Presentation on theme: "EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM"— Presentation transcript:
1 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 = Eq(1)M = 0These requirements are both necessary and sufficient conditions for equilibrium.Back
2 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.Types of supportsThere 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.
3 BODY BODY Action on body Types of Supports Action on bodyTypes of Supports(a) Flexible cable ,belt ,chain, ropeBODYBODYTForce exerted by cable is always a tension away from the body in the direction of cableContact forces are normal to the surfaces(b) Smooth surfacesFF
4 (c) Roller supportContact 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.)
5 ( d )pinned Support / hinged support RhθRvRThis 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.
7 (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.
8 (a) Simply supported beam TYPES OF BEAMSA 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.ABHAMAVAspanspan(a) Simply supported beam(b) Cantilever beam
9 (d) Propped 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.ABHAMAVAspan(c) Overhanging beam (right overhang)(d) Propped Cantilever beam
10 (f) Two Span continuous 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.HAHBHAMAMBspanVCVAVBVAVB(f) Two Span continuous beam(e) Fixed beam
11 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.
12 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 beamIf the number of reaction components is less than the number of non-trivial equilibrium equations available then such a beam is an unstable beam.
13 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 forcesand 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
14 Problems(1)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.20kN25kN30kN3045AB4m3m3m2m
15 Problems Solution HA B 20kN 25kN 30kN 30 45 4m 3m 2m VAVBFBD Fx = HA – 25cos 30 – 30cos = 0 Fy = VA – 20 – 25 sin30 – 30sin45 +VB = 0MA = × sin30× sin 45×10+ VB ×12=0
26 Rc Solution VD VB VA FBD of top beam FBD of bottom beam 40kN 20kN C D
27 Solution For top Beam : Fy = 0 Rc –40 –20+VD=0 40kNSolution20kN2m3.33mRC0.67VDFor top Beam : Fy = Rc –40 –20+VD=0 MD = Rc × × × 3.33=0Solving the above eqnsRC=37.77kN; VD=22.23kN
28 Solving the above eqns VA=22.66kN; VB=15.10kN (Contd.)For bottom beam : Fy = VA –37.77–VB=0 MB = VA× ×3=0Solving the above eqnsVA=22.66kN; VB=15.10kNRC=37.77kN2mVA3mVB
29 (6) A ladder of length 5m has a weight of 200N (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.
34 Solution (contd.) Fx = 0 60 –HA=0 Fy = 0 VA+VC=0 MA = 0 VCx2-60 ×1.5=0 BC3mVC60kNFBDHAVA2m
35 Solving the aboveAns: VA = - 45kNVC = 45 kNHA = 60kNBCRA=75 kN60kNVC36.90FBD after finding reactns# ve sign for VA indicates ,reaction is downwards and not upwards as assumed initially.HAVA
36 (8) A roller (B) of weight 2000N rests as shown in the fig (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.CDBA30°4m1m
37 Solution: RB FBD of beam CD 2000N RAB FBD of Roller Hc 300 Vc 500N 1m 30°2.5mDVD
38 Solving above eqns : RBCD=2309.4N; Solution:2000NRABFBD of RollerRBCD300FBD of Roller : Fy = RBCD cos 300 –2000=0 Fx = RAB – RBCD sin 300 =0Solving above eqns : RBCD=2309.4N;RAB=1154.7N
39 Solving the above eqns: VD=783.33N; VC=1716.67N Fx = 0 For bottom beam : Fy = VD –500+Vc –2309.4cos30=0 MC = 0-VD × 5cos × 2.5 × cos × 1=0Solving the above eqns: VD=783.33N; VC= N Fx = 0sin 30 –HC = Hc= N2309.4N300HcVc500N1m1.5mFBD of beam CD30°D2.5mVD
40 (9) Compute the reactions for the bent beam shown in the figure at A and F. 300Nm50 N/mBCDA45°F4m3m3m4m
45 Problems for practice(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= ˚ )12kN/m20 kN12kN/m4kN/m4kN/m30kN3A4BC40kNm1m2m1m1m2m1m1m2m
46 (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˚)string2.5mB60˚A200N2.5m2.5m
47 (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.)15kN18kN/m10kN/mx2.0m1.0m1.4m0.63.0m
48 (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˚)AL mθBWC0.5L2W