Structures and Materials- Section 1 Statics

Section 1

Statics

© Loughborough University 2010. This work is licensed under a Creative Commons Attribution 2.0 Licence.

An introduction to the module is given, including forces, moments, and the important concepts of free-body diagrams and static equilibrium. These concepts will then be used to solve static framework (truss) problems using two methods: the method of joints and the method of sections.

Contents• Statics• Idealisation of Structural System• Forces• Moments• Newton’s Laws• System of Units• Sign Convention• Supports or Boundary Conditions• Free Body Diagram• Force and Moment Equilibrium• Statically Determinate Structures• Degree of Determinacy• Section 1 Statics- Plane Truss• Structural Analysis of Plane Truss• Determinacy of Truss• Finding Reactions• Method of Joints• Solving Plane Truss• Section 1 Statics - Computer Methods• Computer Methods – Method of Joints• Computer Methods – Solution• Section 1 Statics- Method of Sections• Method of Sections• Worked Example - Methods of Sections• Solution• Credits & Notices

Statics

• Statics is concerned with the equilibrium of bodies under the action of forces.

• Equilibrium describes a state of balance between all forces acting on a body.

• Hence if a body is constrained to be static then the sum of the forces acting on it sum to zero. In this section on Statics we will deal with the analysis of structures that are in static equilibrium.

• If there is no equilibrium then we will have motion of the structure

Idealisation of Structural System

• Rigid bodies:a) Structures made of strong materialsb) Relatively low level of external loadc) Negligible amount of deformationd) Neglecting deformation won’t affect analysis

• Characteristics of the rigid bodies:a) Transmit only axial forcesb) Size of contact area immaterialc) All axial forces regarded as point forces

Forces

What is a Force?A force is a measure of the action of one body on another (push or pull). Force has magnitude, direction and a point of application (It is usually represented as a vector).

Forces also have a line of action. For a rigid body, a force can be applied anywhere along a given line of action for the same effect.

ForcesResolving Forces:

Because a force has direction it will have components that act in three directions for a 3D system. (Two components for 2D system).

z

y

x

F

x

z

y

Force components:

2 2 2

direction: cos

direction: cos

direction: cos

x x

y y

z z

x y z

x F F

y F F

z F F

F F F F

External forces: Force that is applied to a structure externally. Internal Forces: Forces transmitted internally to structure.Reaction forces: External forces that exist to maintain equilibrium.

Moments

What is a moment?

A moment is a force acting about (around) an axis.

z

y

x

Fr

d

M

Example:Force F is parallel to x axis in the x-y plane. ‘d’ is the perpendicular distance between line of action of force and the z axis about which it is acting. Magnitude of moment (M) about z axis isM = F x d

Notes:Force arrow has one arrow head.Moment arrow has two arrow heads.

Because we will generally deal with 2D systems e.g. x-y plane then all moments will be about z axis

Newton’s Laws of Motion

Law 1 (Equilibrium)

A body remains at rest or continues to move in a straight line with a constant speed if there is no unbalanced force acting on it.

Law 2

The acceleration of a body is proportional to the resultant force acting on it and is in the direction of this force.

Law 3 (Reactions)

The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and act along the same line of action (collinear).

System of Units

Base units: length, mass and time

Length Mass Time

SI Units Meter (m) Kilogram (kg) Second (s)

Imperial Units Foot (ft) Slug (slug) Second (s)

Force: Newton (N)

1 N = 1 kg . 1 m s-2

1 Newton is force required to give a 1 kg mass acceleration of 1m s-2

1 Pound is force required to give a 1 slug mass acceleration of 1ft s-2

Sign Convention

For a given problem define an axis system e.g. x-y plane

For known external forces the direction of force arrow defines positive direction. For unknown external forces the force arrows should be drawn in the direction of the axis system.

For internal forces tension is defined as positive and compression as negative.

SIGN CONVENTION – CONSISTENCY IS ESSENTIAL

x

y

Supports or Boundary Conditions

A static structure is prevented from displacing or rotating by external forces acting at certain locations. The external forces at these locations are known as boundary conditions.

Three common supports and associated reactions:

Rx

Ry

Ry

Built in or fixed support

Pin or hinged support

Roller support

x

yRx

Ry

Mz

Free Body Diagram

In order to determine the forces that are acting either internally or externally to a structure it is necessary to isolate either part of or the whole structure. This isolation is achieved by using the concept of

Free Body Diagrams (FBD).

The FBD shows an isolated part (or whole) of the structure with all the forces (directions and locations) that act on that isolated part. This isolated part is the free body.

In static problems all structures are in static equilibrium. If the whole structure is in equilibrium then any part of that structure must also be in equilibrium. We can use Newton’s1st Law to determine any unknown forces.

The free body diagram is the single most important step in the solution of problems in statics

Force and Moment EquilibriumThe condition of force equilibrium requires the sum of all forces in any direction, e.g. x, y or z axis, to be zero as expressed by:

0 0 0X y zM M M

0 0 0X y zF F F The condition of moment equilibrium requires the sum of all moments with respect to any axis e.g. x, y or z axis, to be zero as expressed by:

In a 3-dimensional system we have 6 equations of equilibrium and can therefore determine 6 unknowns. In a 2-dimensional system, e.g. x-y plane there are no forces in z direction and no moments about x or y axes as these all act outside of the x-y plane. We therefore have only 3 equations of equilibrium and can determine 3 unknowns only.

0 0 0X y zF F M

Statically Determinate Structures

If we use the principles of statics we will obtain a number of equations of equilibrium. If the number of equations equals the number of unknowns then the structure is said to be statically determinate (just stiff).

If the number of equations obtained is less than the number of unknowns then the structure is said to be statically indeterminate(over stiff)

If the number of equations obtained is more than the number of unknowns then the structure is said to be mechanism (under stiff) and this structure is likely to collapse under loading.

Degree of determinacyFor a simple (pin jointed) truss there is a relationship between the number of its members m, number of joints j and number of reaction forces r. For a 3D truss we can write three equilibrium equations for each joint namely

0 0 0X y zF F F Hence there are 3j equations to determine all the unknown internal forces of which there are m and the number of unknown reactions of which there are r.

3 just stiff

3 over stiff

3 under stiff

m r j

m r j

m r j

For a 2D system there are 2j equations so,

2 just stiff

2 over stiff

2 under stiff

m r j

m r j

m r j

Example 1

A loaded truss, as shown in the figure, is in equilibrium. The known loads areP1 (horizontal at joint B) and P2 (vertical at joint C). Obtain the following:

i. Check the determinacy of the truss. ii. Moment equilibrium equation with respect to joint Aiii. Determine the support reactions iv. Force equilibrium equations for joint B

P2

A

B

C D

P1

l

(i) Check Determinacy of the Truss

A

B

C D

P1

P2

AX AY

DY

Number of members 5

Number of reactions ?

Number of joints j 4

Support at Joint A is a pin support.

two reactions R and R

Support at Joint D is a roller support.

one reaction R

Hence 3

2

m

r

r

m r

5 3 2.4

truss is statically determinate

j

RAX

RAY RDY

x

y

l

(ii) Moment Equilibrium Equation with Respect to Joint A

1 2

1 2 1 2

For moment equilibrium about point A

(axis normal to x-y plane at point A)

0

2 tan tan 0

(Rearrange to give unknown value

tan)

2 tan 2 tan 2

AZ

AZ DY

DY

DY

M

M R P P

R

P P P PR

l l l

l + l

l

A

B

C D

P1

P2

RAX

RAY RDY

x

y

l

Treat 2d truss as a free body.

(iii) Determine Support Reactions

1 1

2

1 2

1 22 2

2 1

For equilibrium of the free body:

0 0 0

0

0

0

we know 2 tan 2

2 tan 2

2 2 tan

X y z

X AX AX

y AY DY

AZ

DY

AY DY

AY

F F M

F P R R P

F R R P

M

P PR

P PR P R P

P PR

A

B

C D

P1

P2

RAX

RAY RDY

x

y

l

Notes: We could have taken moments about any point e.g. point B or C or D

Treat 2d truss as a free body. There are 3 reactions so need 3 eqns. to find 3 unknowns.

(iv) Force Equilibrium Equations for Joint B

1

For equilibrium of the joint B:

0 0

sin sin 0

cos cos cos 0

cos cos cos 0

X y

X BD BA

y BA BC BD

y BA BC BD

F F

F P P P

F P P P

OR

F P P P

Free body diagram for joint B

BP1

PBDPBCPBA

FBD shows ALL forces acting on free body. Here 3 members are connected to joint B.(These members have an internal force)And an external force P1.

x

y

A

B

C D

P1

P2

RAX

RAYRDY

l

Section 1

Statics – Plane Truss

Structural Analysis of Plane Truss

A planar space truss consisting of a number of pin-jointed straight members is subjected to two external loads as shown in the figure. The members are rigid and carry only axial internal forces.

Determine the internal forces in all members. (Note: angle BAC=)

A

B

C E

D

F

P2

P1

Structural Analysis of Plane Truss

We introduce two methods to solve this problem:

1) Method of joints2) Methods of sections

1. Methods of joints

This method consists of:

– Checking determinacy of truss

– Finding reactions at supports by considering the equilibrium condition of the whole truss

– Drawing a free body diagram for each joint

– Writing equilibrium equations for each joint

Determinacy of TrussSolution for Method of joints

Determinacy of the truss:

Reactions: One reaction RFX from roller support at point FTwo reactions RAY and RAX from pin support at point A

Total of three reactions r = 3 Number of members, m = 9Number of joints j = 6

Using “m+r=2j” to check determinacy 9+3=2x6 i.e. truss is statically determinate.

A

B

C E

D

F

P2

P1

RAX

RAY RFY

A

B

C E

D

F

P2

P1

RAX

RAY RFY

x

y

Finding Reactions

1 1

2

1 2

Horizontal force equilibrium of the whole truss

0 0

Vertical force equilibrium of the whole truss

0 0

Moment equilibrium of the whole truss

0

. .

x AX AX

Y AY FY

AZ

FY AC CE EF BC

P R P R P

P R R P

M

R P P

l l l .l

1 2

2

1 2 2 1 22

2 1

0

.

Using 0

. .

CE AC

BC CE ACFY

AC CE EF

AY FY

BC CE AC AC CE EF BC CE ACAY

AC CE EF AC CE EF

EF BCAY

AC CE EF

P PR

R R P

P P P P PR P

P PR

l l

.l l l

l l l

.l l l l l l .l l l

l l l l l l

.l .l

l l l

A

B

C E

D

F

P2

P1

RAX

RAY RFY

A

B

C E

D

F

P2

P1

RAX

RAY RFY

Method of JointsConsider each joint in turn:Lets start with joint A by drawing a free body diagram of the joint:

A

B

C E

D

F

P2

P1

RAX

RAY RFY

A

B

C E

D

F

P2

P1

RAX

RAY RFY

A

RAX

RAY

PAB

PAC

Free body for Joint A shows that there are two reaction forces acting at the joint and two internal forces acting.In this method we can only write 2 equilibrium equations not 3. The reason is that in the method of joints all forces pass through the same point for the joint under consideration.

Joint A

A

RAX

RAY

PAB

PAC

Vertical force equilibrium for free body

0

sin 0 sin

( is a known)sin

Horizontal force equilibrium for free body

0

cos 0

cos

We have found two

Y

BCAY AB

AB

AYAB AY

x

AX AB AC

AC AX AB

P

R P

RP R

P

R P P

P R P

l

l

internal forces - and -

by considering equilibirum of thejoint AAB ACP P

x

y

Joint B

1


0

sin sin 0

sin sin ( and unknown)


0

cos cos 0

cos cos

Y

BA BC BE

BC BE BA BC BE

x

BD BE BA

BD BE BA

P

P P P

P P P P P

P

P P P P

P P P P

1 ( and unknown)

We have three unknown forces.

We need further information to solve.

We need to consider equilibrium of other joints.

BD BEP P

BP1

PBAPBE

PBD

PBC

Note that force PAB=PBA due to Newton’s 3rd Law.

Note that angle DBE = .

Joint C


0

0


0

0 ( is known)

Notice how easy it was to establish the force

This was because there

Y

CB

x

CE CA CE CA CA

CB

P

P

P

P P P P P

P

was only one force acting vertically

Hence Newtons 1st Law says that this foce must be zero

CPCE

PCB

PCA

Joint D


0

sin 0 ( and unknown)


0

cos 0 ( and unknown)

Y

DF DE DF DE

x

DF DB DF DB

P

P P P P

P

P P P P

D

PDF

PDB

PDE

Joint E

2

2


0

sin 0

sin ( and unknown)


0

cos 0 ( , and unknown)

Y

ED EB

ED EB ED EB

x

EF EC EB EB EC EF

P

P P P

P P P P P

P

P P P P P P

E

P2

PEB

PEF

PED

PEC

Joint F


0

sin 0

( is a known)sin


0

cos 0

cos

Y

FY FD

FYFD FY

x

FD FE

FE FD

P

R P

RP R

P

P P

P P

F

RFY

PFE

PFD

Solving Plane TrussAt start unknown forces are: PAB, PAC, PBC, PBD, PBE, PCE, PDE, PDF, PEF.

Joint A: Solved for PAB and PAC

Joint C: Solved for PCB and PCE

Joint F: Solved for PDF and PEF

Using results of above joints we can solve for the remaining unknowns

PDF leads to solving for PBD and PDE at joint DAnd then we can go to either joint E or B to solve for PBE

A

B

C E

D

F

P2

P1

RAX

RAY RFY

A

B

C E

D

F

P2

P1

RAX

RAY RFY

Section 1

Statics – Computer Methods

Computer Methods – Method of Joints

A

B

C E

D

F

8m

6m

8m

6m 6m

6m

4 kN 4 kN


In this method we do not calculate the reactions first. We write down eachequilibrium equation for each joint in turn i.e. 2j equations (12 in this case):

x direction y direction

Joint A: 0.6 0 0.8 0

Joint B: 0.6 0.6 0 0.8 0.8 0

Joint C: 0 4 0

Joint D: 0.6 0 0.8 0

Joint E: 0.6 0 0.8 4 0

AX AB AC AY AB

BD BE AB BA BC BE

CE CA CB

DF DB DF DE

EF EC EB ED EB

R P P R P

P P P P P P

P P P

P P P P

P P P P P

Joint F: 0.6 0 0.8 0FE FD FD FYP P P R

We have 12 equations and 12 unknowns so what do we do next?


We can rewrite each equation in turn and show all the unknown forces in eachof these equations. I wont do that here for all junctions so lets just look at jointA only:

Joint A - x direction: 0.6 0

0.6 1. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0

Joint A - y direction: 0.8 0

0.8 0. 0. 0. 0. 0. 0. 0. 0.

AX AB AC

AB AC BC BD BE CE DE DF EF AX AY FY

AY AB

AB AC BC BD BE CE DE DF EF

R P P

P P P P P P P P P R R R

R P

P P P P P P P P P

0. 1. 0. 0AX AY FYR R R

Notice that including all the other forces has not changed the equationsbecause they all have a zero coefficient. Note the order of forces has changedand where the coefficient is ‘1’ then this is included in the written equation.


Instead of writing down all twelve equations I can write a matrix of thecoefficients only multiplied by vector of unknown forces and add in the knownexternal loads. (Remember the equations all add up to zero).

0.6 1 0 0 0 0 0 0 0 1 0 0

0.8 0 0 0 0 0 0 0 0 0 1 0

0.6 0 0 1 0.6 0 0 0 0 0 0 0

0.8 0 1 0 0.8 0 0 0 0 0 0 0

0 1 0 0 0 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0.6 0 0 0 0

0 0 0 0 0 0 1 0.8 0 0 0 0

0 0 0 0 0.6 1 0 0 1 0 0 0

0 0 0 0 0.8 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0.6 1 0 0 0

0 0 0 0 0 0 0 0.8 0 0 0 1

0 0

0 0

0 0

0 0

0 0

4 0

0 0

0 0

0 0

4 0

0 0

0 0

AB

AC

BC

BD

BE

CE

DE

DF

EF

AX

AY

FY

P

P

P

P

P

P

P

P

P

R

R

R

A The coefficient matrix of the unknown forces (square matrix) (A-1 is inverse)x Column vector of UNKNOWN forces (member forces + reactions)y Column vector of KNOWN external forces.

(A) (x) (y)


The matrix is square because the truss is statically determinate. The number of rows is 2 x the number of joints, i.e. 12; and the number of columns is equal to the number of unknown forces i.e. 12, that is 9 member forces and 3 reactions.

Because the matrix is square we can use a numerical procedure to determine the unknown forces. First I will summarise the approach and then utilise it to find the unknown forces.

A The coefficient matrix of the unknown forces (square matrix) (A-1 is inverse)x Column vector of UNKNOWN forces (member forces + reactions)y Column vector of KNOWN external forces.

-1

0

Rearranging

Multiplying both sides by A

Since

Finally

-1 -1

-1

-1

Ax y

Ax -y

A Ax A -y

A A I

x A -y

Computer Methods – Solution

0.6 1 0 0 0 0 0 0 0 1 0 0

0.8 0 0 0 0 0 0 0 0 0 1 0

0.6 0 0 1 0.6 0 0 0 0 0 0 0

0.8 0 1 0 0.8 0 0 0 0 0 0 0

0 1 0 0 0 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0.6 0 0 0 0

0 0 0 0 0 0 1 0.8 0 0 0 0

0 0 0 0 0.6 1 0 0 1 0 0 0

0 0 0 0 0.8 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0.6 1 0 0 0

0 0 0 0 0 0 0 0.8 0 0 0 1

A x y

0

0

0

0

0

4

0

0

0

4

0

0

AB

AC

BC

BD

BE

CE

DE

DF

EF

AX

AY

FY

P

P

P

P

P

P

P

P

P

R

R

R


0 0 20 30 0 30 20 15 0 15 0 0

0 0 24 18 36 18 24 9 36 9 36 0

0 0 0 0 0 36 0 0 0 0 0 0

0 0 12 9 0 9 24 18 0 18 0 0

0 0 20 15 0 15 20 15 0 15 0 0

0 0 24 18 0 181

36

AB

AC

BC

BD

BE

CE

DE

DF

EF

AX

AY

FY

P

P

P

P

P

P

P

P

P

R

R

R

-1x A y x

0

0

0

0

0

24 9 36 9 36 0 4

0 0 16 12 0 12 16 12 0 24 0 0 0

0 0 20 15 0 15 20 30 0 30 0 0 0

0 0 12 9 0 9 12 18 0 18 36 0 0

36 0 36 0 36 0 36 0 36 0 36 0 4

0 36 16 24 0 24 16 12 0 12 0 0 0

0 0 16 12 0 12 16 24 0 24 0 36 0

5

3

4

3

0

3

4

5

3

0

4

4


0 5

0 3

0 4

0 3

5 30 0

4 4

0 3

0 4

0 5

4 0

0 4

0 4

AB

AC

BC

BD

BE AB AC B

CE

DE

DF

EF

AX

AY

FY

P

P

P

P

P P kN P kN P

P

P

P

P

R

R

R

-1A

4

3 0 3

4 5 3

0 4 4

C

BD BE CE

DE DF EF

AX AY FY

kN

P kN P P kN

P kN P kN P kN

R R kN R kN

Section 1

Statics – Method of Sections

Methods of Sections

This method consists of:

– Drawing a free body diagram for a section of the structure

– The section must ‘cut’ the members whose internal force you want to determine

– Writing equilibrium equations for each section

A

B

C E

D

F

P2

P1

RAX

RAY RFY

A

B

C E

D

F

P2

P1

RAX

RAY RFY

Methods of Sections

When using the method of sections we will still need to determine:

– Determinacy of truss

– Reactions at supports (by considering the equilibrium condition of the whole truss)

(Remember that the whole truss is a free body with the unknown forces being the external loads)

A

B

C E

D

F

P2

P1

RAX

RAY RFY

A

B

C E

D

F

P2

P1

RAX

RAY RFY

Method of SectionsInstead of drawing a free body for each joint we now draw a free body for a section (part of the structure) by “cutting” members e.g. cut BD, BE and CE

– Free body shows only those forces that are acting on section. – These forces acting are the external forces and forces in cut members.– If we want to determine PBE then we need free body that cuts member BE.– Above section can determine unknown internal member forces PBD, PBE & PCE

– We can write down 3 equilibrium equations for each section.– Unlike method of joints forces normally do not pass through the same point

A

B

C

BPP1

RAXAX

RAYAY

PBD

PBE

PCEA

B

C E

D

F

P2

RFY

A

B

C

D

F

P2

P1P1

RAXRAX

RAYRAY RFY

cut Free body diagram

Method of SectionsOne cut will produce two free body diagrams. Generally only need to consider one of these.

A

B

C

BPP1

RAXAX

RAYAY

PBD

PBE

PCE

E

D

F

P2

RFY

F

P2

RFY

PEC

PEB

PDB

A

B

C E

D

F

P2

RFY

A

B

C

D

F

P2

P1P1

RAXRAX

RAYRAY RFY

cut

Left Free body

Right Free bodyFor each free body:– All external forces are shown– Cut members externalise their

member forces so PBD, PBE and PCE are shown to be acting on both free body diagrams.

Method of Sections

A

B

C

BPP1

RAXAX

RAYAY

PBD

PBE

PCE

If using this method to find all internal forces then we need to draw sufficient sections to determine all the unknown forces. The sections chosen are completely arbitrary. However you will want to minimise the number of sections you draw and also have sufficient information find all the unknowns.

We have nine members. How many sections do we need?

A

B

C E

D

F

P2

RFY

A

B

C

D

F

P2

P1P1

RAXRAX

RAYRAY RFY

cut

Method of SectionsWe have already determined the reactions so lets cut the two members that join at the support

This first section is essentially the same as using method of joints for Joint A.

Resolving horizontally and vertically will give two equations and enable us to solve for the two unknowns PAB and PAC.

AA

RAXRAX

RAYRAY

PAB

PAC

A

B

C E

D

F

P2

RFY

A

B

C

D

F

P2

P1P1

RAXRAX

RAYRAY RFY

cut

Method of SectionsSo far we know 2 out of the 9 unknowns. To find further member forces we cutthe truss in a new location for example as shown.

We have two unknowns PCB and PCE and so again by resolving horizontally andvertically will give two equations and enable us to solve for the two unknowns.

AA

RAXRAX

RAYRAY

PAB

PCE

PCB

A

B

C E

D

F

P2

RFY

A

B

C

D

F

P2

P1P1

RAXRAX

RAYRAY RFY

cut

Method of Sections

A

B

C E

D

F

P2

RFY

A

B

C

D

F

P2

P1P1

RAXRAX

RAYRAY RFY

Now we know 4 out of the 9 unknowns. A new cut is made to determine newmember forces.

We have two unknowns PBD and PBE and so we can solve for them. (Notice that

external load P1 is acting on selected section).

cut

A

B

C

BPP1

RAXAX

RAYAY

PBD

PBE

PCE

Method of Sections

A

B

C E

D

F

P2

RFY

A

B

C

D

F

P2

P1P1

RAXRAX

RAYRAY RFY

Now we know 6 out of the 9 unknowns. A new cut is made to determine newmember forces.

This new section enables us to solve for unknowns PED and PEF

cut

A

B

C E

P2

A

B

C

P2

P1P1

RAXRAX

RAYRAY

PBD

PED

PEF

Method of Sections

A

B

C E

D

F

P2

RFY

A

B

C

D

F

P2

P1P1

RAXRAX

RAYRAY RFY

Now we know 8 out of the 9 unknowns. A new cut is made to determine the lastunknown member force PFD

This new section enables us to solve for the last unknown PFD

cut

F

RFY

F

RFY

PFD

PFE

Method of SectionsThe choice of sections has been arbitrary. We could have chosen the sectionbelow. Would this have been a good choice to start with? Justify your answer.

A

B

C E

D

F

P2

RFY

A

B

C

D

F

P2

P1P1

RAXRAX

RAYRAY RFY

cut

B DB DP1P1

PBA PBC PBE PDE PDF

Worked Example - Methods of Sections

A

B

C E

D

F

8m

6m

8m

6m 6m

6m

4 kN 4 kN

Find all the member forces.

Solution

We will use the known results that:

i. The truss is statically determinate (m+r=2j)

ii. Reaction forces have been determined (RAX=0kN, RAY=4kN and RFY=4kN

A

B

C E

D

F

8m

6m

8m

6m 6m

6m

4 kN 4 kN

Section 1 – PAB and PAC

Lets take our first section as shown. The section cuts members AB and ACand therefore the only unknown forces that act on section are PAB and PAC. Remember we do not want to cut more than three members whose forces are unknown. The horizontal reaction is zero and therefore not shown for section drawing.

A

B

E

D

F

4 kN

A

B D

F

4 kN

4 kN

4 kN

0 kN

x

y

PAB

PAC

4 kN

Section 1 – PAB and PAC

Vertical force equilibrium for section

0

84 sin 0 sin

10

45

0.8

Horizontal force equilibrium for section

0

6cos 0 cos

10

( 5)0.6 3

Y

AB

AB

x

AB AC

AC

P

P

P kN

P

P P

P kN

x

y

PAB

PAC

4 kN

Section 2 – PBC and PCE

So far we know 2 out of the 9 unknowns. With our new section we can cut new

members BC and CE. The section cuts three members in total so has these

three member forces plus the reaction at joint A act on the section and external

load at joint C. External load must be included as it is applied at joint C.

A

B

C E

D

F

RFY

A

B

C

D

F

RFY4 kN 4 kN4 kN

AA

-5 KN

PCE

PCB

4kN

4kN

Section 2 – PBC and PCE


0

4 ( 5)sin 4 0 sin 0.8

( 5) 0.8 4


0

( 5)cos 0 cos 0.6

( 5)0.6 3

Y

CB

CB

x

AC

CE

P

P

P kN

P

P

P kN

x

y

-5 KN

PCE

4kN 4kN

PCB

Section 3 – PBD and PBE

Now we know 4 out of the 9 unknowns. With our 3rd section we can cut newmembers BD and BE. The section cuts three members in total so has these three member forces plus the reaction at joint A act on the section and external load at joint C.

A

B

C E

D

F

RFY

A

B

C

D

F

RFY4 kN 4 kN4 kN

A

B

C

BPBD

PBE

3 KN

4 KN4 KN

Section 3 – PBD and PDE


0

4 4 sin 0

0


0

3 cos 0

3

Y

BE

BE

x

BD BE

BD

P

P

P kN

P

P P

P kN

x

y

A

B

C

BPBD

PBE

3 KN

4 KN 4 KN

Section 4 – PDE and PEF

Now we know 6 out of the 9 unknowns. With our 4th section we can cut newmembers DE and EF. This new section does not have any external loads except the reaction at joint F acting on it.

A

B

C E

D

FA

B

C

D

F

4 kN 4 kN4 kN 4 kN

A

B

C EA

B

C

-3 KN

PED

PEF

4 KN4 KN 4 KN

Section 4 – PDE and PFE


0

4 4 4 0

4


0

( 3) 0

3

Y

DE

ED

x

FE

EF

P

P

P kN

P

P

P kN

x

y

A

B

C EA

B

C

-3 KN

PED

PEF

4 KN 4 KN 4 KN

Section 5 – PDF

Now we know 8 out of the 9 unknowns. With our 5th section we can cut new member DF.

A

B

C E

D

FA

B

C

D

F

4 kN 4 kN4 kN 4 kN

FF

PFD

3 KN

4 KN

Section 5 – PDF


0

4 sin 0

45

0.8

Y

FD

FD

P

P

P kN

We have now determined all the internal forces.

x

y

FF

PFD

3 KN

4 KN

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Structures and Materials- Section 1 Statics

Education

action of forces

forces directions

aspoint forces

reaction forces

s force

force f

unknown external forces

direction of force arrow