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Solid Mechanics 1. Shear force and bending moment diagrams Internal Forces in solids Sign conventions Shear forces are given a special symbol on yV12 and zV The couple moment along the axis of the member is given xM T = =Torque y zM M = =bending moment. Solid Mechanics We need to follow a systematic sign convention for systematic development of equations and reproducibility of the equations The sign convention is like this. If a face (i.e. formed by the cutting plane) is +ve if its outward normal unit vector points towards any of the positive coordinate directions otherwise it is ve face A force component on a +ve face is +ve if it is directed towards any of the +ve coordinate axis direction. A force component on a ve face is +ve if it is directed towards any of the ve coordinate axis direction. Otherwise it is v. Thus sign conventions depend on the choice of coordinate axes. Shear force and bending moment diagrams of beams Beam is one of the most important structural components. Beams are usually long, straight, prismatic members and always subjected forces perpendicular to the axis of the beam Two observations: (1) Forces are coplanar Solid Mechanics (2) All forces are applied at the axis of the beam. Application of method of sections What are the necessary internal forces to keep the segment of the beam in equilibrium? xyzF PF VF M = = = 000 The shear for a diagram (SFD) and bending moment diagram(BMD) of a beam shows the variation of shear Solid Mechanics force and bending moment along the length of the beam. These diagrams are extremely useful while designing the beams for various applications. Supports and various types of beams (a) Roller Support resists vertical forces only (b) Hinge support or pin connection resists horizontal and vertical forces Hinge and roller supports are called as simple supports (c) Fixed support or built-in end Solid Mechanics The distance between two supports is known as span. Types of beams Beams are classified based on the type of supports. (1) Simply supported beam: A beam with two simple supports (2) Cantilever beam: Beam fixed at one end and free at other (3) Overhanging beam (4) Continuous beam: More than two supports Solid Mechanics Differential equations of equilibrium [ ]xF = + 0 yF (= + 0 V V V P xV P xVPx + + == = 0 xV dVPx dxlim= = 0 [ ]AP xM V x M M M = + + =20 02 P xV x MM P xVx + =+ =20202 Solid Mechanics xM dMVx dxlim= = 0 From equation dVPdx = we can write DCXD CXV V Pdx = } From equation dMVdx = D CM M Vdx = } Special cases: Solid Mechanics Solid Mechanics Solid Mechanics Solid Mechanics ( ) ( ) x 0 2 1 1A BVVV ; V === =5 055 5( ) ( )( )( )( )B CxV . xV . xV ; V. xx . + == + = = + = =2 6 2 25 30 7 5 2 05 30 7 5 225 525 7 5 2 05 33( ) ( )C DxVVV ; V + == += + = +6 8 3 35 30 30 10 01515 15( ) ( )D ExVVVV ; V + + =+ == = = 8 10 4 45 30 30 10 20 05 055 5x ( ) ( )x ( ( )x ( ) ( )x ( ) ( ) 0 2 1 12 6 2 26 8 3 38 10 4 4 Solid Mechanics Problems to show that jumps because of concentrated force and concentrated moment ( ) ( )A BxM xM xM ; M + == += + =0 2 1 110 5 05 1010 0( ) ( )( ) ( )( ) ( )Ex .Cxx. xM x x. xM x xM .M== + + == + = +=225 3362 6 2 27 5 210 5 30 2 027 5 210 5 30 2241 6640( ) ( ) [ ]( ) ( ) ( )CxDxx C DM x x x xMM== + + + + == += 686 8 3 310 5 30 2 30 4 10 6 20 02010[ ] ( ) ( )( ) ( ) ( ) ( )Exx D EM x x x x xM= + + + + ==88 10 4 410 5 30 2 30 4 10 6 20 20 8 00 Solid Mechanics We can also demonstrate internal forces at a given section using above examples. This should be carried first before drawing SFD and BMD. [ ] x A B 0 2 Solid Mechanics ABVVVV ====5 0555A BM xM xM ; M + == = =10 5 010 510 0[ ] x B C 2 6( )( )( )B CV . xV . xV ; V. xx . + == + = = + ==5 30 7 5 2 07 5 2 5 3025 525 7 5 2 05 33( ) ( )CEBxM x x .xMM x . .xM + + ==== ===2210 5 30 2 7 5 026405 33 41 6620[ ] x C D 6 8C DVVV , V + === =5 30 10 30 01515 15 Solid Mechanics [ ] x D E 8 10D EVVV , V + + == = = 5 30 10 30 20 055 5 Solid Mechanics [ ][ ]x Axy AyAyF RF RR kNM M .M k m + = = ( + = + = = = + == 0 00 60 90 0300 60 90 4 5 0285( )( )V xV x+ + == = = =30 60 30 3 030 3 9030 3 9090 900( )B AB AM MM M = = + = = 6060 60 285225 Solid Mechanics ( )C BC BM MM M = = + = += 9090 225 90135( )D CD CM MM M = = + = + =135135 135 135 0yAy CyAy CyFR RR R ( ) ( + = + =+ =0200 240 0440 1[ ]ACyCyAyMRR kNR kN = + == = 0200 3 240 4 8 0195245V xV xVV+ == = = =245 200 30 030 4530 8 45 240 45195 Solid Mechanics * M .M .M + = =245 3 90 1 5245 3 90 1 5600[ ]Ay ByA ByByByAyR RM RRR kNR kN+ = = + + + = + + ===320 32 2 18 8 4 064 16 4 01220 Solid Mechanics Problem: [ ]( )xAxy Ay Dy Ay DyFRF R R R R + == ( = + + = + = 000 60 50 0 110 1( )C AC AM MM M = = + = + =5050 8 25 17V xV xxx / .+ == == =20 8 08 208 20 020 8 2 5[ ]A DyDyAyM . RR kNR kN = + == = = 0 60 1 5 50 4 5 029058552 Solid Mechanics ( )yBF V xV x x m ( = + + = | |= |\ .0 52 20 020 52 0 3[ ]( )MxM xxM x x m =+ == 2202052 022052 0 32yB CFVV kN x m ( = + + =| |= |\ .052 60 08 3 4[ ] ( )( )B CM M x x .M x x . x m = + =| |= |\ .0 52 60 1 5 052 60 1 5 3 4 Solid Mechanics B EBM M .M . . = = +1 61 6 67 6x / . m == =20 52 052 20 2 6dMVdxdVPdx= = [ ] ( ) ( )( ) ( ) ( )M M x x . xM x x . x x = + + == 0 52 60 1 5 50 4 052 60 1 5 50 4 4 5( )yFVV kN x ( = + + == 052 60 50 058 4 5 Solid Mechanics D CD CM MM M = = += =585858 58 0C BC BM MM M = = += + =888 66 58B EB EM M .M . M . . = = + = +=1 61 6 1 6 67 666x / . == =20 52 052 20 2 6dMVdxdVPdx= = B AM M Vdx = } Solid Mechanics 2. Concept of stress Traction vector or Stress vector Now we define a quantity known as stress vector or traction as =,,RnAFTAlim0 units aP N / m 2 and we assume that the quantity ,RAMAlim00 (1) nT, is a vector quantity having direction of RF , (2) nT, represent intensity point distributed force at the point "P" on a plane whose normal is n (3) nT, acts in the same direction as RF , Solid Mechanics (4) There are two reasons are available for justification of the assumption that ,RAMAlim00 (a) experimental (b) as A 0, RF , becomes resultant of a parallel force distribution. Therefore RM = 0, for | force system. (5) nT,varies from point to point on a given plane (6) nT, at the same point is different for different planes. (7) n nT T = , , will act at the point P (8) In general Components of nT, R n t s F F n v t v s = + +, Solid Mechanics = = + +,,R n t snA A A AF F v v T n t sA A A Alim lim lim lim0 0 0 0 n nn nt ns T n t s = + +, where = = == = == = =n nnnAt tntAs snsAF dFNormal stresscomponentA dAv dvShear stresscomponentA dAv dvAnother shear componetA dAlimlimlim000 Normal StressShear stress n nndF dA = t ntdV dA = Notation of stress components The magnitude and direction of nT, clearly depends on the plane m-m. Therefore, stress components magnitude & direction depends on orientation of cut m-m. (a) First subscript- plane on which is acting (b) Second subscript- direction Solid Mechanics Rectangular components of stress Cuts to the coordinate planes will give more valuable information than arbitrary cuts. = = + +,,yR x zxA A A AvF F v T i j kA A A Alim lim lim lim0 0 0 0 x xx xy xz T i j k = + +, where xxxAyzxy xzA AFNormal stressAvvShear stress; Shear stressA Alimlim lim = == = = =00 0 Solid Mechanics =x xxdF dA y xydv dA = z xzdv dA = Similarly, = = + +,yR x zyA A A AFF v v T i j kA A A Alim lim lim lim0 0 0 0 = + +,y yx yy yz T i j k = + +,z zx zy zz T i j k xx and xy will act only on x-plane. We can see x and xy only when we take section to x-axis. The stress tensor ( ( ( = ( ( ( xx xy xzjj yx yy yzzx zy zzRectangular stresscomponents This array of 9 components is called as stress tensor. It is a second rank of tensor because of two indices Components a point P on the x-plane in x,y,z directions Solid Mechanics These 9 rectangular stress components are obtained by taking 3 mutually planes passing through the point P Stress tensor is an array consisting of