EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction. Thus.
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EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM
When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction. Thus the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations for two dimensional force system
Fx = 0; Fy = 0 Eq(1)
M = 0These requirements are both necessary and sufficient conditions for equilibrium.
Back
Supports: A structure is subjected to external forces and transfers these forces through the supports on to the foundation. Therefore the support reactions and the external forces together keep the structure in equilibrium.
There are different types of supports. Some of them are a) Roller Support b) Hinged or pinned support c) Fixed or built in support Some supports are shown in the figure along with the reactions that can be mobilised.
Types of supports
Types of Supports Action on body
(a) Flexible cable ,belt ,chain, rope
BODYBODY
T
Force exerted by cable is always a tension away from the body in the direction of cable
(b) Smooth surfacesContact forces are normal to the surfaces
FF
(c) Roller support Contact force is normal to the
surface on which the roller moves. The reaction will always be perpendicular to the plane of the roller . Roller support will offer only one independent reaction component.(Whose direction is known.)
( d )pinned Support / hinged support
This support does not allow any translatory movement of the rigid body. There will be two independent reaction components at the support. The resultant reaction can be resolved into two mutually perpendicular components. Or it can be shown as resultant reaction inclined at an angle with respect to a reference direction.
RvR
Rh
θ
Rv
M
RH
(e) Fixed or Built-in Support
M
(contd .)This type of support not only prevents the translatory movement of the rigid body, but also the rotation of the rigid body. Hence there will be 3 independent reaction components of forces. Hence there will be 3 unknown components of forces, two mutually perpendicular reactive force component and a reactive moment as shown in the figure.
TYPES OF BEAMS A member which is subjected to predominantly transverse loads and supported in such a way that rigid body motion is prevented is known as beam. It is classified based on the support conditions. A beam generally supported by a hinge or roller at the ends having one span(distance between the support) is called as simply supported beam. A beam which is fixed at one end and free at another end is called as a cantilever beam.
(a) Simply supported beam
span
MA
VA
B
span
AHA
(b) Cantilever beam
If one end or both ends of the beam project beyond the support it is known as overhanging beam.
A cantilever with a simple support anywhere along its length is a propped cantilever.
(c) Overhanging beam (right overhang)
MA
VA
B
span
A
HA
(d) Propped Cantilever beam
A beam which is fixed at both ends is called a fixed beam.
A beam with more than one span is called continuous beam.
VC
(f) Two Span continuous beam
VAVB
HA HB
MA
VA
span
VB
(e) Fixed beam
MB
HA
Statically determinate beam and statically indeterminate beam:Using the equations of equilibrium given in Eq(1) ,if all the reaction components can be found out, then the beam is a statically determinate beam ,and if all the reaction components can not be found out using equations of equilibrium only, then the beam is a statically indeterminate beam.In the above fig (a),(b)and (c) are statically determinate beams, where as (d),(e) and (f) are statically Indeterminate beams.
If the number of reaction components is more than the number of non-trivial equilibrium equations available then such a beam is a statically indeterminate beam.
If the number of reaction components is equal to the number of non-trivial equilibrium equations available then such a beam is a statically determinate beam
If the number of reaction components is less than the number of non-trivial equilibrium equations available then such a beam is an unstable beam.
Determination of Beam reactions
Since three equilibrium equations are available, for a planar structure a maximum of three unknown independent reaction components can be determined using these equations.
Step I: Draw the free body diagram of the structure showing the given loadings and the reactions at the supports.
Step 2: Apply the equations Fx = 0, Fy = 0, M = 0.
Assuming some directions and senses for unknown forces and moments.
Step 3: solve for unknown reactions. If any of them is positive, it is along the sense initially assumed while drawing the FBD. If it is negative, it is opposite to the initially assumed sense
Problems
A beam AB of span 12m shown in the figure is hinged at A and is on rollers at B. Determine the reactions at A and B for the loading shown.
(1)
AB
20kN 25kN 30kN
30 45
4m 3m 3m 2m
Problems
Fx = 0 HA – 25cos 30 – 30cos45 = 0
Fy = 0 VA – 20 – 25 sin30 – 30sin45 +VB = 0
MA = 0 -20×4 - 25 sin30×7 - 30 sin 45×10+ VB ×12=0
Solution
HA B
20kN 25kN 30kN
30 45
4m
3m
3m 2m
VA VB
FBDFBD
Problems
RA= 48.21 kN
Solution(contd.)
VA
HA
RA
22
AAA VHR
HA=42.86kN, VA=22.07kN, VB=31.64kN
= 27.25
A
A
H
V1tan
(2) Find the Support reactions for the given beam loaded as shown in the figure.
60°
2m
40kN/m
A
60kN0
.5m
5m
1 m
B
Solution
[Ans: RB=140kN VA=10
HA=61.24 RA= 62.05kN
= 9.3]
60°2m
40kN/m
A60kN
1m
B
RBH=RBCos30
RB
RBv = RBCos60
C
HA
VA
Fx = 0 HA + 60 – RB Cos30 = 0
Fy = 0 VA + RB Cos60 – 40 x 2 = 0
MA = 0 -30 - 40×2×4 + RB Cos60×5 = 0
30kNm
HA
VA
Solution
FBD
2m
RA
(3) Find the Support reactions for the given beam loaded as shown in the figure.
80kN/m
100kN
3m1m
30kN 0.5m
2m
AB
Solutions
80kN/m
100kN
3m1m
30kN
2m
A
B
VA
HA
VB
15kNm
FBD
[ Ans: VB= 112.5kN VA =37.5kN HA= – 100kN
RA= 106.8kN = 20.56]
1m
120kN
A
6 m
C B
15kNm
30kN
2m
RA
HA
VA
Fx = 0 HA + 100 = 0
Fy = 0 VA + VB – 30 –120 = 0
MA = 0 - 30×2 - 15 - (120)x5 + VBx6 = 0
HA
VA
VB
100kN
FBD
(4) Find the Support reactions for the beam loaded as shown in the figure.
3m 2m 2m
20kN23kN30kN15kN/m
Fx = 0 HA = 0
Fy = 0 VA –45 –30 –23 –20 = 0
MA = 0 MA –45x1.5 –30x3 –23x5 –20x7=0 [ Ans: VA = 118kN
MA =412.5kNm]
Solution
2m 2m
20kN23kN30KN45kN
MA
VA
HA
1.5m 1.5mFBD
A
;
(5)
2m 3m 1m 2m
AC B
D
10KN/m
20KN/mFind reactions at A,B,C and D
Solution
2m 3m 1m 2m
A
C
B
D
10kN/m10kN/m
10kN/m
Rc
3m 1m 2m
Solution
2m 3mVA
C
Rc
40kN20kN
1.33m
VD
VB
2.0m
FBD of top beam
FBD of bottom beam
C D
A B
0.67m2m
Solution
For top Beam :
Fy = 0 Rc –40 –20+VD=0
MD = 0 -Rc × 6 +40 × 4 +20 × 3.33=0
Solving the above eqns
RC=37.77kN; VD=22.23kN
20kN
2m
0.6
7
RC VD
3.33m
40kN
(Contd.)For bottom beam :
Fy = 0 VA –37.77–VB=0
MB = 0 -VA× 5 +37.77 ×3=0
Solving the above eqns
VA=22.66kN; VB=15.10kN
2m 3m
RC=37.77kN
VAVB
(6) A ladder of length 5m has a weight of 200N. The foot of the ladder rests on the floor and the top of it leans against the vertical wall. Both the wall and floor are smooth. The ladder is inclined at 60 with the floor. A weight of 300N is suspended at the top of the ladder. Find the value of the horizontal force to be applied at the foot of the ladder to keep it in equilibrium.
FBD OF LADDER
300N
HB
VA
HA
200N2.
5m
2.5m
600
Solution
Fy = 0 VA – 200 – 300=0 ::VA=500N
MA = 0
HB x 5 sin60 – 200 ×2.5 cos 60 – 300 ×5cos60=0
:: HB=230.94N
Fx = 0 HA –HB=0
HA=230.94N(Ans.)
300N
HB
VA
HA
200N
2.5m
2.5m
600
Solution
(7) Find the reactions at the supports A and C of the bent 20
kN
/m
B C
3m
2m
A
Solution 2
0 k
N/m
B C
3m
2mVA
HA
FBD
X
Y
VC
Solution (contd.)
B C
3m
2m
VA
HA
FBD
60kN
Fx = 0 60 –HA=0
Fy = 0 VA+VC=0
MA = 0
VCx2-60 ×1.5=0 VC
Solving the above
Ans: VA = - 45kN
VC = 45 kN
HA = 60kN
FBD after finding reactns
RA=75 kN
36.90
B C
VA
HA
60kN VC
# ve sign for VA
indicates ,reaction is downwards and not upwards as assumed initially.
(8) A roller (B) of weight 2000N rests as shown in the fig. on beam CD of weight 500N.Determine the reactions at C and D. Neglect the weight of beam AB.
C
D
BA
30°4m
1m
Solution: 2000N
RAB
R B
VD
FBD of Roller
D
30°
2.5m
1m
300
500N
Hc
Vc
1.5m
FBD of beam CD
Solution: 2000N
RAB
300 R BCD
FBD of Roller
FBD of Roller :
Fy = 0 RBCD cos 300 –2000=0
Fx = 0 RAB – RBCD sin 300 =0
Solving above eqns : RBCD=2309.4N;
RAB=1154.7N
For bottom beam :
Fy = 0 VD –500+Vc –2309.4cos30=0
MC = 0
-VD × 5cos30 + 500 × 2.5 × cos30-2309.4 × 1=0
Solving the above eqns: VD=783.33N; VC=1716.67N
Fx = 0
2309.4 sin 30 –HC =0 Hc=1154.7 N
FBD of beam CD D
30°
2.5m
1m
300
500N
Hc
Vc
1.5m
2309.4N
VD
(9) Compute the reactions for the bent beam shown in the figure at A and F.
B C D
3m 3m4m
50 N/m
45°A
F
300Nm
4m
Solution
MF = 0 – VA × 14 +200 × 5 – 300=0
VA=50N
FX=0 HF=0
FY=0 VA +VF=200;
VF=200 – 50=150N18kN
9kN
26.570
RA=20.12kN
45°A
B C D
F
3m 3m4m 4mVA
FBD
200 N
2m
HF
VF
300Nm
(10) Determine the support reactions for the trees shown
4m 4m 4m
A 3KN
G3KN
F 3KN
B C D E
4m
4m 4m 4m
A 3KN
G 3KN
F 3KN
B
C D EHB
HA
VA
Solution
FBD
4m
MA = 0 HB × 4 – 3 × 4 – 3 × 8 – 3 × 12=0
HB=18kN
FX=0 : –HA+HB=0
HA=18kN
FY=0 VA –3 –3 –3=0;
VA=9kN
18kN
9kN26.570
RA=20.12kN
4m 4m 4m
A 3KN3KN
3KN
HB
HA
VA4m
4kN/m
12kN/m
(1)Find the reactions at A,B,C and D for the beam loaded as shown in the figure(Ans.RA=RB =34kN;RC=28.84kN;
MC=-140kNm ; θC=-33.69 ˚ )
Problems for practice
12kN/m
4kN/m
20 kN
30kN
1m 2m 1m 1m 2m 1m 1m 2m
AB
C
34
40kNm
(2)A uniform bar AB of weight 50N shown in the figure supports a load of 200N at its end. Determine the tension developed in the string and the force supported by the pin at B.(Ans. T=529.12N;RB=807.15N, θB=64.6˚)
2.5m2.5m
200N
2.5m
A
B
60˚
string
(3)Find the position of the hinged support (x),such that the reactions developed at the supports of the beam are equal..
(Ans.x=2m.)
2.0m 1.4m1.0m 3.0m0.6
15kN18kN/m
10kN/m
x
(4)A right angled bar ABC hinged at A as shown in fig carries two loads W and 2W applied at B &C .Neglecting self weight of the bar find the angle made by AB with vertical(Ans:θ =18.44˚)
0.5L2W
θ
B
L m
WC
A
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