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1360 Structural Design and Analysis > Semester 1 > Flashcards

Semester 1 Flashcards

1
Q

Variable Actions

Environmental actions

Hydrostatic actions

A

variable actions are Live loads associated with use of the building eg people and furniture

whereas environmental actions are accidental actions eg impacts or explosions

hydrostatic actions are pressure exerted by a fluid (or soil) at equilibrium

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2
Q

How does lateral pressure change in relation to depth?

A

LP increases in proportion to depth measured from surface due to the increasing weight of the retained material

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3
Q

What is a FBD?

A

A free body diagram, used to simplify and help solve a statics problem.

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4
Q

How to draw a FBD (5)

A

ground

add known forces

show orientation of principle coordinate system

add resistance forces

draw dimensions of structure

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5
Q

Types of load

A

Unit load

Uniformally distributed load (weight central)

Partially Distributed Load

Varying partially distributed load

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6
Q

how to calculate VPDL

A

Rectangle with triangle on top. With w1 being smaller side height and w2 being taller side height.

W1 = w1(X2 - X1)
W2 = 0.5(W2-W1)(X2-X1)
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7
Q

External Supports and how many resistance forces

A 
Rollers (1 VR)
Pins (2R, V+H)
Encastre (3R, V+H+M)
Link 
Beam to column joint 
Connections in trusses
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8
Q

Link (external support)

A

Allows rotations and translation perpendicular to the direction of the link.

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9
Q

Internal Pins

A

Internal pin - only allows rotation

Internal roller - same as pin but situations between two beams

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10
Q

three types of stress in beams

A

Tensional - beam being pulled apart

Compressional - beam being pushed together

Shear stress - beam being pushed past each other, eg LHS pushed down RHS pushed upwards

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11
Q

Bending moment def.

A

The reaction induced in a structural element when an external force or moment is applied, causing it to bend.

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12
Q

Two ways of beam failure under loading.

A

Shearing the beam across its cross section

Bending by an excessive amount (causing tension in bottom of beam and compression on top)

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13
Q

Bending moments equation

A

MC = total ( external forces ) x Lever arm

MC = Va x L/2, Va being vertical reaction

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14
Q

How to calculate applied bending moment at any point of a beam

A
  1. Cut the beam at point of interest
  2. Bin, ignore everything to either the left or right of cut.
  3. One end in, Working from one end in sum moments about the cut
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15
Q

How to find shearing force

A

Cut at C
consider left or right

(total)V= 0 ( Vertical forces are in equilibrium)

e.g Va - (2 x P1) = 0
Va = 2P1

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16
Q

Sagging or hogging?

A

Sagging- Compression top, Tension bottom

Hogging - Tension top, Compression bottom

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17
Q

Shear force diagrams

A

Calculate shear force throughout beam and plot on graph, downwards forces causing sagging are positive.

start just after first force, using V = 0, then that force is constant until next applied load.

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18
Q

Bending moments diagram

A

Multiply shear force by distance to find bending moments

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19
Q

Bending moments known shapes

A

Concentrated Point load - straight lines w/ changes of direction at point of application of loads

Uniformly Distibuted Loads - Parabolic curves

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20
Q

Statically determinate structure

A

If a structure has less than or equal to 3 unknown reactions it can be analysed using equations of equilibrium and is externally statically determinate

if a structure is externally redundant/ externally statically indeterminate, we can’t yet analyse these structures

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21
Q

what does m + r = 2j tell us in pin jointed frames

A
  • minimum number of constraints, yet doesn’t show how to configure them
  • necessary condition for a statically determinate structure but it’s not sufficient
  • there’s a need to check the reactions restrain the rigid body freedoms and that sections of the structure and not mechanisms
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22
Q

Types of trusses (6)

A 
pratt 
howe 
warren 
parker 
king post (simplest)
queen post truss
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23
Q

What do trusses do?

A

increase load carrying capacity while reducing material consumption. has a light open appearance and many shapes can be made from it.

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24
Q

assumptions w/ trusses (5)

A

Members are connected at nodes (ends) only

Connected by frictionless pins (don’t resist moments)

Pin jointed truss structure loaded only at nodes

The weight of the member may be neglected

All bars are two forces members, weight of members are neglected and members work in either tension or compression

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25
Q

what m + r >/= 2j tells us

A

m + r < 2j, statically unstable

m + r > 2j, statically indeterminate but stable

m + r = 2j, statically determinate and stable

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26
Q

what do the constants mean in m + r = 2j

A 
m = members (lines)
r = support reactions 
j = nodes (corners)
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27
Q

Method by section steps (4)

A
  1. check all equilibrium equations
  2. Cut the truss through members where forces are to be found.
  3. Draw the FBD of that part of the truss which has fewest unknowns
  4. write equilibrium equation to solve for unknown forces
28
Q

Method of joints steps (5)

A
  1. Draw FBD of whole pin jointed frame
  2. Apply equations of equilibrium
  3. Determine the reactions
  4. Draw FBD of EACH JOINT in turn
  5. Determine the unknown forces
29
Q

St Venant’s Principle

A

if self equilibriated loads are applied on one end of a long cylinder, the strain produced in the body are much larger near the loaded end than at points further away.

30
Q

Assumptions made when developing the expression for direct stress

A
  1. Material has continuous and solid structure
  2. Material is homogeneous and isotropic
  3. There are no internal forces prior to loading
  4. The effect of a system of forces acting on the body is equal to the sum of the effects of those forces applied in succession and in any order.
  5. St Venants Principle applies
31
Q

Tensile strain equation

A

change in L/L

32
Q

Difference between brittle and ductile materials

A

Brittle materials like glass and concrete fail without yielding

Ductile materials like plastic or steel fail after yielding

33
Q

Strain and stress sign conventions

A

tensile stresses and strains (positive) produce increase in length

compressive stresses and strains (negative) produce decrease in length

34
Q

Poissons ratio

A

Elongation in the x axis of an object regulars in a reduction in y axis

35
Q

Stress equation

A

applied force / area of cross section

36
Q

centroid of section algebraic equation

A

y(bar) = [(int)ydA]/A

same for x bar replaying y with x

37
Q

Centroid of standard rectangle

A

Middle
x(bar)= b/2
y(bar) = h/2

38
Q

centroid of a Standard iscocoles triangle

A 
x(bar) = b/2
y(bar)= h/3
39
Q

centroid of standard scalene triangle w no lines of symmetry

A 
x(bar) = (x coord of point) + b/3
y(bar) = h/3
40
Q

centroid of standard circle

A

x(bar) = y(bar)= d/2

41
Q

Second moment of area for rectangular section

A

I = (bh^3)/12

42
Q

Second moment of area for a standard circle

A

I(x or y) = (pi.r^4)/4

43
Q

standard second moment of area for a square

A

I (x or y) = (a^4)/12

44
Q

Standard second moment of area for a hollow circle

A

I = (pi(d^4-d1^4))/64

45
Q

radius of curvature 3 main equations

A

Length of AB before bending
= R x thetre

Length of AB after bending
= (R+y)thetre

Change in length
= y x thetre

R being distance between the neutral axis and the where the extended lines cross over

y being distance between bottom of beam and neutral axis

46
Q

stress equations

A

weird 0 with flick is the symbol

= E x e (e bring strain) (E being modulus of elasticity for the material)

= F/A

47
Q

Tension and Compression equation

A

since T = C

T = C = ((stress max)bd)/4

48
Q

moment of resistance equation for rectangular section

A

M = (stress max x b x d^2)/6

49
Q

Moment of resistance definition and it’s unit

A

the internal moment provides the beams resistance to the external bending moment resulting from the external load

measured in KN/m

50
Q

engineers theory of bending equation? idk

A

M/I = stress/y

M being moment of resistance
I being second moment of area about Na
stress being in any fibre at distance y from NA

51
Q

what 3 characteristics can be altered to increase resistance to shear and bending

A

shape
size
material

52
Q

why do we find the centroid and what are the symbols

A

it’s the centre of mass/gravity, the neutral axis passes through it

53
Q

how to find centroid of section for t and i section

A

split section into multiple sections
use a table to input the formula

y/x(bar) = (a1x1 + a2x2 + a3x3)/(a1+a2+a3) use y instead of x for y bar

54
Q

neutral axis equations

A

Ixx = (integral of) y^2 da

Iyy = (integral of) x^2 da

55
Q

which second moment of area/ inertia is the neutral axis for a section with two axis of symmetry

A

if the applied bending moments is in the vertical axis Ina = Ixx

if the applied bending moment is in the horizontal axis Ina = Iyy

56
Q

which second moment of area is on the section with only one axis of symmetry

A

it’s perpendicular to the axis of symmetry and runs through the centroid

57
Q

when do i use parallel axis theorem?

A

when it isn’t a standard shape

58
Q

what is parallel axis theorem

A

when there’s a vertical line of symmetry

ixx = icc + Ah^2

icc being the second moment of area for the part, ixx being the second moment of area for the whole

59
Q

what is the elastic section modulus

A

direct measure of strength in the beam

60
Q

elastic section modulus equation for any non rectangular section

A

z = I/ymax

i being inertia (second moment of area)
ymax being distance from NA to top/bottom

61
Q

elastic section modulus for rectangular section equation

A

Zyy = (bd^2)/6

62
Q

Moment of resistance for non rectangular equation

A

M/I = stress/y

63
Q

Moments of resistance equation with elastic section of modulus subbed in

A

M = stress max x Z

64
Q

Moment of resistance equation for rectangular section

A

M = (stress max x b x d^2)/6

65
Q

maximum value for bending moments for simply supported beam with distributed load

A

(wl^2)/8

w being force downwards
l being length of beam

1360 Structural Design and Analysis (1 decks)

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