structures and maaterials (lecture 2)

7/29/2019 Structures and MaAterials (Lecture 2)

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The University of Nottingham, School of Civil Engineering

Structures and Materials 1 (H21SM1)by B.S. Choo & D.J.Ridley-Ellis

Page No. 1

Coverage

Revisit pin and rigid connections

Look at support conditions

Review equilibrium

Introduce static determinacy

External

Internal

Analysis of pin jointed trusses

Method of joints

Method of sections

(Tension coefficient method)

(Graphical method)

Example problems

Static determinacy


Lecture 2Pin jointed structures


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Page No. 3

Rigid connections

Simple representation

of above rigid joint

x

y

z

x y 0

z 0Mathematical description

of degrees of freedom

The end of the beam, where it is attached to the column

cant rotate, and it cant move upwards, downwards or

sideways.


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Page No. 4

Support conditions for a beam

Pinned supports transfer forces, but not moments

Load

PinPin

The simplest support condition for a beam is to have a

pinned support at each end. In order to avoid extra forcesdue to changes in length, we can to make one support a

roller bearing. this is called a simply supported beam.

Load

PinPin + roller

Plan view


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Page No. 5

Support conditions for a beam

Rigid (or encastr) supports transmit forces and moments

Load

Encastr

A beam with an encastr support at each end is known as a

fixed end beam. A beam supported at one end only (this

must be an encastr support) is known as a cantilever.

Plan view

Fixed end beam Cantilever

Note that

Encastr


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Page No. 6

Equilibrium

Free body diagram

PP

P P

Externally applied

axial force

Free body Free body

Internal forces in equilibrium with external force P

But note that a long, thin (slender) member in compressionmay buckle.


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Page No. 7

Equilibrium

In structures, we normally study bodies at reststatics

Consider a particle subject to two forces

F1

F2

F1

F2

R

The two forces are equivalent to a single force R. If the

same particle is subject to a third force R1 that is equal

and opposite to R then it is in equilibriumat rest.

F1

F2R1


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Page No. 8

Equilibrium

In a 3-dimensional world there are 6 equations of equilibrium

x

y

z

Fy

Fx

Fz

Mz

My

Mx

Resolve forces in the directions of the three Cartesian axes

Resolve moments about the three Cartesian axes

MMRather than more usual to draw


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Page No. 9

Equilibrium

x

y

Fy

Fx

In a 2-dimensional, or plane system, world there are 3

equations of equilibrium.

z

Mz

Fx 0

Fy 0

Mz 0

Load

PinPin + roller

RA,Horiz

RA,Vert RB,Vert

If there are 3 unknown reactions, we have enough equations

to calculate them using equilibrium.

A B


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Page No. 11

Static determinacy for reactions

4 unknown reactions

Externally statically indeterminate

3 unknown reactions

Externally statically determinate

5 unknown reactions


3 unknown reactions


25 kN


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Page No. 13

Static determinacy for reactions

3 unknown reactions


3 unknown reactions


3 unknown reactions



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Page No. 14

Pin jointed trusses

Made from members connected together with pin joints.

Based on triangles.

If forces are applied at the joints (nodes) the members are

under axial forces only (if they are straight).

Examples:

A Warren truss

A Howe truss

A Pratt truss


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Page No. 15

Pin jointed trusses

How many unknowns?

4 reaction forces

3 member forces

Externally staticallyindeterminate

but if we knew the reactions we could calculate themember forces from equilibrium at the nodes.

The structure is internally statically determinate.

The triangular shape allows pinjointed trusses to take the load.

This structure would fall over

under the applied force. Such a

structure is known as a

mechanism.


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Page No. 16

Internal static determinacy

In a 2D truss with M members, there are M unknown

member forces.

If there are J connections, we can resolve forces

horizontally and vertically at each (no moments) giving 2J

equations.

However, 3 equations are required to satisfy the conditions

of global equilibrium.

Fx 0 Fy 0 Mz 0

So if M = 2J - 3 we can find the member forces using

equilibriumthe structure is internally statically

determinate.

In a 3D truss (a space truss), there are 3 equations per

joint and 6 equations for global equilibrium.

So if M= 3J - 6 the structure is internally staticallydeterminate


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Page No. 17

Internal static determinacy

M = 3, J = 3 3 = 63

Internally statically determinate

M = 3, J = 3 3 = 63



M = 15, J = 9

15 = 18 - 3

M = 27, J = 14

27 > 28 - 3Internally statically indeterminate



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Page No. 18


Calculation of reactions at supports using equilibrium as for

beams shown previously.

4 main methods for finding member forces using equilibrium:

Method of joints (Megson 4.6)

Resolving horizontal and vertical forces at each node

Finds all member forces one by one

Method of sections (Megson 4.7)

Cutting the structure and resolving horizontally, vertically and

for moments

Useful for finding a member force in the middle of a structure

without having to calculate lots of other member forces

Tension coefficient method (Megson 4.8)

Alternative form of method of jointsuseful for 3D trusses

Graphical method (Megson 4.9)

Pen and paper vector methodfor those without a calculator

This lecture will explain the method of joints and the

method of sections.

Assumptions:

Deflections are small no changes to geometry

Members are able to resist the forces


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Page No. 19

R2

R1

Method of joints

F1 F2

F3

R2R1 R3

F3

Fm1

m1

m2

m3

m4

Fm2

Fm1 Fm3

Fm4

Step 1: Resolve globally to find reactions (R1, R2 & R3).

Fx 0 Fy 0 Mz 0

Step 2: Pick a support node* and

resolve locally to find member

forces at node (Fm1 & Fm2).

Fx 0 Fy 0

Step 3: Pick an adjacent node and

resolve locally to find remaining

member forces at node (Fm3 &Fm4). Repeat step 3.

Fx 0 Fy 0

* Or any node where 2 unknown forces meet


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Page No. 21

Method of joints example

20kN

Fm1

Fm2

46.7kNStep 2: Pick a support node and

resolve locally to find member

forces at node.

Fx 0

Fm2 + Fm1 cos45 20 0 kN

45o

-66.0 kN

Fm3

Fm420 kN

45o

Fy 0

Fm1 sin45 + 46.7 0 kN Fm1 -66.0 kN i.e. compression

Fm2 66.7 kN i.e. tension

Step 3: Pick an adjacent node and

resolve locally to find remaining

member forces at node.

Fx 0 Fm4 + 20 -66 sin45 Fm4 -66.7 kN (comp)

Fy 0 Fm3 66 cos45 0 Fm3 46.7 kN (tens)


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Page No. 23

Method of sections

F1 F2

F3

R2R1 R3

m1

m2

m3


Fx 0 Fy 0 Mz 0

Step 2: Cut the structure at point of interest and resolve

globally to find unknown released member forces (Fm1,

Fm2 & Fm3). Fx 0 Fy 0 Mz 0

F1 F2R3

m1

m2

m3

Fm1

Fm2

Fm3


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Page No. 24

Method of sections example

100kN

R2R1 R3

Find the force in members m4, m5and m6


Fx 0

R2 -20kN (ie 20kN to the left)

Fy 0

R1 + R3 100 0kN

Mz 0 (about roller support)

R1

6 + 20 1 + R2

0 100 3 0 kNm

6R1 + 20 300 kNm

R1 46.7 kN and R3 53.3 kN

1m45o

20kN

m1

m2

m3

m4

m5

m6


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Page No. 26

True or false?

a) This structure is externally statically indeterminate

b) This structure is externally statically indeterminate

c) This structure is externally statically indeterminate


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Page No. 28

True or false?

g) Member A is in compression

A

h) Member A is in compression

i) Member A is in compression

A

A


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Page No. 29

True or false answersa) False. 3 external reactions and 3 equations of

equilibrium so structure is externally statically

determinate.

b) False. 3 external reactions and 3 equations of

equilibrium so structure is externally statically

determinate.c) True. 4 external reactions and 3 equations of equilibrium

so structure is externally statically indeterminate.

d) True. 3 external reactions and 3 equations of equilibrium

so structure is externally statically determinate.

e) True. The roller support does not prevent the beam

rotating about the other support.f) False. 9 members and 6 joints. 9 2 6 3 so

internally statically determinate. Note there are 4

external reactions and 3 equations of equilibrium so

structure is externally statically indeterminate.

g) True. The beam is bending downwards (sagging) so the

top chord will be in compression.

h) False. The beam is bending downwards (sagging) so the

bottom chord will be in tension.

i) False. This member cannot possibly be under load.

Consider equilibrium at the left-hand node.


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Page No. 30

Final puzzle

How many unknowns?

How many reaction forces?

How many member forces?

Would you be able to find them using equilibrium?

Is the structure externally statically determinate?

Is the structure internally statically determinate?

1 m

3 m

3 m

2 kN


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