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Unit-IV
COPLANAR NON-CONCURRENT FORCE SYSTEMS
By Prof. G. Ravi
Overview of System of forces It is well known that a system of coplanar forces can occur in different configurations some of the possibilities are
- Coplanar, Collinear, Concurrent
- Coplanar and Concurrent
- Coplanar and Non Concurrent To determine the resultant of any system of forces we adopt the principle of Resolution and Composition. The following figures depict the principles involved. . Composition of system of forces
2 2
tan
i
i
i i
x R y
x y
f
f
R f f
∑
∑
∑ ∑ =^ −
Equilibrium: Equilibrium is the status of the body when it is subjected to a system of forces. We know that for a system of forces acting on a body the resultant can be determined. By Newton’s 2 nd^ Law of Motion the body then should move in the direction of the resultant with some acceleration. If the resultant force is equal to zero it implies that the net effect of the system of forces is zero this represents the state of equilibrium. For a system of coplanar concurrent forces for the resultant to be zero, hence
Equilibriant : Equilbriant is a single force which when added to a system of forces brings the status of equilibrium. Hence this force is of the same magnitude as the resultant but opposite in sense. This is depicted in Fig 4.
Free Body Diagram: Free body diagram is nothing but a sketch which shows the various forces acting on the body. The forces acting on the body could be in form of weight, reactive forces contact forces etc. An example for Free Body Diagram is shown below.
f 0
f 0
i
i y
x
= ∑
∑
Example 2: A string or cable is hung from a horizontal ceiling from two points A and D. The string AD, at two points B and C weights are hung. At B, which is 0.6 m from a weight of 75 N is hung. C, which is 0.35 m from D, a weight of wc is hung. Compute wc such that the string portion BC is horizontal.
T NT N
T
T T
FBD
AB BC
AB
BC AB
sin 75 0
f 0
cos 0
f 0
of B
1
y
1
x
i
i
∑
∑
W N
T W
T N
T T
FBD
c
CD c
CD
BC CD
- 57
sin 0
f 0
- 85
cos 0
f 0
of C
2
y
2
x
i
i
=
− =
=
=
− + =
=
∑
∑
θ
θ
Example 3: A block of weight 120N is kept on a smooth inclined plane. The plane makes an angle of 320 with horizontal and a force F allied parallel to inclined plane. Compute F and also normal reaction.
Example 4: Three smooth circular cylinders are placed in an arrangement as shown. Two cylinders are of radius 052mm and weight 445 N are kept on a horizontal surface. The centers of these cylinders are tied by a string which is 406 mm long. On these two cylinders, third cylinder of weight 890N and of same diameter is kept. Find the force S in the string and also forces at points of contact.
N N
F N
Sin
NR
Sin
F
Sin
R
o o o
- 76
F 598 N
F 598N
of A
BA
AC
FBD
R N
F N
f
f
FBD
D
BC
y
x i
i
of B
∑
∑
- Compute moment arm
- Also compute x- intercept as
- And Y intercept as
Example 1: Compute the resultant for the system of forces shown in Fig 2 and hence compute the Equilibriant.
R d (^) R =^ ∑ Mo
R =^ ∑ ∑ fx io X M
=^ ∑
x i R o f y M
28.8 KN
44.8 - 32 cos 60 o
∑ fxi^ =
KNM
M
f
o o^ o
yi
- 4 ( 3 ) 32 cos 60 ( 4 ) 32 sin 60 ( 3 )
R 44.6KN
- 34.11KN
8 - 14.4- 32 sin 60
R^ o
o
∑
∑
ς
α
y^6228 ..^3482. 164 m
x^6234 ..^34111. 827 m
d 4462 .. 6434 1. 396 m
R
R
R
Example 2: Find the Equilibriant for the rigid bar shown in Fig 3 when it is subjected to forces.
- Resultant and Equilibriant
Equilibrium: The concept of equilibrium is the same as explained earlier. For a system of Coplanar Non concurrent forces for the status of equilibrium the equations to be satisfied are
The above principles are used in solving the following examples.
R^ o
y
x f KN
f i
i
=
∑
∑
α
- 1462 KNM
ς+ (^) ∑ MA =− + −
∑ f^ x (^) i =^0 ;^ ∑ fyi =^0 ;∑ Mo =^0 ;
SUPPORT REACTIONS IN BEAMS: Beams are structural members which are generally
horizontal. They are subjected to lateral forces which act orthogonal to the length of the member.
There are various types of mechanisms used for supporting the beams. At these supports the
reactive forces are developed which are determined by using the concept of equilibrium. The
different types of supports are depicted in the table below.
SUPPORT REACTION NO.OF REACTIONS
ROLLER
HINGE
FIXED
VA
TYPES OF LOADS ACTING ON BEAMS: There are various types of forces or loads which
act on beams. They are (a) Concentrated or point load (b) Uniformly distributed load (UDL) (c)
Uniformly varying load (UVL) (d) Arbitrary distributed load. The methodology of converting
UDL, UVL to equivalent point load is shown in the Fig below.
Some example problems of determining support reactions in beams are illustrated next.
Example 6: Determine the support reactions for the beam shown in Fig 9 at A and B.
Example 7: Determine the support reactions for the beam shown in Fig 10 at A and B.
Review
- Coplanar system of Forces.
- Concurrent, Non Concurrent.
- Resultant, Equilibrium.
- Concept of Equilibrium.
- Examples.
- Analysis of Trusses
H KN
H
f
A
A
xi
- 32
∑ =
45 KN; 10 KN
− × + − + − − =
∑
∑
B A
B
A
A B
A B
y
V V
V
M
V V
V V
f (^) i
ς
62 ;
24 ;
12 ;
( 10 ) 20 ( 6 ) 0
0 ;
- 866 20
0 ; 20 cos 30 0
- 5
sin 30 0
0 ; 0
H KN
R KN
V KN
V
M
V R
f V R
H R
H R
f
A
B
A
A
B
A B
o y A B
A B
A B
x
i
i
=
=
− =
− + =
= − + =
=
− =
=
∑
∑
∑
ς
ANALYSIS OF PLANE TRUSSES: Trusses are special structures which are formed by joining different members. Trusses are used as part of roofing systems in industrial buildings, factories workshops etc. Prominent features of trusses are
- Trusses are articulated Structures.
- The basic Geometry used in a truss is a triangle.
- Every member is pin connected at ends.
- Trusses carry loads only at joints. Joints are junctions where members meet.
- Self weight is neglected.
- The forces in various members of the truss are axial in nature.
A typical figure of a plane truss and the scheme by which truss configuration is arrived at is shown by the following figures.
Plane Trusses
Truss configuration
- A truss is said to be perfect if m= 2 j – 3 where m Members; j Joints
• PDB = 23.21 KN
- ∑fyi=
- -10+PCD = 0
- PCD = 10 KN
- ∑fxi=
- -PBD – PBC cos θ =
- PBC = -29.02 KN
- ∑fyi=
- VB +PBC sin θ = 0
- 17.41 – 29.02 sin θ = 0
Sl.No Member Force Nature 1 AC 16.52 C 2 AD 23.21 T 3 CB 29.02 C 4 CD 10 T 5 DB 23.21 T
Example 2 : Analyse the truss shown in figure and hence compute member forces.
- ∑fxi=
- HA-10+10=0; HA = 0
- ∑fyi=
- VA+ VB – 20= 0
- VA+ VB= 20
- ζ + ∑MA = 0
- 10(4)-20(3)+10(4)+VE(6)=
- VE = 10 KN;
- VA =10 KN;
- Symmetrical o Geometry ; o Loads
- ∑fxi = 0
- PAC=
- ∑fyi = 0
- PAB + 10 =
- PAB = - 10KN
- tan θ = 4/
- θ=53.13o
- ∑fxi = 0
- -10 + PBD+PBC cos θ =
- PBD +0.6PBC =
- ∑fyi = 0
- -PBA− PBC sin θ =
- -(-10)-0.8 PBC = 0
- PBC= 12.5 KN
- PBD =2.5 KN
- ∑fxi = 0
- PAC cos 300 + PAD = 0
- ∑fyi = 0
- 1.67+PAC sin 300 = 0
- PAC = -3.34 KN
- PAD = 2.89 KN
- ∑fxi = 0
- -PDC cos 600 -2.89 +PDB = 0
- ∑fyi = 0
- PDC sin 600 – 5 = 0
- PDC = 5.77 KN
- PDB = 5.77 KN
- ∑fxi = 0 – PBC cos 300 –5.77 = 0
- PBC = -6.66 KN
Sl. No Member Force Nature 1 AC -3.34 KN C 2 AD 2.89 KN T 3 BC 6.66 KN C 4 BD 5.77 KN T 5 CD 5.77 KN T
- Method of Sections: Another method of analysis of trusses is method of sections wherein which the concept of equilibrium of a system of coplanar non concurrent forces is made use of. The concept of free body diagram is an important part in this method. This method will be very useful when only few member forces are required. The equation of moment equilibrium becomes an important tool in this method. The method is illustrated in following figure.
PROCEDURE FOR METHOD OF SECTIONS
- Step 1: Compute support reactions (if need be).
- Step 2: Place the section to cut not more than three members.
- Step 3: Write FBD, unknown forces away from section(T).
- Step 4: Use equilibrium concept to get member forces This procedure is used for analyzing some examples as shown below.
