulus about the y axis Fy= yield stress Iyy= second moment of area for y z= distance from y axis to extreme top and bottom fibres if the cross section
52
Iyz = - Σ yzdA
Product second moment of area Iyz
53
σxx = [MyIzz+MzIyz/ IyyIzz-Iyz^2] z + [MzIyy+MyIyz/IyyIzz -Iyz^2] y
Generalised bending formula (works for asymmetric bending aswell)
54
IGyy = ΣIAyy + ΣAz^2
IGyy = second moment of area calculated at the centroid of the entire section IAyy = second moment of area calculated indicating a value calculated for one of the rectangular areas of the cross section Z= distance in z direction going from the centroid of the entire section to the centroid of the rectangular areas (Can be done y and z switched)
55
IGyz = ΣIAyz - ΣAyz
IGyz = second moment of area calculated at the centroid of the entire section IAyz = second moment of area calculated for one of the rectangular areas of the cross section A= Area y= distance in the y going from the centroid of the entire section to the centroid of the rectangular areas z= distance in the z going from the centroid of the entire section to the centroid of the rectangular areas z