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ME101: Engineering Mechanics

Lecture 6

10th January 2011

Equilibrium of Rigid Bodies



Find the reaction at the fixed end ‘E’

DF = 7.5 m

Beer/Johnston; 4.4

A B

20k

N

20k

N

C

2.25

3.75

E

D

4.5

F1.8

20k

N

20k

N

1.8 1.8 1.8

4.5150 0 90

7.5

x x xF E E kN = + × = ⇒ = −∑(sign change)

6.04 20 150 0 200

7.5 y y yF E E kN = − × − × = ⇒ =∑

150kN

A B

20k N 20k N

C

2.25

3.75

E4.5

F1.

8

20k N 20k N1.

8

1.

8

1.

8

E x

E y

20 7.2 20 5.4 20 3.6 20 1.8

6.0150 4.5 0

7.5

E

E

M

M

= × + × + × + ×

− × × + =

∑

180 E kNm= +∑



Cylinders A and B weigh 500 N each and cylinder C weighs 1000 N.

Compute all contact forces.

2m

1m1m

3m



• Six scalar equations are required to express the conditions for the

equilibrium of a rigid body in the general three dimensional case.

∑ =∑ =∑ =

∑ =∑ =∑ =000

000

z y x

z y x

M M M

F F F

• These e uations can be solved for no more than 6 unknowns which

Equilibrium of a Rigid Body in Three Dimensions

generally represent reactions at supports or connections.

• The scalar equations are conveniently obtained by applying the vector

forms of the conditions for equilibrium,

∑ ∑ =∑ ×== 00 F r M F O



Equilibrium of a Rigid Body in Three Dimensions



Reactions at Supports and Connections for a Three-Dimensional

Structure



Types of 3D equilibrium

7



SOLUTION:

• Create a free-body diagram for the

sign.• Apply the conditions for static equilibrium

to develop equations for the unknown

reactions.

Sample Problem 4.8

A sign of uniform density weighs 1200-N and is supported by a ball-and-socket joint at A and by two cables.

Determine the tension in each cable and the reaction at A.



( )k jiT

k jiT

r r

r r T T

BD

BD

B D

B D BD BD

3

2

3

1

3

2

6.3

4.22.14.2

−+−=

−+−

=

−

−=

Sample Problem 4.8

• Create a free-body diagram for the sign.

Since there are only 5 unknowns, the sign is

partially constrain. It is free to rotate about

the x axis. It is, however, in equilibrium for

the given loading.

( )k jiT

k jiT

r r

r r T T

EC

EC

E C

E C EC EC

7

2

7

3

7

6

7

236

++−=

++−=

−

−

=



( )

( ) ( ) 0N1200m1.2

0:

0N1200:

0:

0N1200

7

2

3

2

73

31

7

6

3

2

=−×+×+×=

=+−

=−++

=−−

=−++=∑

EC E BD B A

EC BD z

EC BD y

EC BD x

EC BD

jiT r T r M

T T Ak

T T A j

T T Ai

jT T AF

Sample Problem 4.8

• Apply the conditions for

static equilibrium to developequations for the unknown

reactions.

0m.N1440771.08.0:

0514.06.1:

=−+

=−

EC BD

EC BD

T T k

T T j

( ) ( ) ( )k ji A

T T EC BD

N100.1N419N1502

N1402N451

−+=

==

Solve the 5 equations for the 5unknowns,



2D, 3D force system

• Rectangular components

• Moment

Equilibrium equations

ΣFx = 0; ΣFy = 0; ΣMA = 0

ΣFx = 0; ΣMA = 0; ΣMB = 0 2D

Summary

11

• ar gnon s eorem

• Couple

• Force-couple system

• Resultant

• Principle of moment

ΣMA = 0; ΣMB = 0; ΣMc = 0

ΣF = 0 (or) ΣFX = 0; ΣFY = 0; ΣFZ = 0

ΣM = 0 (or)

ΣMX = 0;

ΣMY = 0;

ΣMZ = 0

3D



'

10'

c y

z A derrick as shown is supporting a

1000lb load. The vertical beam has

a ball-and-socket connection into

the ground at d and is held by guy

wires ac and bc. Neglect the

weight of the members and guy

wires, and find the tensions in the

guy wires ac, bc, and ce.

1000 lb3'

8'

8'

10'

a

b

d

e




Lecture 7

10th January 2011

Analysis of Structures



• For the equilibrium of structures made of several

connected parts, the internal forces as well the external

forces are considered.

• In the interaction between connected parts, Newton’s 3rd

Law states that the forces of action and reaction

between bodies in contact have the same magnitude,

same line of action, and opposite sense.

Introduction

• Three categories of engineering structures are considered:

a) Frames: contain at least one multi-force member,

i.e., member acted upon by 3 or more forces.

b) Trusses: formed from two-force members, i.e.,

straight members with end point connections

c) Machines: structures containing moving parts

designed to transmit and modify forces.



• A truss consists of straight members connected at

joints. No member is continuous through a joint.

• Most structures are made of several trusses joinedtogether to form a space framework. Each truss

carries those loads which act in its plane and may

be treated as a two-dimensional structure.

Definition of a Truss

• Bolted or welded connections are assumed to be

pinned together. Forces acting at the member ends

reduce to a single force and no couple. Only two-

force members are considered.

• When forces tend to pull the member apart, it is in

tension. When the forces tend to compress the

member, it is in compression.




Members of a truss are slender and not capable of

supporting large lateral loads. Loads must be applied at

the joints.



• A rigid truss will not collapse under

the application of a load.

• A simple truss is constructed by

successively adding two members and

one connection to the basic triangular

Simple Trusses

truss.

• In a simple truss, m = 2n - 3 where

m is the total number of members

and n is the number of joints.



• Dismember the truss and create a freebody

diagram for each member and pin.

• The two forces exerted on each member are

equal, have the same line of action, andopposite sense.

• Forces exerted by a member on the pins or

Analysis of Trusses by the Method of Joints

and equal and opposite.

• Conditions of equilibrium on the pins provide

2n equations for 2n unknowns. For a simple

truss, 2n = m + 3. May solve for m member

forces and 3 reaction forces at the supports.

• Conditions for equilibrium for the entire truss

provide 3 additional equations which are not

independent of the pin equations.



• Forces in opposite members intersecting in two

straight lines at a joint are equal.

• The forces in two opposite members are

equal when a load is aligned with a third

member. The third member force is equal to

the load (including zero load).

• The forces in two members connected at a

o n are equa e mem ers are a gne an

zero otherwise.

• Recognition of joints under special loading

conditions simplifies a truss analysis.



• An elementary space truss consists of 6 members

connected at 4 joints to form a tetrahedron.

• A simple space truss is formed and can be

extended when 3 new members and 1 joint are

added at the same time.

• In a simple space truss, m = 3n - 6 where m is the

Space Trusses

• Equilibrium for the entire truss provides 6

additional equations which are not independent of

the joint equations.

number of members and n is the number of joints.

• Conditions of equilibrium for the joints provide 3n

equations. For a simple truss, 3n = m + 6 and the

equations can be solved for m member forces and

6 support reactions.



SOLUTION:

• Based on a free-body diagram of the

entire truss, solve the 3 equilibrium

equations for the reactions at E and C .

• Joint A is subjected to only two unknown

member forces. Determine these from the

Sample Problem 1

Using the method of joints, determine

the force in each member of the truss.

.

• In succession, determine unknown

member forces at joints D, B, and E from

joint equilibrium requirements.

• All member forces and support reactionsare known at joint C . However, the joint

equilibrium requirements may be applied

to check the results.



Sample Problem 1



SOLUTION:

• Based on a free-body diagram of the entire truss,

solve the 3 equilibrium equations for the reactions

at E and C .

( )( ) ( )( ) ( )m3m6kN5m12kN10

0

E

M C

−+=

=∑

Sample Problem 1

= kN50 E

∑ == x x C F 0 0= xC

∑ ++−== y y C F kN50kN5-kN100

↓= kN35 yC



• Joint A is subjected to only two unknown

member forces. Determine these from the

Sample Problem 1

.

534

kN10 AD AB F F == C F

T F

AD

AB

kN5.12

kN5.7

=

=

• There are now only two unknown memberforces at joint D.

( ) DA DE

DA DB

F F

F F

5

32=

=

C F

T F

DE

DB

kN15

kN5.12

=

=



• There are now only two unknown member

forces at joint B. Assume both are in tension.

( )kN75.18

kN12kN50 5

4

5

4

−=−−−==∑

BE

BE y

F

F F

C F BE kN75.18=

33 −−−

Sample Problem 1

kN25.26

...55

+=

−−−

BC

BC x

F T F BC kN25.26=

• There is one unknown member force at joint

E . Assume the member is in tension.

( )

kN75.43

kN75.18kN1505

3

5

3

−=

++==∑

EC

EC x

F

F F

C F EC kN75.43=



• All member forces and support reactions are

known at joint C . However, the joint equilibrium

requirements may be applied to check the results.

( ) ( )

( ) ( )checks 075.4335

checks 075.4325.26

5

4

5

3

=+−=

=+−=

∑

∑

y

x

F

F

Sample Problem 1




Lecture 8

17th January 2011




Method of sections

• In method of joints, we need only two equilibrium equations, as

we deal with concurrent force system

• In method of sections, we will consider three equilibriumequations, including one moment equilibrium eqn.

• Force in almost any desired member can be obtained directly from

29

an analysis of a section which has cut the member• Not necessary to proceed from joint to joint

• Not more than three members whose forces are unknown should

be cut. Only three independent equilibrium eqns. are present

• Efficiently find limited information



A

BC

D

EF

A

BC

D

EF A

Methodology for method of sections

30

L• The external forces are obtained initially from method of joints,

by considering truss as a whole

• Assume we need to find force in BE, then entire truss has to be

sectioned across FE, BE, BC as shown in figure; we have only 3

equilibrium equations.

• AA – section across FE, BE, BC; Forces in these members are

initially unknown

R1

R2



Section 1 Section 2

31

• Now each section will apply opposite forces on each other

• The LHS is in equilibrium with R1, L, three forces exerted on the cut

members (EF, BE, BC) by the RHS which has been removed

• In this method the initial direction of forces is decided by moment aboutany point where known forces are present

• For eg., take moment about point B for the LHS, this will give BE, BC

to be zero; Then moment by EF should be opposite to moment by R1;

Hence EF should be towards left hand side - compressive



• Now take moment about ‘F’ => BE should be opposite to R1

moment; Hence BE must be up and to the right; So BE is tensile

• Now depending on the magnitudes of known forces, BC direction

has to be decided, which in this case is outwards i.e., tensile

32

Σ MB = 0 => FORCE IN EF; BE, BC = 0

Σ Fy = 0 => FORCE IN BE; BC, EF = 0

Σ MF = 0 => FORCE IN BC; EF, BE = 0Section 1 Section 2



Section AA and BB are

possible

33

Convenient



Important points

• In method of sections, an entire portion of the truss is considered a

single body in equilibrium

• Force in members internal to the section are not involved in the

analysis of the section as a whole

• The cutting section is preferably passed through members and not

through joints

34

• Either portion of the truss can be used, but the one with smallernumber of forces will yield a simpler solution

• Method sections and method of joints can be combined

• Moment center can be selected through which many unknownforces pass through

• Positive force value will sense the initial assumption of force

direction



• When the force in only one member or the

forces in a very few members are desired, themethod of sections works well.

Analysis of Trusses by the Method of Sections

• o eterm ne t e orce n mem er , pass a

section through the truss as shown and create

a free body diagram for the left side.

• With only three members cut by the section,

the equations for static equilibrium may be

applied to determine the unknown member

forces, including F BD.



• Compound trusses are statically

determinant, rigid, and completely

constrained.32 −= nm

• Truss contains a redundant member

and is statically indeterminate.

Trusses Made of Several Simple Trusses

−nm

• Necessary but insufficient condition

for a compound truss to be statically

determinant, rigid, and completely

constrained,

nr m 2=+

non-rigid rigid

32 −< nm

• Additional reaction forces may be

necessary for a rigid truss.

42 −< nm



SOLUTION:

• Take the entire truss as a free body.

Apply the conditions for static equilib-

rium to solve for the reactions at A and L.

• Pass a section through members FH ,

Sample Problem 8.1

Determine the force in members FH , GH , and GI .

GH , and GI and take the right-hand

section as a free body.

• Apply the conditions for static

equilibrium to determine the desiredmember forces.



SOLUTION:

• Take the entire truss as a free body.

Apply the conditions for static equilib-

rium to solve for the reactions at A and L.

Sample Problem 8.1

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )

↑=

++−==

↑=

+−−

−−−==

∑

∑

kN5.12

kN200

kN5.7

m25kN1m25kN1m20

kN6m15kN6m10kN6m50

A

A LF

L

L

M

y

A



• Pass a section through members FH , GH , and GI

and take the right-hand section as a free body.

Sample Problem 8.1

( )( ) ( )( ) ( )

kN13.13

0m33.5m5kN1m10kN7.50

0

+=

=−−

=∑

GI

GI

H

F

F

M

• pp y t e con t ons or stat c equ r um to

determine the desired member forces.

T F GI kN13.13=

8 mtan 0.5333 28.07

15 m

FG

GLα α = = = = °



( )( ) ( )( ) ( )( )

( )( )

kN82.13

0m8cos

m5kN1m10kN1m15kN7.5

0

07.285333.0m15

m8tan

−=

=+

−−

=

°====

∑

FH

FH

G

F

F

M

GL

FG

α

α α

C F FH kN82.13=

Sample Problem 8.1

( )

( )( ) ( )( ) ( )( )

kN371.1

0m10cosm5kN1m10kN1

0

15.439375.0m8

m5tan

3

2

−=

=++=

°====

∑

GH

GH

L

F

F

M

HI

GI

β

β β

C F GH kN371.1=

I H Sh



I.H. Shames FBD - 1

FBD - 2

41

From FBD-2ΣMB = 0 => -(10)(500)+30 (789)- FAC Sin 30 (30) = 0

FAC = 1244.67 N

From FBD -1

ΣFx = 0 => FDA Cos 30 – (1244.67) cos 30 – 1000 sin 30 = 0 ;

FDA = 1822 N

ΣFy = 0=> (1822)Sin 30 + (1244.67) sin 30 +FAB – 1000 Cos 30 = 0;

FAB

= -667 N




Lecture 9

18th January 2011




• Frames and machines are structures with at least one

multiforce member. Frames are designed to support loads

and are usually stationary. Machines contain moving parts

and are designed to transmit and modify forces.

• A free body diagram of the complete frame is used to

determine the external forces acting on the frame.

• Internal forces are determined b dismemberin the frame

Analysis of Frames

and creating free-body diagrams for each component.

• Forces between connected components are equal, have the

same line of action, and opposite sense.

• Forces on two force members have known lines of action

but unknown magnitude and sense.

• Forces on multiforce members have unknown magnitude

and line of action. They must be represented with two

unknown components.



• Some frames may collapse if removed from

their supports. Such frames can not be treated

as rigid bodies.

• A free-body diagram of the complete frame

indicates four unknown force components which

can not be determined from the three e uilibrium

Frames Which Cease To Be Rigid When Detached From

Their Supports

conditions.

• The frame must be considered as two distinct, but

related, rigid bodies.

• With equal and opposite reactions at the contact

point between members, the two free-bodydiagrams indicate 6 unknown force components.

• Equilibrium requirements for the two rigid

bodies yield 6 independent equations.

F i d FBD



Force representation and FBD

• Representing force by rectangular components

• Calculation of moment arms will be simplified

• Proper sense of force is necessary; Some times arbitrary assignment is done;

Final force answer will yield correct force direction

• Force direction should be consistently followed

R. Ganesh Narayanan 45



SOLUTION:

• Create a free-body diagram for the

complete frame and solve for the support

reactions.• Define a free-body diagram for member

BCD. The force exerted by the link DE

has a known line of action but unknown

Sample Problem 8.2

Members ACE and BCD are

connected by a pin at C and by the

link DE . For the loading shown,

determine the force in link DE and thecomponents of the force exerted at C

on member BCD.

magnitude. It is determined by summing

moments about C .

• With the force on the link DE known, the

sum of forces in the x and y directions

may be used to find the force

components at C .

• With member ACE as a free-body,

check the solution by summing

moments about A.

Sample Problem 8 2



SOLUTION:

• Create a free-body diagram for the complete frame

and solve for the support reactions.

N4800 −==∑ y y AF ↑= N480 y A

( )( ) ( )mm160mm100N4800 B M A +−==∑

Sample Problem 8.2

→= N300 B

x A BF +==∑ 0 ←−= N300 x A

°== −07.28tan

150

801α

Note:

Sample Problem 8 2



• Define a free-body diagram for member

BCD. The force exerted by the link DE has a

known line of action but unknown

magnitude. It is determined by summing

moments about C .

( ) ( ) ( ) ( ) ( ) ( )0 sin 250 mm 300 N 80 mm 480 N 100 mm

561 N

C DE

DE

M F

F

α = = + +

= −

∑

Sample Problem 8.2

DE

• Sum of forces in the x and y directions may be used to find the force

components at C .

( ) N300cosN5610

N300cos0

+−−=

+−==∑

α

α

x

DE x x

C

F C F

N795−= xC

( ) N480sinN5610

N480sin0

−−−=

−−==∑

α

α

y

DE y y

C

F C F

N216= yC



• With member ACE as a free-body, check

the solution by summing moments about A.

( )( ) ( )( ) ( )

( )( ) ( )( ) ( )( ) 0mm220795mm100sin561mm300cos561

mm220mm100sinmm300cos

=−−−+−=

−+=∑

α α

α α x DE DE A C F F M

(checks)

Meriem/Kraige3m 2m



Find the horizontal and vertical

components of all the forces;

neglect weight of each member

FBD of full frame

Ay

Ax

A

B

C

D

E

F

3m 2m

1.5m

0.5m

1.5m

1.5m

R =0.5m

50

0.4 x 9.81 = 3.92

ΣMA

= 0 => 5.5 (-0.4) (9.81) + 5Dx

= 0 => Dx

= 4.32 kN

ΣFx = 0 => -Ax + 4.32 = 0 => Ax = 4.32 kN

ΣFy = 0 => Ay – 3.92 = 0 => Ay = 3.92 kN

Dx

400 kg

FBD of individual members



3.92

4.32

3.92

Bx

By

Cx

A

3.92

3.92

3.92

3.92

F

E

Ex

Ey

E

EyEx

Bx

By

B

3.92

3.92

Cy

3m 2m

.

D C

Cx

Cy

Apply equilibrium equn. And solve for

forces

A

B

C

D

EF

400 kg

1.5m

0.5m

1.5m

1.5m

R =0.5m

Machines



• Machines are structures designed to transmit

and modify forces. Their main purpose is to

transform input forces into output forces.

• Create a free-body diagram of the complete

Machines

• Given the magnitude of P, determine themagnitude of Q.

machine, including the reaction that the wire

exerts.

• The machine is a nonrigid structure. Use

one of the components as a free-body.

• Taking moments about A,

Pb

aQbQaP M A =−==∑ 0



R f b k



1. Vector Mechanics for Engineers – Statics & Dynamics, Beer &

Johnston; 7th edition

2. Engineering Mechanics Statics & Dynamics, Shames; 4th

edition3. Engineering Mechanics Statics Vol. 1, Engineering Mechanics

Dynamics Vol. 2, Meriam & Kraige; 5th edition

Reference books

54

STATICS – MID SEMESTER – DYNAMICS

Tutorial: Thursday 8 am to 8.55 am

4. Schaum’s solved problems series Vol. 1: Statics; Vol. 2:Dynamics, Joseph F. Shelley

Lecture Slides Statics 2011 Lecture6 7 8 9 [Compatibility Mode]

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