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1 Strength of Material ZONE TECH C H TE E N ZO  2 E - Book ZONE TECH CHAPTER - 1 (STRESS AND STRA...

1 Strength of Material ZONE TECH C H TE E N ZO  2 E - Book ZONE TECH CHAPTER - 1 (STRESS AND STRAIN) Strength of Material :- SOM is that branch of science which deals with the behaviour of deformable body which are subjected to stress and strain Difference between SOM and Engineering mechanics is SOM Mechanics H (1) SOM deals with deformable body (1) Mechanics deals with rigid body (2) Energy will store in body i.e. strain energy (2) Energy will not store in body C Strain :- Strain is a measure of deformation which represents relative displacement with respect to original length or natural length [separation between molecules of body] Strain is represented as the ratio of change in dimension to the original dimension if load is TE acting along the length of body then Final length - Original length Strain = Original length L Strain = L E P P P P N L LL ZO Stress :- When a body is subjected to external force then deformation tendency will develope in a body. To resist this deformation molecules of body will provide a internal resistive force, this internal resistive force per unit area is known as stress. X ×× × ×× ×× ×× ×× P ×× Fi Fi ×× P × ×× ×× ×× × × X Free Body Diagram  3 Strength of Material ZONE TECH Internal Re sistance Force Fi Stress ()   Original Cross Sectional Area A At equilibrium : fi  P Fi P Hence = = A A Stress is a tensor quantity which means it is represented with the help of direction, magnitude and plane of a body. Type of Stress :- H Actual or True stress( C Stress Nominal or Converntional or Engineering stress Actual stress:- TE It is defined as the load divided by actual area. It is represented with the help of  T For tension case Actual  E For tension case Aactual < Ainitial E Engineering stress:- It is defined as the load divided by original area. It is represented with the help of E N For compression case Actual  E For compression case Aactual > Ainitial ZO A' P A P Relation between True stress and Engineering stress T  Engg  1    For tension (+) sign is used and vice versa. Relation between True strain and Engineering strain   True  l n 1  Engg  Engineering Stress - Strain curve for mild steel OR Tensile Test Diagram  4 E - Book ZONE TECH This test is always performed on mild steel because it exhibit all phenomenon like yielding, ultimate etc. For this test dog bone or dumble shape specimen is used. Specimen may be circular or non circular. For circular section gauge length  5.D0 For rectangular section gauge length  5.65 A 0 Deformation in gauge length is measured with the help of Extensometer. This test is performed on Universal Testing Machine (UTM). H strain - 0.002 C strain - 0.001 max stress concent TE lg  Guage Length Stress-Strain Curve E partial plastic zone elastic region E str ain N r ain sof ten B C' st den g in or n ing A har eck in g ×F C D  yield plateau ZO O E where Point A (Proportional Limit) :- Hook's law is valid upto proportional limit where stress  strain Point B (Elastic Limit) - Material will regain its original shape after unloading upto this point Point C' (Upper Yield Point) :- Upper Yield point depends on rate of loading Point C (Lower Yield Point) 5 Strength of Material ZONE TECH C-D Yield Plateau - In this zone strain will occur at constant stress or load D-E strain hardening zone :- This zone occurs due to reorientation of molecules Point E (Ultimate point) - Maximum stress is known as ultimate stress E-F strain softening zone :- It is also known as necking region that means reduction of area at cross section of the body. H Point F Fracture point or Break Point or Failure Point or Rupture Point C Engineering and True Stress - Strain curve for mild steel in tension TE ×  T  T  Curve  T  E 1    × So T  E Under Tension  E   E  Curve E N Engineering and True Stress - Strain curve for mild-steel in compression ZO E  E  Curve T  E 1    So E  T Under Compression   T   T  Curve  6 E - Book ZONE TECH CHAPTER - 2 (PROPERTIES OF MATERIAL) Important Properties :- (I) Elasticity :- Ability of material by virtue of which material will regain its original shape after unloading is known as elasticity and such type of materials are known as an elastic material. H (II) Ductility :- The property of material due to which material can be coverted into a thin wire under uniaxial tensile load without fracture is known as ductility and such type of materials C are known as ductile material. Those material which under goes a large plastic deformation before fracture are known as ductile material TE Eg - Mild Stel, Al, Cu, etc c   t   Ductile materials are weak in shear Most Ductile material is platinum With increase in temperature ductility of material increases but strength decreases. E With increase in carbon content ductility will decrease and strength of material will increase. N (III) Brittle Material :- Those material which have very less plastic deformation [almost negligible] before fracture are known as brittle material. Brittle materials are weak in tension. Ex:- Concrete , Glass, Rubber ZO For brittle material c     t ×  × e P  7 Strength of Material ZONE TECH (IV) Malleability :- Property of material due to which material can be converted into thin sheet under uniaxial compression is known as malleability. Ex:- Most malleable material is Gold 'Hg' is malleable but not ductile, (V) Hardness :- Resistance of any material agency skretches or abraision is known as hardness Hardness is measured with the help of following test:- (a) Rockwell Hardness Test (b) Brinell Hardness Test (c) Vicker's Hardness Test H (VI) Resilence :- Energy stored in a material upto elastic limit is known as resilence. Area under the load v/s deformation curve represent resilence in a body. Resilence is C applicable upto elastic limit. (a) Proof Resilence :- Maximum energy which can be stored in a material upto elastic limit is known as proof resilence. Maximum value of resilence is known as proof resilence. TE elastic limit PM P2 1  Proof Resilence 1 P P1 2  Resilence 2 E 1 2   N (b) Modulus of Resilence :- Max energy per unit volume which can be stored in a material upto elastic limit is known as modulus of resilence OR ZO It is represented by proof resilence per unit volume OR Area under stress - Strain curve upto elastic limit represents modulus of resilence elastic limit  MOR  Proof Resilence = (volume × MOR)  8 E - Book ZONE TECH (VII) Toughness :- Resistance of material against impact loading is known as toughness OR Total energy stored upto fracture point in material is known as toughness. OR Area under load v/s deformation curve upto fracture point is known as toughness U fracture point × ness H Load g h To u Deformation C (a) Modulus of Toughness :- Max energy stored per unit volume upto fracture point is known as modulus of toughness Area of stress - strain curve upto frqacture point represents modulus of toughness. Toughness = Volume × MOT TE (VIII) Proof Stress :- This term is used for those ductile materials which do not have clearly defined yield point Proof stress is determined by offset method in which a offset line is constructed drawn at 0.2% strain E proof stress N  ZO  0.2% = 0.002 Note 1:- Loading and unloading curve - When a material is loaded upto elastic limit then after unloading elastic material regain its original shape (during unloading material follow the same path as loading) Refer fig.1 Note 1:- If a material is loaded beyond elastic limit then after unloading material does not follow the same curve as loading but unloading curve is parallel to initial tangent of loading curve. This effect/phenomenon is given by Bauchinger hence known as Bauchinger effect. Refer fig. 2 and 3  9 Strength of Material ZONE TECH elastic elastic limit limit n g di tic loa as el g d in g  g  in d in lo a ad l oa g lo in un ad lo un ET  P e  (permanent (recoverable H strain) strain) Fig.1 Fig.2 Fig.3 C (IX) CREEP :- It is a time dependent permanent deformation of material when it is subjected to constant load for a long period of time. It is also called a "Time Dependent Strain". TE E,A,L E,A,L E P E P P P N During loading After unloading E - Elastic deformation ZO P - Permanent deformation / creep deformation Creep is mainly function of following factor:- (a) Magnitude of load (b) Temperature (c) Time Note :- Temperature at which creep deformation become significant is known as homologous temp  Melting point  Homologous temp   2     10 E - Book ZONE TECH Different stages of Creep:- ng si r ea c Creep stage II in ate stage I r Phenomenon stage III constant rate E H P = const time t=0 C (X) Fatigue :- When a material is subjected to a cyclic loading then material fails at lower stress and such type of failure is known as fatigue failure. TE (XI) Endurance Limit :- It is the value of stress at which a material has almost negligible probability of failure under infinite no of (107) loading cycles Ex:- Mild steel - 181 - 186 MPa end. log10S S1 S2 E Endurance Limit N 6 N1 N2 (10 Cycle) log10N ZO Endurance limit is determined by S-N curve Type of Stress :- stress normal stress shear stress () () (Acts perpendicular to plane) (Acts parallel to plane) Direct normal Flexural or Direct shear Torsional stress bending stress stress shear stress (Uniform (Linear (Parabolic (Linear variation) variation) variation) variation)  11 Strength of Material ZONE TECH CHAPTER - 3 (ELASTIC CONSTANTS) Elastic Constants :- Elastic constants are used to determine the relation between engineering stress and engineering strain. For homogenous and isotropic material the number of elastic constants are four i.e. H (1) Modulus of Elasticity or Young's Modulus (E):- As per hooke's law upto proportional limit under uniaxial stress condition Normal stress is directly proportional to normal strain. C Normalstress    E = Normalstrain    E=   TE (2) Modulus of Rigidity or Shear Modulus (G) :- As per hooke's law upto proportional limit under biaxial stress condition shear stress is directly proportional to shear strain. Shear stress    G = Shear strain    E  G=  N x  ZO l  x Shear strain () =  Bulk Modulus (K) - As per hooke's law under hydrostatic condition of stress normal stress is directly proportional to volumetric strain.   K=   V  V   V    dv Volumetric strain = V= V 12 E - Book ZONE TECH Note:- If the value of   0.5, then volumetric strain will be zero and such type of material is known as incompressible.   1 Poisson's Ratio   or m  :- Upto proportional limit it is used to determine lateral strain and   volumetric strain. It is unitless quantity. Lateral strain  Longitudinal strain D P P H L  L L C L Longitudinal Strain = Strain in the direction of applied load L Lateral Strain = D D TE Strain in the perpendicular direction of applied load 1  lateral strain D / D or  =  (-) sign represents the nature of lateral strain which is opposite to m long.strain L / L longitudinal Range –1    0.5 E Generally Range  0    0.5 Note:- If the value of Poisson ratio is zero, then lateral strain become zero and infinite change N will occur in longitudinal strain. Type of Material Poissons Ratio Cork 0 ZO Glass 0 - 0.02 Concrete 0.1 - 0.2 Cast Iron 0.2 - 0.3 Copper 0.355 Steel 0.27 - 0.3 Mild steel 0.3 Al 0.33 Au(Gold) 0.44 Rubber, wax, Paraffin 0.5 Generally with increase in ductility, poisson's ratio increases exception "Rubber"  13 Strength of Material ZONE TECH Relationship Between Elastic Constants :- (i) E = 2G (1+ ).... (i) (ii) E = 3K (1 – 2).... (ii) E From (i)  = –1 2G  E  1 From (ii)  =  1– 3K  2   H  E  E By equating both- 1 – 3K   G – 2   E E E  3K  G  C 3  So that 3 G 3K 3KG 9KG (iii) E TE 3K  G Homogenous Material :- When a material exhibit same elastic properties at any point in a given direction then the material is known as homogeneous material. Hetrogenous Material :- Those material in which composition is not uniform through out the body are known as heterogenous material. Isotropic Material :- Those material which have same properties in all direction at a particular E point are known as isotropic material Anisotropic Material :- Those materials which have different properties in different direction at a particular point are known as anisotropic material. N Orthotropic Material :- Those material which have different properties in three mutually perpendicular direction are known as orthotropic material ZO Note :- Type of Material No of Elastic Constants No of Independent elastic constants (i) Homogenous & Isotropic 4 (E, G, K,  2 (with help of E and ) (ii) Homogenous & orthotropic 12 9 (iii) Homogenous & anisotropic  21 Examples:- (i) Isotropic:- Ceramic, wood, slate (mineral) (ii) Orthotropic:- One layer material plywood, graphite and many crystals. (iii) Anisotropic:- Fibre, Reinforced composite, etc. Different Type of Materials :- 14 E - Book ZONE TECH       linear elastic non-linear elastic Elasto plastic H   C  linear elastic with  TE Perfect plastic or Ideal  Ideal fluid  strain hardening rigid plastic   ideal rigid (time dependent) E   N visco- plastic ZO Hook's Law :- According to this law for a homogenous and isotropic material, within elastic limit stress is directly proportional to strain   Elongation of a prismatic bar subjected to tensile load 'P' P P L,E,A If elongation of bar = L 15 Strength of Material ZONE TECH L then = L P Stress in bar =  = A ACC. to Hook's Law :-  P L   A L H PL L AE Elongation of a prismatic bar due to its self weight C If unit weight of bar is  TE A L,A,C dx x E B For strip at distance 'x' from N C/S area of strip = A Length of strip = dx ZO Total Force acting on strip = .A.L = Weight of body Px = ( × A × x) Px dx Px then elongation of strip at x Px.dx d = A.E Axdx x d =  dx AE E 16 E - Book ZONE TECH L L x Total elongation of bar = l =   dx  d  0  E dx 0 L   x2  L =   E  2  0 .L.L .L2 L   Or 2E 2E H .L.L A  .A.L   L WL L     2E A 2AE 2AE WL C L = 2AE Where W - Total weight of bar = ( × A × L) TE Elongation of a uniform tappered bar having circular C/S due to external load 'P'. E D2 D1 P Dx P N A x dx ZO B Consider a strip at distance x from 'A'  D 2 – D1  Dia of strip = Dx = D1 +  L x    D2 – D1  Dx = Kx =  where K  L    Length of strip = dx force acting on strip = Px = P Then elongation of strip 17 Strength of Material ZONE TECH Dx P P dx H Pdx Pdx  d = Ax E  D 2  E  x 4 C 4Pdx d = E  D  Kx 2 1 L TE4pdx Then total elongation of l =  d   E  D  kx  dx 0 1 2 L   D  Kx  –2 1 1  4P  1   1 l = E  K D  kx      1   0   –1 K  E 4P   1  1  = Ek  –  D  KL   D    1   1 N   4P  1 1   ZO  =  D – D1    D – D1  D1  E  2   D1   2    L   L   4PL  1 1  = E  D – D   – D  D  2 1  2 1 4PL  D2 – D1  = E  D – D   D D  2 1  1 2  4PL l = ED D 1 2 18 E - Book ZONE TECH Elongation of a composite bar :- A1 , E 1 (1) P P A2 , E 2 (2) H L L Since both bars are attached with the help of rigid plate. Hence final elongation of both bars C will be equal 1 = 2 =  If P1 = force carried by bar (1) TE P2 = force carried by bar (2) P1L P2 L then 1 = A1E1 2 = A E 2 2 P1 + P2 = P.... (1) from 1 = 2 E P1 L PL = 2 A1E1 A 2 E2 P1 A1E1.... (2) N = P2 A 2 E 2 from equation (1) and (2)  A1 E1  ZO P1 =  A E  A E   P  1 1 2 2   A 2E2  P2 =  A E  A E   P  1 1 2 2  Total elongation =  =1 = 2  AE1  P    L 1 = P1L   A1E1  A 2 E 2  A1 E1 A1E1 PL 1 = A E  A E 1 1 2 2 PL 1 = 2 =  A E  A E 1 1 2 2  19 Strength of Material ZONE TECH Note :- If given composite bar is replaced by a single bar having modulus of elasticity equation, cross section area A1 + A2, length 'L' which have same alongation under same tensile load A1, E 1 (1) P P P P Eeq, (A 1 + A )2 A2, E 2 (2) H L L L L C Elaongation of equivalent bar PL 1 =  A  A  Eeq For 1 = 2 1 2 TE PL PL   A1  A 2  E eq  A1E1  A 2 E 2  (A1 + A2) Eeq = A1E1 + A2E2 A1E1  A 2 E 2 Eeq = E A1  A 2 Expression for volumetric strain under Tri-axial loading:- N When load is applied in x direction only ZO x x x Strain in x direction will be  x  E –y –z   x x  x Strain in y direction will be  y  –  x  – E –  x Strain in z direction will be  z  E 20 E - Book ZONE TECH When load is applied in y direction y y H y Strain in y direction will be  y = E  y C Strain in x direction will be x = – E  y Strain in z direction will be  z = – TE E When load is applied in z direction only z E z N z Strain in z direction will be  z = E z Strain in x direction will be  x = – ZO E z Strain in y direction will be  y = – E When load is applied in all three directions i.e. for 3-D loading condition y z x x z y 21 Strength of Material ZONE TECH x  y z Total strain in x-direction will be x = - - E E E y  x  Total strain in y-direction will be  y = - - z E E E z  x  y Total strain in z-direction will be z = - - E E E Then Volumetric Strain will be  V   x +  y+  z H x  y  z     y x z  z x     y    EV =  E – E – E    E – E – E    E – E – E         x  y  z  1 – 2  C EV = E V x  y  z   1 – 2  EV = V  Ex  Ey  Ez  E TE For incompressible body, vol. strain = 0  = 0.5 Thermal Stresses:- Total stress developed on the cross-section at a component is equal to mechanical normal stress + thermal stress. Case I:- (Free expansion) - No Reaction E N ZO "Free Expansion of rectangular Body" Thermal deformation in bar:-  L  l1 - l  (  th)XX  .T. l  b  b1 - b  (  th)ZZ  .T.b  t  t1 - t  (  th)YY  .T.t where   linear coefficient of thermal expansion Thermal strain in bar:- (  th )XX  .T (  th )YY  .T (  th )ZZ  .T Hence (  th )XX  (  th )YY  (  th )ZZ   T 22 E - Book ZONE TECH Volumetric Strain in bar:- V V    x   y  z  3 th   3 . T V Volume Increase become  V   3.. T  l. b. t   ____________m3 To maintain dimensional stability in presence of temperature variation Invar (Ni and steel) is selected as the material because  of Invar is equal to 1.2 × 10-6 °/C Case II:- Completely restricted expansion in one direction. H  , E,  T  d C L  Total x  0 TE  Total y/z  . T  . . T  Total  . T  H  Reaction force:- (  Total)  th + ad  zero  RL    . TL     AE   0 E   R  . T.E.A R N    . T.E A Partially Restricted expansion of a bar:- ZO A , E RA  R 1 R1  R L  R1 R1   Change in length of bar Change in length  (  Total)X  th + (  axial def)X   X  R 1L    . T.L     AE       23 Strength of Material ZONE TECH th.L . T.L   E  . T.L    th   L  E     th    th E  L  Thermal stress P R H   th     . T.E X A A  thX  . T.E C Hence When temperature increases th is compression nature  th is tensile nature TE Temperature decreases th is tensile nature  th is compression nature Thermal stress in compound bar:- Al Steel E When temperature will rise  Compress in both shaft When temperature will drop  Tensile in both bar N Thermal stress in composite bar:- (1) When temperature will increase ZO Steel  Al   Steel Al T Al (  )   Compression thermal stress Steel (  )   Tensile thermal stress T Al (  )   Tension thermal stress Steel (  )   Compression thermal stress  24 E - Book ZONE TECH CHAPTER - 4 (CENTROID & MOMENT OF INERTIA) CENTROID AND MOMENT OF INERTIA Centre of gravity and Centroid:- The centre of gravity of a body is that point through which the resultant of system of parallel H force formed by the weight of all practice passes Centroid:- Centre of area for the plane x-sectional area like circle, rectangle, triangle etc. It is represented by C or G. C Centroid for same standard shape/Areas:- (I) Triangle TE h h/3 b E h y 3 N 1 Area = bh 2 ZO (II) Rectangle x h y b b x 2 h y 2 Area = b × h  25 Strength of Material ZONE TECH (III) Circle x D y  2 Area = D 4 H D xy 2 (IV) Semi - Circular Disc C TE y O D = 2R 1  D2  Area = 2  4    E 4R y 3 N (V) ZO x h y 2 y = kx =0 b 2  Area =  3  b  h    3 x b 8 3 y h 5  26 E - Book ZONE TECH (VI) y 2 kx y= h x y =0 x b H 1  Area =  3  b  h    C 3 x b 4 y 3 10 h TE Note :- Centroid for any non-symmetric body:- A1X1  A 2 X 2 .........  A n X n x E A1  A 2 ......  A n A1 y1  A 2 y 2 .........  A n y n y N A1  A 2 ...........  A n Area - Moment of Inertia :- Second order moment of area about avg axis is known as area - moment of inertia ZO y dA' x dA y x x y 2 Ixx =  y dA Ixx – MOI about xx – ax1 2 Iyy =  x dA Iyy – MOI about yy – ax1  27 Strength of Material ZONE TECH Note :- Product of Moment of Inertia :- Product of MOI about x - x and y - y axis - y dA x y H x x y C Ixy =  yx dA Ixy =  xy dx dy Those axis about which product of inertia is zero are known as principal axis. TE Note :- Unit of area MOI = mm4, m4, cm4 mass MOI Unit of = kgm2   mr 2 Rectangle :- Consider a strip at distance 'y' from x - x axis E N D dy y ZO x x B Area of strip = dA = Bdy MOI of strip about x-axis = dI = y2dA = y2(Bdy) Total MOI of rectangle about x-axis D Ixx =  dI   y2 Bdy 0 D  y3  Ixx B = 3   0 BD3 Ixx = 3  28 E - Book ZONE TECH Parallel Axis Theorem :- Moment of inertia with respect to any axis parallel to the centroidal axis is equal to hte moment of inertia with respect to centroidal axis plus the product of area of figure and square of the distance between axis. G CG G r H K K IKK = IGG + Ar2 C IGG = MOI about C.G axis axis passing through C.G IKK = MOI about K.K axis TE A = Total area Y = distance between G - G & K - K axis Ex. BD3 IXX = 3 E IGG = ? Acc. to 11 axis theorem :- N C.G G D G ZO r = D/2 x x B IXX = IGG + Ar2 2 BD3 P = IGG + BD ×   3  2 BD3 BD3 IGG = – 3 4 BD3 IGG = 12  29 Strength of Material ZONE TECH Perpendicular Axis Theorem :- Moment of inertia about an axis perpendicular to the plane and passing through the intersection of the other two axis X-X and Y-Y contained by plane is equal to the sum of Moment of Inertia about XX and YY. y y x x x C.G H z y C Izz = Ixx + Iyy TE where ; Ixx = MOI about x-axis Iyy = MOI about y-axis Izz = MOI about z-axis y E 50mm 100mm k 20mm k N r x x ZO 100mm 70mm k 20mm k 100mm y For given I-section calculate MOI about x - axis and y - axis (1) MOI about X-axis ;- Ixx = (Ixx)web + (Ixx)flange... (1) + (Ixx)flange 3 (Ixx)web =  20 100   1666666.67 12  30 E - Book ZONE TECH (Ixx)flange...(1) = [Ixx + Ar2]flange...(1)  100  20 3  =   100  20  60 2   12    = 7266666.67 (Ixx)flange...(1) = (Ixx)flange...(2) Ixx = Ixx)web + Ixx)flange...(1) + Ixx)flange...(1) = [1666666.67 + 7266666.67 × 2] H = 16.2 × 106 mm4 Method-2 :- C y E A TE F L B k x x E H G I C D N y Ixx = (Ixx)ABCD = (Ixx)EFGH = (Ixx)IJKL ZO 3 3 3 100 140  40 100  40 100  =   12 12 12 Ixx = 16.2 × 106 Iyy = (Iyy)ABCD – (Iyy)IJKL – (Iyy)EFGH 3  140 100  12 Iyy = (Iyy)web + (Iyy)flange...(1) + (Iyy)flange...(2) 3 3 3 100  20  20 100  20 100     12 12 12 Iyy = 03.4 × 106 mm4 31 Strength of Material ZONE TECH Standard Results :- y D x x k k H B y BD3 DB3 BD3 C Ixx = Iyy = Ikk = 12 12 3 TE C.G H x x k k E B BH3 BH3 N Ixx = Ikk = 36 12 y ZO y D x x x z y D 4 D 4 Ixx = Iyy = Izz = Ixx + Iyy Izz = 64 32  32 E - Book ZONE TECH CHAPTER - 5 (BENDING MOMENT AND SHEAR FORCE DIAGRAM) S.F.D And B.M.D Different type of supports :- (I) Roller Support :- H or C only vertical reaction will be there R R (II) Hinge Support :- TE H R E Moment reaction will be zero Only horizontal & vertical reaction will be there N (III) Fixed Support H ZO M R Horizontal & vertical reaction will be there Moment reaction will also there (IV) Guided Roller M (so no rtation) R  33 Strength of Material ZONE TECH Types of Beam :- (1) Simply Supported Beam :- Will not permit thermal expansion so we will not use this H Horizontal displacement can't be resist C TE Best combination for SSB so one end is hinged & another is rollar support (2) Fixed Beam (3) Cantilever Beam E (4) Propped Cantilever N ZO (5) Over hanging beam (b) Double side overhang (a) single side (6) Continuous beam  34 E - Book ZONE TECH Type of Loading :- UVL (uniformly varying load) conc moment point load / UDL (uniformly couple (kN-M) conc. load distributed load) uniformly distributed moment H Calculation of Support Reaction :- y C P HA A B a b x RA L TE RB z By using Static Equilibrium :-  fx = 0 –HA = 0 HA = 0  fy = 0 RA + RB – P = 0 RA + RB = P.... (1) E  Mz  0 ,  MA  0 P(a) – RBL = 0 Pa = RBL N Pa RB = L From equation.... (1) ZO Pa RA + =P L P – Pa RA = L Pb RA = L Ex. Calculate the reaction in given arrangement M HA A B RA RB 35 Strength of Material ZONE TECH  fx  0 , HA = 0  fy  0 , R + R = 0 A B  MA  0 M – RBL = 0 RB = M/L RA = RB RA = –M/L H where, (–ve) sign represents that direction of 'RA' will be opposite C P HA A B C a TE b L–x x RA RB 0 Mx Hx Vx P E x O b FBD Vx Pa Pb (L – x) RB = RA = L N L 0 Shear Force :- Shear force at any section is the algebric sum of total transvers forces which are acting to left side or to right side of the given section ZO Bending Moment :- Bending moment at any section is the algebric sum of total moment due to transverse forces or couples to the left side or to right side of given section. For above figure in left side portion -  fy  0 Pb Vx = RA  Vx  L  M0  0 Mx = RAX Pb Mx = x L  36 E - Book ZONE TECH For right side portion :-  fy  0  RB + Vx = P Vx = P – RB  Pa   Vx =  P –  L   M0  0  Mx + P [(L – x) – b] = RB (L – x) H Mx = RB (L – x) – P [(L – x) – b] C S.F.D (Shear Force Diagram) :- A curve which represents the shear force on different section along the length of member is known as shear force diagram. TE B.M.D (Bending Moment Diagram) : - A curve which repressents the bending moment of diff sections along the length of member is known as bending moment diagram. Axial Thrust Diagram :- E A curve which represents axial threst on diff sections along the length of member is known as axial thrust diagram. N Steps to Draw SFD and BMD :- Step - I ZO Calculate, Support Reactions. P HA A B C a b L RA RB HA = 0 Pb RA = L Pa RB = L  37 Strength of Material ZONE TECH Step - II: Sign Convention :- Mx Hx x Vx Vx Vx = (+ve) Vx = +ve Mx (sagging) = +ve Mx (sagging) = +ve H Mx Mx Mx Mx C Step - III :- TE Write the equation for shear force and bending moment for different segment P A B C E RA x RB For segment AC (0 < x < a) N where 'x' from 'A' ZO Mx A O x Vx Pb RA = L For equilibrium :- Pb  f y  0 V = RA = x L  M0  0 Mx = RA × x Pbx Mx = L 38 E - Book ZONE TECH At Pt A, x = 0 Pb VA = L MA = 0 at Pt c, x = a Pb (Vc)left = L Pb Pab (Mc)left = (a) = L L H For segment CB (a < x < L) where x from A C Mx P TE A O a Vx x Pb RA = L for equilibrium :- E  fy  0  Vx + P = RA Pb Pa Vx = – P= L L N  M0  0 Mx + P(x – a) = RAx ZO Pb Mx = x – P(x – a) L at Pt C, x = a –Pa (Vc)right = L Pab (Mc)right = L at Pt B, x = a –Pa (VB) = L Pab (MB) = (L) – P(L – a) L = Pb – Pb = 0  39 Strength of Material ZONE TECH Step-IV :- Draw SFD and BMD ;- P HA A B a b RA RB Pb + SFD H L Pa – L C 1° Pa b/LTE 1° + BMP O Properties of SFD and BMD ;- dM (I)  V (SF) 'A' to 'C' E dx dV dv  –W 0 dx dx N dv = 0 where : M = B.M V = C (const) dM ZO V = S.F C dx W = loading M = C(x') + c2 Loading SFD BMD n0 (n + 1)0 (n + 2)0 Point load Const (0°) linear (1°) (const) UDL 1° 2° (linear) UVL 2° 3° (II) Point at which a concentrated force is acting, there will be a sudden change in SFD and slope of BMD. (III) A point at which a couple is acting there will be a sudden change in BMD  40 E - Book ZONE TECH Note :- (1) A point at which SFD changes its sign, bending moment will be max (max sagging) or min (max hagging) at that point + + – (2) A point at which BMD changes its sign is known as point of contraflexure sometimes it is also known as point of inflexion H Step for SFD and BMD without using equation :- Step-I C Calculate support reactions Step -II Calculate shear force and bending moment on critical section (point) TE Critical points are (a) Point load, starting and end point of UDL and UVL point at which moment is acting support reactions, poit at which SFD change its sign. Step-III :- Determination of Shear Force. Case-I : If left portion of given section is considers E Shear force at given section = total upward force to left of section. Ex. N P A B a b RA RB ZO Pb Pa RA = , RB =  L Case-II If right portion of given section is considered Shear force at given section = total downward force to the right of section Step-4 : Determination of bending section is considered Bending momet at given section = total clockwise moment due to forces & couples to the left side of section. Case-II : If right portion of given section is considered Bending moment at given section = total anticlock wise moment due to forces & couples to the right side of section. Step-5 : By using properties of SFD and BMD draw the SFD & BMD 41 Strength of Material ZONE TECH P A B a b RA RB 0° (const) Pb L + SFD H Pa – L C 1° Pa b/c 1° + O TE BMP Mathematical equaltion for above problem 1 2 P 3 4 E A B 1 N RA a 2 3 b 4 RB VA)R = total upward force on left of (1).... (1) ZO Pb VA = RA = L VC)l = total upward force on left of (2).... (2) Pb RA = L VC)R = total upward force on left portion of (3).... (3) RA = – P Pb –Pa = –P= L L VB)l = total downward force on right portion (4).... (4) –Pa –RB = L  42 E - Book ZONE TECH Bending Moment :- BM)A = BM at section (1).... (1) total clockwise momet due to forces & couples to the left side of section MA = 0 BC)l = BM at section (2).... (2) Total clockwise moment due to forces acting to left side of section Pb Pab MC)l = RA × a = a = L L MC)R = B.M at section (3).... (3) Total clockwise moment due to force acting to left side of (3).... (3) H = RA × a – P(0) = RA × a Pab MC)R = L C MB = B.M at section (4).... (4) Total anticlockwise moment due to forces acting on right side of (4).... (4) MB = MB (0) = 0 TE Note :- Case for double side over hanging beam with UDL load for most economical section (+ve) B.M at mid span should be equal to (–ve) B.M at supports w 2 w  – 4a 2   a 2 8   2 E 1 2  – 4a 2   a 2 4    2  8a 2 N   2 2a [less strength requires at this length] ZO a l a L RA RB L = l + 2a for most economical section   2 2a L – 2a  2 2a  L a 22 2  L a  0.207L 22 2   2 2a L a  0.207L 22 2  43 Strength of Material ZONE TECH Ex. A horizontal beam AD, 10 metres long carries a uniformly distributed load of 160 N/m together with a concentrated load of 400 N at the left end A. The beam is supported at a point B which is 1 m from A and at C which is one the right-hand, half of the beam and x metres from the end, D. 400 N 160 N/m A D B E C 1m Vb Vc x 5m 5m H Determine the value of x, if the mid-point is a point of contraflexure and for this arrangement draw S.F. and B.M. diagrams. Sol. Fig. shows the beam carrying the load system mentioned in the problem. Let Vb and Vc be C the reaction at the supports B and C. Resolving the forces on the beam vertically TE Vb + Vc  400 + 160 × 10  2000 N  Vb + Vc  2000 N It is given the B.M. at the mid-point E  0. Taking moments about E of the forces on L.H.S. of E, 52 Vb  4  400  5  160   0 2 E  Vb  1000 N Since, Vb + Vc  2000 N N We have Vc  1000 N Taking moments about E of the forces on the RHS of E, ZO 52 1000(5  x)  160   0 2  x  3m Fig. shows the loaded beam showing the actual position of the support C. S.F. analysis S.F. just on RHS of A  –400 N S.F. just on LHS of B  –400 – 160 × 1  –560 N S.F. just on RHS of B  –560 + 1000  +440 N S.F. just on LHS of C  1000 – 400 – 160 × 7  –520 N S.F. just on RHS of C  –520 + 1000  +480 N S.F. at D  0 44 E - Book ZONE TECH 400 N 160 N/m A D B E C 1m 6m 3m 440 N 480 N H G D A B C 400N C 560 N 520 N 3.75 m A 2.5 m B F TE 125 Nm E C D 480 Nm 720 Nm =0 5m 5m E Section of zero shear. Let G be the section of zero shear. If this section is x metres from A, N equating the S.F. to zero, 1000 – 400 – 160x  0  x  3.75 m ZO B.M. of analysis B.M. at A  0 160  12 B.M. at B  400  1  2  –480 Nm 160  3.75 2 B.M. at G  1000 × 2.75 – 400 × 3.75   125 Nm 2 32 B.M. at C  160   720 Nm 2 B.M. at D  0 Points of contraflexure between B and C Let a point of contraflexure be x metres from A 45 Strength of Material ZONE TECH Equating the B.M. to zero, x2 1000(x  1)  400x  160   0 2  2x2 – 15x + 25  0  (2x – 5)(x – 5)  0  x  2.5 m and 5 m. C H TE E N ZO  46 E - Book ZONE TECH CHAPTER - 6 (BENDING STRESS & SHEAR STRESS IN BEAM) BENDING STRESSES bending = along transverse axis load H twisting = along longitudinal axis load y (transverse dire)) C x (long. axis) TE z (transverse axis) When a moment is applied about a trasverse axis of beam then bending will generate in beam due to this bending compressive and tensile stress (normal stresses ) will generate which are known as bending stress or flexural stress. Bending Equation Assumptions :- E Material is homogenous, isotropic and obey hooks law Beam is initially straigt and prismatic N Plane section before bending remains plane after bending also i.e strain in any fibre is proportional to the distance of that fibre from neutral axis. Resultant of applied loading is in plane of symmetry ZO Modulus of elasticity is same in tension and compression Beam is subected to bending only Note :- Neutral axis? While bending, one fibre exists such that the length of which remains unchanged. This layer is neutral layer. The line of intersection of the neutral layer with plane of cross-section of the beam is called neutral axis. It represents the location of zero flexural stress in the member subjected to bending. Neutral fibre - A fibre in which total strain is zero (i.e. no change in length of that fibre) is known as neutral fibre A line which passes through the inersection of neutral fibre and any c/s is known as neutral axis of that cross-section 47 Strength of Material ZONE TECH M Compression A M A' A A' Tension b H N.A d Neutral fibre N.A C TE Bending equation Derivation A B C D y E P Q A B N CD = PQ O ZO  R A B c' d' P' y Q' 48 E - Book ZONE TECH C'D' = CD (at NA - No strain, No stress) C'D' = R P'' = (R + y) PQ = CD' = C'D' = R Strain in fibre at distance 'y' from NA P'Q'– PQ = PQ  R  y   – R = R H y = R  y C where R Radius of curvature 1 Curvature R TE  y  ..... 1 E R Stress in fibre at distance 'y' from N.A E N.A y N dA y  E ZO R Force resisted by strip dm = y × df = y ×  × dA yE dM = y  dA R Total moment carrying capacity of M   dM se