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:
Structures made of strong materials
Relatively low level of external load
Negligible amount of deformation
Neglecting deformation won’t affect analysis
Characteristics of the rigid bodies:
Transmit only axial forces
Size of contact area immaterial
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.

Forces
Resolving Forces:
Because a force has direction it will have components that act in three directions for a 3D system. (Two components for 2D system).
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. z y x F  x  z  y Force components:

Moments
What is a moment?
A moment is a force acting about (around) an axis.
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 is M = 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 z y x F r d M

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: R x R y R y Built in or fixed support Pin or hinged support Roller support x y R x R y M z

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 Equilibrium
The 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:
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.

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

Example 1
A loaded truss, as shown in the figure, is in equilibrium. The known loads are
P 1 (horizontal at joint B) and P 2 (vertical at joint C). Obtain the following:
Check the determinacy of the truss.
Moment equilibrium equation with respect to joint A
Determine the support reactions
Force equilibrium equations for joint B
P 2 A B C D P 1 l  

(i) Check Determinacy of the Truss R AX R AY R DY A B C D P 1 P 2 x y l  

(ii) Moment Equilibrium Equation with Respect to Joint A Treat 2d truss as a free body. A B C D P 1 P 2 R AX R AY R DY x y l  

(iii) Determine Support Reactions 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. A B C D P 1 P 2 R AX R AY R DY x y l  

(iv) Force Equilibrium Equations for Joint B Free body diagram for joint B 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. B P 1   P BD P BC P BA x y A B C D P 1 P 2 R AX R AY R DY l  

Section 1 Statics – 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 P 2 P 1

We introduce two methods to solve this problem:
1) Method of joints
2) Methods of sections
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 Truss
Solution for Method of joints
Determinacy of the truss:
Reactions: One reaction R FX from roller support at point F
Two reactions R AY and R AX from pin support at point A
Total of three reactions r = 3 Number of members, m = 9
Number of joints j = 6
Using “m+r=2j” to check determinacy
9+3=2x6 i.e. truss is statically determinate.
x y

Finding Reactions

Method of Joints
Consider each joint in turn:
Lets start with joint A by drawing a free body diagram of the joint:
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  x y

Joint B B P 1 P BA P BE P BD P BC  Note that force P AB = P BA due to Newton’s 3 rd Law. Note that angle DBE =  .

Joint C C P CE P CB P CA

Joint D  D P DF P DB P DE

Joint E E P 2 P EB P EF P ED P EC 

Joint F  F R FY P FE P FD

Solving Plane Truss
At start unknown forces are: P AB , P AC , P BC , P BD , P BE , P CE , P DE , P DF , P EF .
Joint A: Solved for P AB and P AC
Joint C: Solved for P CB and P CE
Joint F: Solved for P DF and P EF
Using results of above joints we can solve for the remaining unknowns P DF leads to solving for P BD and P DE at joint D And then we can go to either joint E or B to solve for P BE

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

Computer Methods – Method of Joints In this method we do not calculate the reactions first. We write down each equilibrium equation for each joint in turn i.e. 2j equations (12 in this case): We have 12 equations and 12 unknowns so what do we do next?

Computer Methods – Method of Joints We can rewrite each equation in turn and show all the unknown forces in each of these equations. I wont do that here for all junctions so lets just look at joint A only: Notice that including all the other forces has not changed the equations because they all have a zero coefficient. Note the order of forces has changed and where the coefficient is ‘1’ then this is included in the written equation.

Computer Methods – Method of Joints Instead of writing down all twelve equations I can write a matrix of the coefficients only multiplied by vector of unknown forces and add in the known external loads. (Remember the equations all add up to zero). 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)

Computer Methods – Method of Joints 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.

Computer Methods – Solution

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

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)

Method of Sections
Instead 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 P BE then we need free body that cuts member BE.
Above section can determine unknown internal member forces P BD , P BE & P CE
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 B P P 1 P BD P BE P CE A B C E D F P 2 R FY A B C D F P 2 R FY cut Free body diagram R AX AX R AY AY P 1 P 1 R AX R AX R AY R AY

Method of Sections
One cut will produce two free body diagrams. Generally only need to consider one of these.
cut Left Free body Right Free body
For each free body:
All external forces are shown
Cut members externalise their member forces so P BD , P BE and P CE are shown to be acting on both free body diagrams.
A B C B P P 1 R AX AX R AY AY P BD P BE P CE E D F P 2 R FY F P 2 R FY P EC P EB P DB A B C E D F P 2 R FY A B C D F P 2 P 1 P 1 R AX R AX R AY R AY R FY

Method of Sections
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 B P P 1 R AX AX R AY AY P BD P BE P CE A B C E D F P 2 R FY A B C D F P 2 P 1 P 1 R AX R AX R AY R AY R FY cut

Method of Sections
We 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 P AB and P AC . A A R AX R AX R AY R AY P AB P AC A B C E D F P 2 R FY A B C D F P 2 P 1 P 1 R AX R AX R AY R AY R FY cut

Method of Sections
So far we know 2 out of the 9 unknowns. To find further member forces we cut
the truss in a new location for example as shown.
We have two unknowns P CB and P CE and so again by resolving horizontally and vertically will give two equations and enable us to solve for the two unknowns. A A R AX R AX R AY R AY P AB P CE P CB A B C E D F P 2 R FY A B C D F P 2 P 1 P 1 R AX R AX R AY R AY R FY cut

Method of Sections
Now we know 4 out of the 9 unknowns. A new cut is made to determine new
member forces.
We have two unknowns P BD and P BE and so we can solve for them. (Notice that external load P 1 is acting on selected section). cut A B C E D F P 2 R FY A B C D F P 2 P 1 P 1 R AX R AX R AY R AY R FY A B C B P P 1 R AX AX R AY AY P BD P BE P CE

Method of Sections
Now we know 6 out of the 9 unknowns. A new cut is made to determine new
member forces.
This new section enables us to solve for unknowns P ED and P EF cut A B C E D F P 2 R FY A B C D F P 2 P 1 P 1 R AX R AX R AY R AY R FY A B C E P 2 A B C P 2 P 1 P 1 R AX R AX R AY R AY P BD P ED P EF

Method of Sections
Now we know 8 out of the 9 unknowns. A new cut is made to determine the last
unknown member force P FD
This new section enables us to solve for the last unknown P FD cut A B C E D F P 2 R FY A B C D F P 2 P 1 P 1 R AX R AX R AY R AY R FY F R FY F R FY P FD P FE

Method of Sections
The choice of sections has been arbitrary. We could have chosen the section
below. Would this have been a good choice to start with? Justify your answer.
cut A B C E D F P 2 R FY A B C D F P 2 P 1 P 1 R AX R AX R AY R AY R FY B D B D P 1 P 1 P BA P BC P BE P DE P DF

Worked Example - Methods of Sections Find all the member forces. A B C E D F 8m 6m 8m 6m 6m 6m 4 kN 4 kN

Solution
We will use the known results that:
The truss is statically determinate (m+r=2j)
Reaction forces have been determined (R AX =0kN, R AY =4kN and R FY =4kN
A B C E D F 8m 6m 8m 6m 6m 6m 4 kN 4 kN 

Section 1 – P AB and P AC
Lets take our first section as shown. The section cuts members AB and AC
and therefore the only unknown forces that act on section are P AB and P AC .
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 P AB P AC 4 kN

Section 1 – P AB and P AC x y P AB P AC 4 kN

Section 2 – P BC and P CE
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 R FY A B C D F R FY 4 kN 4 kN 4 kN A A -5 KN P CE P CB 4kN 4kN

Section 2 – P BC and P CE -5 KN P CE 4kN 4kN P CB x y

Section 3 – P BD and P BE
Now we know 4 out of the 9 unknowns. With our 3rd section we can cut new
members 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 R FY A B C D F R FY 4 kN 4 kN 4 kN A B C B P BD P BE 3 KN 4 KN 4 KN

Section 3 – P BD and P DE A B C B P BD P BE 3 KN 4 KN 4 KN x y

Section 4 – P DE and P EF
Now we know 6 out of the 9 unknowns. With our 4th section we can cut new
members 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 F A B C D F 4 kN 4 kN 4 kN 4 kN A B C E A B C -3 KN P ED P EF 4 KN 4 KN 4 KN

Section 4 – P DE and P FE A B C E A B C -3 KN P ED P EF 4 KN 4 KN 4 KN x y

Section 5 – P DF
Now we know 8 out of the 9 unknowns. With our 5th section we can cut
new member DF.
A B C E D F A B C D F 4 kN 4 kN 4 kN 4 kN F F P FD 3 KN 4 KN

Section 5 – P DF We have now determined all the internal forces. F F P FD 3 KN 4 KN x y

This resource was created by Loughborough University and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme. © 2010 Loughborough University. Except where otherwise noted this work is licensed under a Creative Commons Attribution 2.0 Licence . The name of Loughborough University, and the Loughborough University logo are the name and registered marks of Loughborough University. To the fullest extent permitted by law Loughborough University reserves all its rights in its name and marks, which may not be used except with its written permission. The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. Credits

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