Indeterminacy STATIC INDETERMINACY. Where, Static Indeterminacy I=Internal Static Indeterminacy E= External Static Indeterminacy

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1 Indeterminacy After calculating the load on the structure it becomes necessary to find the ways by which the structure can be solved. A structure can be solved by two means 1) Statically 2) Kinematically In Statical method we solve the structure by treating the reaction and internal force as unknown. In Kinematics, we solve the structure by finding the displacement and rotation as unknown. Determinate A structure is said to be determinate if total number of unknown is equal to total number of equilibrium equations. Indeterminate A structure is said to be indeterminate if total number of unknown is more than total number of equilibrium equations. Unstable A structure is said to be unstable if total number of unknown is less than total number of equilibrium equations STATIC INDETERMINACY Where, Static Indeterminacy I=Internal Static Indeterminacy E= External Static Indeterminacy F=Number of unknown reactions U=Equation of static equilibrium available m = 3 for 2-Dimension and m=6 for 3-Dimension R=Any additional equation available. 2-Dimension Beams:- For beam generally the internal Indeterminacy is zero, as no closed loop can be formed in beam For calculating the number of unknown Reaction(1) Roller gives one reaction (2) Hinge gives two reaction (3) Fixed support gives three reaction U=Equation of static equilibrium available =3 1 Example:- Find the Indeterminacy of the beams. F=2+1=3 U=3 E=3-3=0 hence beam is statically determinate. F=2+1+1=4 U=3 E=4-3=1 Beam is indeterminate of 1 degree. F=3+1=4 U=3 E=4-3=1 Beam is indeterminate of 1 degree F=2+2=4 U=3 E=4-3=1 Beam is indeterminate of 1 degree F=3+3=6 U=3 E=6-3=3 Beam is indeterminate of 3 degree

2 F=1+2+3=6 U=3 E=6-3=3 Beam is indeterminate of 3 degree F= =6 U=3 E=6-3=3 Beam is indeterminate of 3 degree F=2+2+2=6 U=3 E=6-3=3 Beam is indeterminate of 3 degree Internal Hinge:-it is denoted by it is just like a pin in the structure, which can transmit the vertical load but unable to transmit the bending moment from one joint to other. Hence, it provide additional equation =n-1 where n is the number of members joined at the hinge. If the number of members jointed are 2 than additional equation provided will be 1. If 3 members are jointed at the hinge than additional equation provided will be 2 and if the number of members joined are 4 than the additional equation provided will be 3. Example:- F=3+1=4 U=3 R=2-1=1 E=4-3-1=0 hence beam is statically determinate 2 F=2+1=3 U=3 R=2-1=1 E=3-3-1=-1 hence beam is unstable F= =6 U=3 R=2-1=1 E=6-3-1=2 F=3+3=6 U=3 R= =2 E=6-3-2=1 2-Dimension Frame:- A frame is a structure comprising of beam and column jointed by rigid joint. Rigid Joint :- A joint is said to be rigid, if it can transfer the axial load, shear force and bending moment from the one member to the other member without any change in the angle between the members. All the member in the flexural frame in this course will be treated like rigid joint. For calculating the number of unknown Reaction(1) Roller gives one reaction (2) Hinge gives two reaction (3) Fixed support gives three reaction U=Equation of static equilibrium available =3 )

3 Example:- Find the Indeterminacy of the Frames:- E=2+2-3=1 I=0 E=2+1-3=0 I=0 E=3+3-3=3 I=0 E=3-3=0 I=0 E=3+3-3=3 I=0 E=3+2-3=2 I = 3 E= =5 I=3*2=6 E= =5 I=3*0=0 E=3+3-3=3 E=3+3-3=3 I=3*0=6 R =2-1=1 I=3*0=6 R =2-1=1 E= =7 I=3*3=9 R =(3-1)+(3-1)=4

4 TREE METHOD Every tress with fixed base is a determinate structure. As shown in Fig. the tree with many branches is same as a structure with a column with fixed support jointed to beam with rigid joint. In this method we divide the structure into tree shape by providing the cuts. The Indeterminacy of the structure = Where, C= Number of cuts and R is no of constraint applied to make the support fixed or joint to be rigid (if joint is not rigid) Example:- Find the Indeterminacy of the Frames E=3+2-3=2 I= 3*2=6 4 E= =8 I=0 E= =7 I=0 E=2+2-3=1 I=3*4=12

5 E= =5 I=3*2=6 R =4-1=3 E= =5 I=3*2=6 R =2-1=1 E= =2 I=3*0=0 R =0 5 Special When the structure is not having external support than,as shown below than the indeterminacy = I=3*4=12 E=-3 At point B horizontal displacement and rotation is permitted. E=3+3-3=3 Two additional equation is available R=2 Link link is structure used to join two different structure. The point where link is joined, will have only rotation but no vertical or horizontal deflection. Hence link gives two additional information. As shown in next diagram.

6 CD is a Link one additional equation at D and one additional equation at C hence E= =3 R=2 Stability of Structure There should be no reactions that are neither concurrent (concurrent mean meeting at a point) nor parallel The structure shown on left is unable as both the roller reaction meets at a point (concurrent) The beam shown on left is unstable as all the reactions are parallel 6 Space Frame In space Frame the fixed support gives 6 reaction ( the hinge gives 3 reaction ( the two side roller gives one reaction the one side roller gives two reactions ( The available equations will be 6 ( For internal indeterminacy the m=6 For using tree method one cut will provide 6 indeterminacy. The internal hinge will be 3*(n-1) additional equations where n is number of member joined at internal hinge. E= =18 I=6*1=6 Using Tree method 4 cuts are required hence

7 E= =18 I=6*1=6 R=3*(3-1)+3*(2-1)=9 When not given treat the roller as two way roller E= =10 I=6*1=6 E= =14 I=2*6=12 R=3*(4-1)+3*(4-1)+3*(2-1)=21 7 E= =14 I=(7-1)*6=36 R=3*(5-1)+3*(5-1)+3*(2-1) =27 E= =24 I=(11-2)*6=54 R=0

8 Kinematic Indeterminacy In Kinematic Indeterminacy we measure the total number of degree of freedom possible at joints. Degree of freedom is displacement and rotation at a joint and it is opposite of reaction. For 2 dimension, Fixed support gives 3 reaction viz is one rotation and 2 translation hence degree of freedom at fixed support is 0, Hinge gives 2 reaction both translation and no rotation reaction hence degree of freedom at hinge is 1 viz is rotation, and roller gives 1 reaction in translation direction hence degree of freedom at roller is 2 viz is one rotation and one translation. Similarly, for a 2 dimension rigid joint or free end the number of degree of freedom at a joint is 3 (one rotation and 2 translation). For internal hinge, the degree of freedom is 4, two rotational and 2 translational. Generally K.I=3*j r + i where j is no. of joint and r is number of reactions and Ii is no. of internal hinge. For 3 dimension, Fixed support gives 6 reaction viz is 3 rotation and 3 translation hence degree of freedom at fixed support is 0, Hinge gives 3 reaction all in translation and no rotation reaction hence degree of freedom at hinge is 3 viz is rotation in all three direction, and roller gives 1 reaction in translation direction hence degree of freedom at roller is 5 viz is 3 rotation and 2 translation, For a 3 dimension rigid join or free end the number of degree of freedom at a joint is 6 (3 rotations and 3 translations) For internal hinge, the degree of freedom is 9, 6 rotational and 3 translational. Ignoring axial deformation if we ignore axial deformation than the kinematic indeterminacy will decrease Kinematic indeterminacy after ignoring the axial deformation is kinematic indeterminacy-number of members in the structure. 8 K.I = 0+3=3 Ignoring axial deformation K.I = 3-1=2 j=2 r=3 K.I=3*2-3=3 Ignoring axial deformation K.I = 3-1=2 K.I = 1+2=3 Ignoring axial deformation K.I = 3-1=2 j=2 r=3 K.I=3*2-3=3 Ignoring axial deformation K.I = 3-1=2 K.I = 0+2=2 Ignoring axial deformation K.I = 2-1=1 j=2 r=4 K.I=3*2-4=2 Ignoring axial deformation K.I = 2-1=1 K.I = 1+2+2=5 Ignoring axial deformation K.I = 5-2=3 j=3 r=4 K.I=3*3-4=5 Ignoring axial deformation K.I = 5-2=3 K.I = 2+1=3 Ignoring axial deformation K.I = 3-2=1 j=3 r=6 K.I=3*3-6=3 Ignoring axial deformation K.I = 3-2=1

9 K.I = 3+2+1=6 Ignoring axial deformation K.I = 6-3=3 j=4 r=6 K.I=3*4-6=6 Ignoring axial deformation K.I = 6-3=3 K.I = 1+4+2=7 Ignoring axial deformation K.I = 7-2=5 j=3 r=3 i=1 K.I=3*3-3+1=7 Ignoring axial deformation K.I = 7-2=5 K.I = =9 Ignoring axial deformation K.I = 9-3=6 j=4 r=4 i=1 K.I=3*4-4+1=9 Ignoring axial deformation K.I = 9-3=6 K.I = =14 Ignoring axial deformation K.I = 14-5=9 j=6 r=6 i=2 K.I=3*6-6+2=14 Ignoring axial deformation K.I = 14-5=9 j=4 r=6 K.I=3*4-6=06 j=4 r=3 K.I=3*4-3=09 Igoring axial deformation K.I = 6-3=3 Ignoring axial deformation K.I=09-3=06 9. j=4 r=4 K.I=3*4-4=08 Ignoring axial deformation K.I = 8-3=5 j=5 r=3 K.I=3*5-3=12 Ignoring axial deformation K.I = 12-4=8 j=6 r=3 i=1 K.I=3*6-3+1=16 Ignoring axial deformation K.I = 16-5=8 j=7 r=11 K.I=3*7-11=10 Ignoring axial deformation K.I = 10-6=4

10 Find the reaction and draw the bending moment diagram for the beam having internal hinges. Taking moment about at left side of the internal hinge. Σ Mc=0 Taking moment about right side of the internal hinge Σ Mc=0 Taking moment about A Σ M A =0 Solving 1 and 2 The BMD is shown in the above Fig. The free body diagram method to solve the above problem is as follows. Find the reaction for the beams shown in below 10 Since C is a hinge, hence bending moment at C is zero on either side of the C. Taking right side of the C Since C is a hinge, hence bending moment at C is zero on either side of the C. Taking right side of the C

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