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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 F x = 0; F y = 0 Eq(1) M = 0 These requirements are both necessary and sufficient conditions for equilibrium. Back

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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

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Types of Supports Action on body (a) Flexible cable,belt,chain, rope BODY T Force exerted by cable is always a tension away from the body in the direction of cable (b) Smooth surfaces Contact forces are normal to the surfaces F F

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(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.)

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( 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. RvRv R RhRh θ

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RvRv M RHRH (e) Fixed or Built-in Support M

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(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.

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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 MAMA VAVA B A HAHA (b) Cantilever beam

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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) MAMA VAVA B span A HAHA (d) Propped Cantilever beam

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A beam which is fixed at both ends is called a fixed beam. A beam with more than one span is called continuous beam. VCVC (f) Two Span continuous beam VAVA VBVB HAHA HBHB MAMA VAVA span VBVB (e) Fixed beam MBMB HAHA

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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.Eq(1) In the above fig (a),(b)and (c) are statically determinate beams, where as (d),(e) and (f) are statically Indeterminate beams.

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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.

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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 F x = 0, F y = 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

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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) A B 20kN25kN 30kN m3m 2m

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Problems Fx = 0 H A – 25cos 30 – 30cos45 = 0 Fy = 0 V A – 20 – 25 sin30 – 30sin45 +V B = 0 M A = × sin30 × sin 45 × 10+ V B × 12=0 Solution HAHA B 20kN25kN 30kN m 3m 2m VAVA VBVB FBD

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Problems R A = kN Solution(contd.) VAVA HAHA RARA H A =42.86kN, V A =22.07kN, V B =31.64kN = 27.25

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(2) Find the Support reactions for the given beam loaded as shown in the figure. 60° 2m 40kN/m A 60kN 0.5m 5m 1 m B

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Solution [Ans:R B =140kN V A =10 H A =61.24 R A = 62.05kN = 9.3 ] 60° 2m 40kN/m A 60kN 1m B R BH =R B Cos30 RBRB R Bv = R B Cos60 C HAHA VAVA F x = 0 H A + 60 – R B Cos30 = 0 F y = 0V A + R B Cos60 – 40 x 2 = 0 M A = ×2×4 + R B Cos60×5 = 0 30kNm HAHA VAVA Solution FBD 2m RARA

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(3) Find the Support reactions for the given beam loaded as shown in the figure. 80kN/m 100kN 3m 1m 30kN 0.5m 2m A B

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Solutions 80kN/m 100kN 3m 1m 30kN 2m A B VAVA HAHA VBVB 15kNm FBD

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[ Ans: V B = 112.5kN V A =37.5kN H A = – 100kN R A = 106.8kN = ] 1m 120kN A 6 m CB 15kNm 30kN 2m RARA HAHA VAVA Fx = 0 H A = 0 Fy = 0 V A + V B – 30 –120 = 0 M A = × (120)x5 + V B x6 = 0 HAHA VAVA VBVB 100kN FBD

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(4) Find the Support reactions for the beam loaded as shown in the figure. 3m 2m 20kN23kN 30kN15kN/m

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Fx = 0 H A = 0 Fy = 0 V A –45 –30 –23 –20 = 0 M A = 0 M A –45x1.5 –30x3 –23x5 –20x7=0 [ Ans: V A = 118kN M A =412.5kNm] Solution 2m 20kN 23kN 30KN 45kN MAMA VAVA HAHA 1.5m FBD A ;

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(5) 2m3m1m2m A C B D 10KN/m 20KN/m Find reactions at A,B,C and D

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Solution 2m3m1m2m A C B D 10kN/m RcRc 3m1m2m

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Solution 2m3m VAVA C RcRc 40kN 20kN 1.33m VDVD VBVB 2.0m FBD of top beam FBD of bottom beam CD AB 0.67m 2m

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Solution For top Beam : Fy = 0 Rc –40 –20+V D =0 M D = 0 -Rc × × × 3.33=0 Solving the above eqns R C =37.77kN; V D =22.23kN 20kN 2m 0.67 RCRC VDVD 3.33m 40kN

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(Contd.) For bottom beam : Fy = 0 V A –37.77–V B =0 M B = 0 -V A × ×3=0 Solving the above eqns V A =22.66kN; V B =15.10kN 2m 3m R C= 37.77kN VAVA VBVB

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(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.

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FBD OF LADDER 300N HBHB VAVA HAHA 200N 2.5m 60 0 Solution

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F y = 0 V A – 200 – 300=0 ::V A =500N M A = 0 H B x 5 sin60 – 200 × 2.5 cos 60 – 300 × 5cos60=0 :: H B =230.94N F x = 0 H A –H B =0 H A =230.94N(Ans.) 300N H B VAVA HAHA 200N 2.5m 60 0 Solution

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(7) Find the reactions at the supports A and C of the bent 20 kN/m B C 3m 2m A

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Solution 20 kN/m B C 3m 2m VAVA HAHA FBD X Y VCVC

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Solution (contd.) B C 3m 2m VAVA HAHA FBD 60kN Fx = 0 60 –H A =0 Fy = 0 V A +V C =0 M A = 0 V C x2-60 ×1. 5=0 VCVC

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Solving the above Ans: V A = - 45kN V C = 45 kN H A = 60kN FBD after finding reactns R A =75 kN B C VAVA HAHA 60kN VCVC # ve sign for V A indicates,reaction is downwards and not upwards as assumed initially.

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(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 B A 30° 4m 1m

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Solution: 2000N R AB RBRB VDVD FBD of Roller D 30° 2.5m 1m N Hc Vc 1.5m FBD of beam CD

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Solution: 2000N R AB 30 0 R BCD FBD of Roller FBD of Roller : F y = 0 R BCD cos 30 0 –2000=0 F x = 0 R AB – R BCD sin 30 0 =0 Solving above eqns : R BCD =2309.4N; R AB =1154.7N

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For bottom beam : Fy = 0 V D –500+Vc –2309.4cos30=0 M C = 0 -V D × 5cos × 2.5 × cos × 1=0 Solving the above eqns: V D =783.33N; V C = N Fx = sin 30 –H C =0 Hc= N FBD of beam CD D 30° 2. 5 m 1m N Hc Vc 1.5m N VDVD

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(9) Compute the reactions for the bent beam shown in the figure at A and F. B CD 3m 4m 50 N/m 45 ° A F 300Nm 4m

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Solution M F = 0 – V A × × 5 – 300=0 V A =50N F X =0 H F =0 F Y =0 V A + V F =200; V F =200 – 50=150N 18kN 9kN R A =20.12kN 45° A B C D F 3m 4m VAVA FBD 200 N 2m HFHF VFVF 300Nm

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(10) Determine the support reactions for the trees shown 4m A3KN G F BCD E 4m

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A3KN G F B CD EHBHB HAHA VAVA Solution FBD 4m

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M A = 0 H B × 4 – 3 × 4 – 3 × 8 – 3 × 12=0 H B =18kN F X =0 : –H A +H B =0 H A =18kN F Y =0 V A –3 –3 –3=0; V A =9kN 18kN 9kN R A =20.12kN 4m A 3KN HBHB HAHA VAVA 4m

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4kN/m 12kN/m (1)Find the reactions at A,B,C and D for the beam loaded as shown in the figure(Ans.R A =R B =34kN;R C =28.84kN; M C =-140kNm ; θ C = ˚ ) Problems for practice 12kN/m 4kN/m 20 kN 30kN 1m2m1m 2m1m 2m A B C kNm

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(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;R B =807.15N, θ B =64.6 ˚ ) 2.5m 200N 2.5m A B 60˚ string

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(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 1.4m 1.0m3.0m kN 18kN/m 10kN/m x

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(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˚) 0.5L 2W θ B L m W C A

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