Course Details
Review of Basic Principles
Introduction
(Structural Analysis – CE21202)
Dr. Puneet Kumar Patra
Department of Civil Engineering
Indian Institute of Technology Kharagpur
puneet.patra@civil.iitkgp.ac.in
January 4, 2023
Course Details
Outline
1
Course Details
Course Plan (Tentative)
Learning Outcome
2
Review of Basic Principles
Free Body Diagrams
Equilibrium Equations
Sample Problems
Plane Trusses
Methods of Joints and Sections
Sample Problems on Trusses
Review of Basic Principles
Course Details
Review of Basic Principles
Review of Basic Principles
Course Details
Review of Basic Principles
Free Body Diagrams I
FBD is the most important step in solution of problems of structural
mechanics
The body is isolated from all other bodies
A complete and accurate account of all forces acting on this body is
considered
The isolation is imaginary
Steps to construct FBD:
1
Decide the body to be isolated
2
Draw the complete external boundary of the isolated body. Ensure complete
isolation
3
Identify all forces acting on the isolated body due to the removal of the
contacting bodies
4
Represent these forces appropriately where they act
Course Details
Review of Basic Principles
Free Body Diagrams II
Action of External Supports:
Figure: Type of contact and action on a body
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Review of Basic Principles
Free Body Diagrams III
Figure: Type of contact and action on a body
Course Details
Review of Basic Principles
Free Body Diagrams IV
Degrees of Freedom and Contact Forces
In two-dimensions, a rigid body can have two translations and one rotation
→ degrees of freedom.
The translations along the x and y axes and rotation about the z axis.
Presence of a contact prevents the motion along one or more degrees of
freedom.
Whichever degree of freedom is not allowed free motion, a constraint force
develops
Most common supports in Structural Engineering
A roller support → prevents the free motion of the normal-to-the-contact
degree of freedom → a contact force develops in this direction
A hinge support → prevents the free motion of the x and y degrees of
freedom → a contact force develops in x direction and another in y direction
A fixed support → prevents the translation and the rotation → two forces
develop as in a hinge support and a moment also develops.
Course Details
Free Body Diagrams V
Sample Free Body Diagrams:
Review of Basic Principles
Course Details
Review of Basic Principles
Equilibrium Equations I
One equilibrium equation for each degree of freedom
X
X
X
Fx = 0;
Fy = 0;
MO = 0
Sum of forces along the x axis = 0
Sum of forces along the y axis = 0
Sum of moment about any point O on or off the body = 0
(1)
Course Details
Review of Basic Principles
Equilibrium Equations II
Categories of equilibrium
System of collinear forces:
P
Fx = 0 is sufficient
P
P
System of concurrent forces:
Fx = 0 and
Fy = 0 are sufficient
P
P
System of parallel forces:
Fx = 0 and
MO = 0 are sufficient
(a) Collinear case
(b) Concurrent case
(c) Parallel case
Course Details
Review of Basic Principles
Equilibrium Equations III
Two and Three Force Members
A structural member subjected to two forces is in equilibrium only when the
forces are equal, opposite and collinear.
A structural member subjected to three forces is in equilibrium only when
the forces are concurrent.
Course Details
Review of Basic Principles
Equilibrium Equations IV
Alternative Equilibrium Equations
Force balance along x axis and moment balance about any two points on the
body:
X
X
X
Fx = 0;
MA = 0;
MB = 0
(2)
A and B must not lie on a line perpendicular to the x direction.
Moment balance along any three points about the body:
X
MC = 0;
X
MA = 0;
X
MB = 0
(3)
All different forms are identical
No new information is gained by solving these additional equations – you
can still solve only three unknowns.
Why can’t you obtain any new additional information? Or solve more
than three unknowns? Prove it.
Course Details
Review of Basic Principles
Problem (a) I
Determine the magnitude of forces C and T .
Since all forces are concurrent; second category of equilibrium:
X
P
Fx = 0
=⇒
P
Fy = 0
=⇒
Fx = 0;
X
Fy = 0
8 + T cos 40◦ + C sin 20◦ − 16
0.766T + 0.342C
T sin 40◦ − C cos 20◦ − 3
0.643T − 0.940C
(4)
=
=
=
=
Solving the two equations give: T = 9.09kN and C = 3.03kN.
0
8
0
3
(5)
Course Details
Review of Basic Principles
Problem (b) I
A cable as indicated in the figure is subjected to a tension T . This cable supports
a 500 kg mass with a pulley system as shown. Each pulley is frictionless and have
negligible mass. Find the total force on the bearing of pulley C .
Correctly draw the free body diagram.
Use the general equations of equilibrium to solve the problem
Assume the radius of each pulley to be r .
Course Details
Review of Basic Principles
Problem (b) II
Figure: The FBD of pulleys at A, B, and C are shown.
For the pulley at A:
X
X
MO = 0 : T1 r − T2 r = 0 =⇒ T1 = T2
(6)
Fy = 0 : T1 + T2 − 500 × 9.81 = 0 =⇒ 2T1 = 500 × 9.81N
(7)
Course Details
Review of Basic Principles
Problem (b) III
For the pulley at B
X
X
MO = 0 : T3 r − T4 r = 0 =⇒ T3 = T4
(8)
Fy = 0 : T3 + T4 − T2 = 0 =⇒ T3 = 1226N
(9)
For the pulley at C :
X
MO = 0 : =⇒ T = T3
(10)
Fx = 0 : 1226 cos 30◦ − Fx = 0 =⇒ Fx = 1062N
(11)
Fy = 0 : Fy + 1226 sin 30◦ − 1226 = 0 =⇒ Fy = 613N
(12)
X
X
q
Therefore the total force at the bearing:
Fx2 + Fy2 = 1226N.
Course Details
Review of Basic Principles
Problem (c) I
Determine the magnitude of the tension force T in the supporting cable and the
magnitude of the force on the pin joint at A. The beam AB is a standard I-beam
with a mass of 95 kg per meter of length
Correctly draw the free body diagram.
The joint at A is a pin joint → two support reactions
The weight of the beam = 4.66 kN acts at the center of the beam AB.
There are three unknowns in the problem – the reactions at A and the
tension T in the cable.
Course Details
Review of Basic Principles
Problem (c) II
Figure: The FBD of the beam is shown.
The moment at A equals 0, which gives T = 19.61kN:
T cos 25◦ ×0.25+T sin 25◦ ×(5−0.12)−10×(5−1.5−0.12)−4.66×(2.5−0.12) = 0
(13)
From the sum of forces in x direction = 0:
Ax − 19.61 × cos 25◦ = 0 =⇒ Ax = 17.77kN
(14)
From the sum of forces in y direction = 0:
Ay + 19.61 × sin 25◦ − 4.66 − 10 = 0 =⇒ Ay = 6.37kN
Therefore the total magnitude of force:
q
A2x + A2y = 18.88kN.
(15)
Course Details
Review of Basic Principles
Plane Trusses I
What is a TRUSS?
A framework comprising members joined at their ends.
Commonly seen in bridges and roofs.
The weight of each truss member is negligible in comparison to the load it
bears.
All members are two force members.
Course Details
Plane Trusses II
Review of Basic Principles
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Plane Trusses III
Review of Basic Principles
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Review of Basic Principles
Plane Trusses IV
(a) Simplest stable truss
(b) Unstable truss
Simple Trusses
The simplest plane truss which is stable looks like a triangle
Three bars are joined by pins at their end.
Typically, complex trusses are made of combination of several simple trusses
Trusses are supported by pinned and roller joints.
Why is the right figure unstable?
Course Details
Review of Basic Principles
Plane Trusses V
(a) Zoomed view of a Joint
(b) Typical Supports – Pinned
and Roller
Truss Joints and Supports
Assume that each joint is a pinned joint.
Assumption works if the centerlines of the members are concurrent.
At each joint, two equations of equilibrium.
One support is a hinged support.
Other support is a roller to provide for thermal expansion and contraction
and deformation due to loads.
Course Details
Review of Basic Principles
Plane Trusses VI
(a) A typical truss
(b) FBD of the truss
(c) FBD of the joint A.
Free Body Diagram of Trusses
Isolate the truss
Neglect the weight and draw the forces due to supports and external loads
Is anything missing at the FBD of the truss?
You can isolate the joint at A – two forces due to the members AF and AB
act along with the reaction from the ground.
Assume some direction of forces in the members while drawing the FBD.
Course Details
Review of Basic Principles
Plane Trusses VII
(a) A typical truss
(b) FBD of the truss
(c) FBD of the joint A.
Free Body Diagram of Trusses
Isolate the truss
Neglect the weight and draw the forces due to supports and external loads
Is anything missing at the FBD of the truss?
You can isolate the joint at A – two forces due to the members AF and AB
act along with the reaction from the ground.
Assume some direction of forces in the members while drawing the FBD.
Course Details
Review of Basic Principles
Plane Trusses VIII
In Summary:
Trusses are structures whose all elements are in either Tension or
Compression
Internal forces are always axial
No element has a force (shear) in a direction perpendicular to its axis
Each joint is a pinned joint
A,B & C are the nodes
AB , BC & AC are the elements
Course Details
Review of Basic Principles
Plane Trusses IX
Why should trusses carry only axial forces? Let us look into the FBD
For member AC the moment equilibrium equation is:
MA : Cy × L = 0
L 6= 0, so we have Cy =0 & we get Ay =0 since, Ay + Cy = 0.
This can be extended to other members as well.
(16)
Course Details
Review of Basic Principles
Plane Trusses X
How do you include the weight of the members (if you have to)?
Notice the FBD of member AC
Let the weight of the member be W .
Transfer the weight identically to the two nodes.
In the FBD of any member in a truss no moment exists, since the members are
connected by pinned joints that allow rotation of 2 members relative to each other
Course Details
Review of Basic Principles
Methods of Joints and Sections I
Find the forces in the members of the shown truss.
Methods of Joints – Looking at the FBD of the joints
Methods of Section – Looking at the FBD of a section of a truss
Course Details
Review of Basic Principles
Methods of Joints and Sections II
Method of Joints
1
Isolate a joint / node
2
Draw the FBD of each joint
3
Look at the equilibrium of each joint / node.
4
As all forces are concurrent at the joint, this joint has two equilibrium
equations
X
X
Fx = 0;
Fy = 0
5
For each joint, therefore, we have 2 equilibrium equations
6
For N joints, we have 2N equilibrium equations
(17)
In this specific problem:
1
There are 3 joints; So there are 6 equilibrium equations
2
Total Number of unknowns = 6; 2 unknown reactions at B, 1 unknown
reaction at C , the axial force in member AB, the axial force in member BC
and the axial force in member AC .
3
Using the 6 equilibrium equations (2 for each joint), the 6 unknowns can be
obtained.
Course Details
Review of Basic Principles
Methods of Joints and Sections III
(b) FBD of each node
(a) Given Problem
For Joint A
FAC
FAB + FAC cos 45◦ = 0 ⇒ FAB + √ = 0
2
P + FAC sin 45◦ = 0
√
⇒ FAC = −P 2
FAB =
−FAC
√
=P
2
(18)
(19)
(20)
(21)
Course Details
Review of Basic Principles
Methods of Joints and Sections IV
(a) Given Problem
(b) FBD of each node
For Joints B and C :
FBA + RBy = 0 =⇒ RBy = −P
(22)
RBx + FBc = 0
(23)
FCB = −FCA cos 45◦ = +P
(24)
RC + FCA sin 45◦ = 0 =⇒ RC = P
(25)
So, RBY = −P
(26)
Course Details
Review of Basic Principles
Methods of Joints and Sections V
Method of Sections
1
Cut the truss into discrete non-overlapping sections
2
Each section behaves as a body
3
Draw the FBD of each section
4
Look at the equilibrium of each joint / node.
5
As the section behaves as a body, we have three equations of equilibrium for
the section:
X
X
X
Fx = 0;
Fy = 0;
MO = 0
(27)
6
For each section, therefore, we have 3 equilibrium equations
7
For N sections, we have 3N equilibrium equations
In this specific problem:
1
The structure can be cut into 2 non-overlapping sections; So there are 6
equilibrium equations
2
Total Number of unknowns = 6
3
Using the 6 equilibrium equations (2 for each joint), the 6 unknowns can be
obtained.
Course Details
Review of Basic Principles
Methods of Joints and Sections VI
When you superpose the two sections you get the actual problem
For Section I, the equations of equilibrium are
P + FAc cos 45◦ = 0
(28)
FAB + FAc sin 45◦ = 0
(29)
Mc = 0 ⇒ PL = FAB L =⇒ FAB = P
(30)
This method is very powerful some times.
Course Details
Review of Basic Principles
Methods of Joints and Sections VII
Figure: FBD of each pinned joint and member
Identify the members in tension and compression
1
Draw the FBD of each member and the pinned joint
2
See if a member is subjected to pull or push forces at the ends
3
Pull → elongation → member is in tension
4
Push → shortening → member is in compression
Course Details
Review of Basic Principles
Problem (d) I
Find the forces in the members of the shown structure.
Use Methods of Joints and the idea that the forces in a two-force member
are along the line joining the two points.
The FBD of joint B is shown – three forces act : FAB , FBC due to the
members AB and BC and RB due to the reaction from the ground.
Course Details
Review of Basic Principles
Problem (d) II
By symmetry
RA = RB =
1
For Joint B
X
L
2
(31)
Fy = 0 : −FBC sin 45◦ + RB = 0
(32)
L
⇒ FBC = √
2
X
Fx = 0 : −FAB + FBC cos 45◦ = 0 ⇒ FAB =
(33)
L
2
(34)
Course Details
Review of Basic Principles
Problem (e) I
Find the forces in the member CL of the shown structure.
This is tedious to solve using the method of joints.
Using the method of section, cut a section at KL-CL-CB, it can then be
solved using the equilibrium of the left section easily
Course Details
Review of Basic Principles
Problem (f ) I
Use method of joints to find all zero force members.
We will begin with joint D as it has the least number of unknowns for a joint
Subsequently, we go for other joints
Course Details
Review of Basic Principles
Problem (f ) II
For Joint D
Figure: FBD of Joint D
X
X
Fy = 0 : FDC sinθ = 0 ⇒ FDC = 0
(35)
Fx = 0 : FDC cosθ + FDE = 0 ⇒ FDE = 0
(36)
Course Details
Review of Basic Principles
Problem (f ) III
For Joint E
Figure: FBD of Joint E
X
Fx = 0 : FED = FEF ⇒ FEF = 0
X
Fy = 0 : FEC = P
(37)
(38)
Course Details
Review of Basic Principles
Problem (f ) IV
For Joint H
Figure: FBD of Joint H
X
X
Fy 0 = 0 : FHB = 0
(39)
Fx 0 = 0 : FHF = FHA
(40)
Course Details
Review of Basic Principles
Problem (f ) V
For Joint G
Figure: FBD of Joint G
X
Fy = 0 : FGA = 0
(41)