Mechanics of structures module2
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This document discusses mechanics of solids, specifically torsion and bending moment. It begins with an overview and objectives of Module II which covers torsion of circular elastic and inelastic bars, as well as axial force, shear force, and bending moment diagrams. It then provides explanations, equations, and examples relating to torsion, including assumptions in torsion theory, the torsion equation, torsional rigidity, power transmission in torsion, and example problems calculating torque and shaft diameters. It also discusses statically indeterminate shafts and torsion of inelastic circular bars.
Mechanics of structures module2
- 1. Mechanics of solidsMechanics of solids Torsion, Bending moment and shear forceBending moment and shear force Dr. Rajesh K. N. Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering Govt. College of Engineering, Kannur 1
- 2. Module IIModule II Torsion - torsion of circular elastic bars - statically indeterminate problems - torsion of inelastic circular bars Axial force, shear force and bending moment - diagrammatic conventions for supports and loading, axialdiagrammatic conventions for supports and loading, axial force, shear force and bending moment diagrams - shear force and bending moments by integration and by singularity functions Dept. of CE, GCE Kannur Dr.RajeshKN 2
- 3. TorsionTorsion Torsional moment (Torque) Twist (Angle of twist) • Circular shafts and non-circular shafts Dept. of CE, GCE Kannur Dr.RajeshKN 3
- 4. Assumptions in torsion theory for circular shaftsAssumptions in torsion theory for circular shafts • Material is uniform throughout • Shaft remains circular after loading • Plane sections remain plain after loading T i i if h h• Twist is uniform throughout • Distance between any two normal sections remain the same after loading • Stresses are within elastic limit• Stresses are within elastic limit Dept. of CE, GCE Kannur Dr.RajeshKN 4
- 5. T ϕ θ m’ r T ϕ θ m O r T L Any cross-section Any radial distance r mm L rφ θ′ = = Angle of twist per unit length L θ → mm L rφ θ G τ φ =Shear strain L θ O r rτ θ ∴ = G rθ φ =But Dept. of CE, GCE Kannur Dr.RajeshKN 5 G L ∴ L φ =But
- 6. T Torsion equation R T θ r τ θ L T O maxG r L R τ θ τ ∴ = = r G L τ θ = Aδ maxr R τ τ⇒ = . .T A rδ τ δ= 2max max . r T A r r A R R τ τ δ δ δ= = Torsional moment on the elemental area = 2max T T r A R τ δ δ= =∫ ∫ R R Total torsional moment on the section, A A R∫ ∫ 2max r A R τ δ= ∫ max J R τ = , Dept. of CE, GCE Kannur Dr.RajeshKN 6 A R R Polar moment of inertiaJ →
- 7. T τ T GθmaxT J R τ = T G J r L τ θ = = T GJ L θ = GJ Torsional rigidity GJ L maxT Tτ J R Torsional section modulusmax max J R J R τ= ⇒ = J R Torsional section modulus 4 dπ ZZ XX YYJ I I I= = +Polar moment of inertia 32 dπ = Dept. of CE, GCE Kannur Dr.RajeshKN 7
- 8. Power transmittedo e t a s tted W T θ= Work done (per second) by torque T making a twist θ1 /second 1.W T θ= 1.P T θ=i.e., Power transmitted 2 60 nT P π =If n is the rotation per minute, k 1 Watt 1 Joule/Second = 1 Nm/s= = 1 HP 0.75 kW= Dept. of CE, GCE Kannur Dr.RajeshKN 8
- 9. Problem 1: A hollow shaft is to transmit a power of 300 kW at 80 rpm. If the shear stress is not to exceed 60 MN/m2 and internal diameter is 0.6 of external diameter, find the internal and external diameters assuming that the maximum torque is 1.4 times the mean torque.g q q 300 kWP = 2 60 nT P π =80 rpmn = 60 2 mean P T T nπ ∴ = = 3 60 300 10 35809.862 Nm 35.81 kNm 2 80 meanT π × × = = = × max 1.4 meanT T∴ = 1.4 35.81 50.134 kNm= × = maxT J R τ = ( ) ( )( )44 4 4 0.6 32 32 J D d D D π π = − = − ( ) 4 0.8704 32 Dπ = Dept. of CE, GCE Kannur Dr.RajeshKN 9 J R 32 32
- 10. 3 6 50 134 10 60 10× × ( ) 4 50.134 10 60 10 2 0.8704 32 D Dπ × × = 3 0.00488914D = 0.169 mD⇒ = 0.6 0.102 md D⇒ = =0.6 0.102 md D⇒ Dept. of CE, GCE Kannur Dr.RajeshKN 10
- 11. Problem 2: A solid circular shaft is to transmit 75 kW power at 200 rpm. If the shear stress is not to exceed 50 MPa, and the twist is not to exceed 10 in 2 m length of shaft, find the diameter of shaft. G = 100 GPa. 75 kWP = 2 60 nT P π =200 rpmn = 3 60 60 75 10 3581 Nm 2 2 200 P T × × = = = ×2 2 200nπ π × maxT Gτ θ 4 D J π 2 50 N mmτ =max J R L = = 32 J = max 50 N mmτ = 6 4 3581 50 10 2 32 DDπ × = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 0.0714 mD⇒ =maxT J R τ = ⇒ Dept. of CE, GCE Kannur Dr.RajeshKN 11 32⎝ ⎠
- 12. T G J L θ = 9 4 3581 100 10 1 180 2 32 D π π × × × ⇒ = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 0.0804 mD⇒ = 32⎝ ⎠ Required diameter of shaft is the greater of the two values 0.0804 m 80.4 mmD∴ = = Dept. of CE, GCE Kannur Dr.RajeshKN 12
- 13. Problem 3: A steel bar of 20 mm diameter and 450 mm length fails at a torque of 800 Nm. What is the modulus of rupture of this steel in torsion? Modulus of rupture in torsion is the maximum shear stress (on the surface) at failure. T R J τ = Hence, modulus of rupture TR J τ = where T is the torque at failure. 3 2800 10 10 509N × × J 2 4 800 10 10 509N mm 20 32 τ π = = ⎛ ⎞× ⎜ ⎟ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN
- 14. Problem 4: A solid brass rod AB (G=39 Gpa, 30 mm dia, 250 mm length) is b d d lid l i i d BC (G 27 G 36 di 320 l h)bonded to solid aluminium rod BC (G=27 Gpa, 36 mm dia, 320 mm length). Determine the angles of twist at A and B. T G TL J L GJ θ θ= ⇒ = A BC 3 3 180 10 320 27 10 164895.92 Bθ × × = × × 0 0.0129 rad 0.74= = Torque, T 180 Nm 4 436 164895.92mm 32 BCJ π × = = 3 3 180 10 250 39 10 79521 56 ABθ × × = × × 0 0.0145 rad 0.831= = 4 430 79521.56mm 32 ABJ π × = = 39 10 79521.56× × 0 00 0.74 0.83 . 711 1 5ACθ = + = Dept. of CE, GCE Kannur Dr.RajeshKN
- 15. Problem 5: Two solid brass rods (G=39 Gpa) AB (30mm dia, 1.2m length) & BC (40 di 1 8 l h) li d i h h D i h(40mm dia, 1.8m length), are applied with torques as shown. Determine the angles of twist between i) A and B; ii) A and C. T G TL J L GJ θ θ= ⇒ = 3 3 400 10 1200 39 10 79521.56 BAθ × × = × × 0.1548 rad= 3 3 800 10 1800 39 10 251327.41 CBθ − × × = × × 0.1469 rad= − 4 430 79521.56mm 32 ABJ π × = =θ θ θ= + 4 440 251327.41 mmBCJ π × = = 32 AB CA BA CBθ θ θ= + 0.1548 0.1469 0.0079 radCAθ = − = Dept. of CE, GCE Kannur Dr.RajeshKN 251327.41 mm 32 BCJ
- 16. Statically indeterminate shaftsy A B T T B C Torque, T A BT T T= + (1) A B 0Aθ = with respect to B 0θ θ− = Torque, T B C 0CB ACθ θ = 0B BC A CAT L T L G J G J ⇒ − =Torque, T Reactive torque, TA Reactive torque, TB BC BC CA CAG J G J (2) Dept. of CE, GCE Kannur Dr.RajeshKN Solve (1) and (2) for &A BT T
- 17. Torsion of inelastic circular bars • Linear elastic range: Stress strain curve is a straight line (Torsion of circular bars beyond elastic range) g g Stress Strain • Corresponding shear stress distribution in the shaft: τ T r J τ = maxτ R Entire cross-section is in the elastic range Dept. of CE, GCE Kannur Dr.RajeshKN
- 18. Materials in the inelastic ranges • Inelastic: Nonlinear stress strain curve g Stress Elastic rebound on unloading • Elastic plastic : Initially in the elastic range then fully plastic Strain • Elastic-plastic : Initially in the elastic range, then fully plastic Stress Elastic rebound on unloading Dept. of CE, GCE Kannur Dr.RajeshKN S Strain
- 19. Shear stress-strain curve Corresponding shear stress distribution in the shaft ss maxτStres Strain R ss maxτ Stres Strain R Dept. of CE, GCE Kannur Dr.RajeshKN
- 20. maxτ maxτFor any stress distribution, . .dT dA rτ= T dT r dAτ= =∫ ∫ R R y , maxτ R . A A T dT r dAτ∫ ∫ Dept. of CE, GCE Kannur Dr.RajeshKN
- 21. Problem 1: A solid steel shaft of 24 mm diameter is so severely twisted that only an 8 mm diameter inner core remains elastic, while the rest of the diameter goes to inelastic range. If the material is elastic-plastic with shear stress-strain diagram h h h id l d id l i h ill i has shown, what are the residual stress and residual twist that will remain at the surface? Take G = 80 GPa. τ 160 MPa 4 mm 12 mm 0.002 γ φ= Dept. of CE, GCE Kannur Dr.RajeshKN
- 22. The inner core is in the elastic rangeThe inner core is in the elastic range. 160 MPa 12 mm4 mm It is required to find the applied torque. ∫ R R ∫ ∫ Applied torque, . . A T r dAτ′ = ∫ 2 0 0 . .2 . .2 .r r dr r drτ π τ π= =∫ ∫ 4 12 160⎛ ⎞ ( ) 4 12 2 2 0 4 160 .2 . 160 .2 . 4 r r dr r drπ π⎛ ⎞ = +⎜ ⎟ ⎝ ⎠ ∫ ∫ ( ) 3 3 16 558 10 Nmm 574 10 Nmm= + × = × Dept. of CE, GCE Kannur Dr.RajeshKN
- 23. Residual stresses on rebound (removal of torque) • Consider the external torque is removed after a portion of the cross-section has gone to inelastic range ( q ) has gone to inelastic range. • After the torque has been removed completely, the inner portion has a tendency to rotate back to original position but the outer portion which has atendency to rotate back to original position, but the outer portion, which has a permanent rotation, prevents this. • Outer portion has a tendency to stay in the permanently set position, butOuter portion has a tendency to stay in the permanently set position, but the inner portion, which has a tendency to rotate back to original position, prevents this. • Hence, stresses are remaining in the shaft even after the removal of torsion. • The rebound is elastic. • If the deformation of the outer portion was elastic, shear stress would have reached a value of τ’ at the surface. Dept. of CE, GCE Kannur Dr.RajeshKN
- 24. • The stress recovery at the surface is τmax . • Hence, as a result of inner portion applying a torsion on the outer portion, a stress remains at the surface, equal to τ’ - τmax , opposite in direction to the li d t i it t th di ti fapplied torque. i.e., opposite to the direction of τmax . • Similarly, all the inner regions at various radial distances have residual stressesstresses. • These residual stresses are obtained as the difference between elastic-plastic stress distribution and the elastic stress distribution of rebound as shown instress distribution and the elastic stress distribution of rebound, as shown in figure. Elastic rebound τ′E g : Elastic plastic material Residual stress maxτ E.g.: Elastic-plastic material Dept. of CE, GCE Kannur Dr.RajeshKN
- 25. To find residual stresses on an elastic rebound, (i.e., on removal of the T R′ 3 2574 10 12× × torque) Shear stress on the surface T R J τ′ = 2 4 574 10 12 211 N mm 24 32 π × × = = ⎛ ⎞× ⎜ ⎟ ⎝ ⎠ (considering elastic rebound) Hence, residual shear stress on the surface 2 211 160 51 N mm= − = maxτ τ′= − 160 MPa Residual stress 211 MPa 12 mm4 mm Dept. of CE, GCE Kannur Dr.RajeshKN
- 26. 0.006γ φ′ ′= = Residual twist on rebound Strain variation is linear along the radius. 4 0.002φ = Residual twist on rebound θ 12mmR = 4mmmax max L RG θ τ =At the initiation of yield, maxT T= When T increases further to T’, L θ increases beyond max , ot L L θ θ′ Strain variation L L L L rG θ τ =After yielding , is invalid from r=4 to 12.Note: Final twist (after yielding, when T=T’) θ′ i.e., . L RG θ τ′ ′ ≠ But, r L r R θ φ φ′ ′ = = On elastic rebound, recovered twist Elastic Rebound (recovery of twist) Dept. of CE, GCE Kannur Dr.RajeshKN T L GJ θ ′ = Twist of a radial line (recovery of twist)
- 27. To find residual twist on an elastic rebound, 3 0.002 4 10 0.5 rad m m− = = × r L r θ φ′ = Final twist per unit length (after yielding, when T=T’) L r 3 rad m 160 4 m 0.5 ra 10 d 8 m 0 = = × × or, L rG θ τ′ = 4 1080L rG L rG θ τ =∵ is valid from r=0 to 4. Elastic twist recovered (considering elastic rebound) 4 9 2 12 4 574Nm 24 80 10 N m 10 m π − = ⎛ ⎞× × × ×⎜ ⎟ ( g ) T L GJ θ ′ = 0.22 rad m= Hence, residual twist 0.5 0.22 0.28rad m= − = 80 10 N m 10 m 32 × × ×⎜ ⎟ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN
- 28. Problem 2: A solid circular shaft 1 2 m long and 50 mm diameter is subjected to a torque ofA solid circular shaft, 1.2 m long and 50 mm diameter is subjected to a torque of 4.6 kNm. Assuming the shaft is made of an elastoplastic material with yield strength in shear of 150MPa and G = 80 Gpa, what are the radius of the elastic core and angle of twist of the shaft? Also, what are the residual stresses andcore and angle of twist of the shaft? Also, what are the residual stresses and residual (permanent) angle of twist of the shaft, after removal of torque? A li d t 150 MPa 25 mmρ To find elastic core. . . A T r dAτ′ = ∫ 2 0 0 . .2 . .2 . R R r r dr r drτ π τ π= =∫ ∫ Applied torque, 25 mmρ ( ) 25 6 2 2 0 150 4.6 10 N.mm .2 . 150 .2 .r r dr r dr ρ ρ π π ρ ⎛ ⎞ × = +⎜ ⎟ ⎝ ⎠ ∫ ∫ ( )3 33 6 2 252 4.6 10 N.mm 150 150 4 3 π ρπρ − × = + Dept. of CE, GCE Kannur Dr.RajeshKN 28 15.743 mmρ =∴
- 29. To find angle of twist γ ′ To find angle of twist. (Final angle of twist in the inelastic (plastic) range) ?rφ = 25 mmR =15.743 mmρ = G L τ θ ρ = (Valid from r = 0 to 15.743 only) Strain variation along radius 3 150 1200 80 10 15.743 L G τ θ ρ × ∴ = = × ×ρ 3 142.92 10 rad− = × Final twist (after yielding, when T=T’) θ′ 0 3 0180 142.9 8. 82 0 11 π − = × × = Elastic Rebound (reco er of t ist) Dept. of CE, GCE Kannur Dr.RajeshKN 29Twist of a radial line (recovery of twist)
- 30. To find residual stresses on an elastic rebound, (i.e., on removal of the T R′ 6 4.6 10 25× × torque) Shear stress on the surface 2 187 52NT R J τ′ = 4 50 32 π = ⎛ ⎞× ⎜ ⎟ ⎝ ⎠ (considering elastic rebound) 2 187.52N mm= Hence, residual shear stress on the surface 2 187.52 150 37.52 N mm= − = maxτ τ′= − 150 MPa Residual stress 187.52 MPa 25 mm15.743 mm Dept. of CE, GCE Kannur Dr.RajeshKN
- 31. To find residual twist on an elastic rebound, Elastic twist recovered (considering elastic rebound) 3 4 9 2 12 4 4.6 10 Nm 1.2m 50 80 10 N m 10 m π − × × = ⎛ ⎞× × × ×⎜ ⎟ T L GJ θ ′ = 0 0.1124 rad 6.44= = Hence residual twist 0 8 18 6 4 74 1 4= − = 80 10 N m 10 m 32 × × ×⎜ ⎟ ⎝ ⎠ Hence, residual twist 8.18 6.4 74 1. 4= = Final twist (after yielding, h T=T’) θ′ when T=T ) Elastic Rebound (recovery of twist) Dept. of CE, GCE Kannur Dr.RajeshKN Twist of a radial line (recovery of twist)
- 32. Problem 3: A solid circular steel shaft 0 6 m long and 32 mm diameter has been twistedA solid circular steel shaft 0.6 m long and 32 mm diameter has been twisted through 60. Steel is elastoplastic with yield strength in shear of 145MPa and G = 77 Gpa. What is the maximum residual stress in the shaft, after removal of torque?torque? To find elastic core. 6 0.1047 rad 180 π θ = × = 145 MPa G L τ θ ρ = (Valid from r = 0 to ρ only) 16 mmρ 10.792 mm= ρ 3 145 600 10.792 mm 77 10 0.1047 L G τ ρ θ × ∴ = = = × × Dept. of CE, GCE Kannur Dr.RajeshKN 32
- 33. Applied torque To find torque applied. . . A T r dAτ′ = ∫ 2 0 0 . .2 . .2 . R R r r dr r drτ π τ π= =∫ ∫ Applied torque, A 0 0 ( ) 10.792 16 2 2 0 10 792 145 .2 . 145 .2 . 10.792 r r dr r drπ π⎛ ⎞ = +⎜ ⎟ ⎝ ⎠ ∫ ∫0 10.792⎝ ⎠ ( )3 33 2 16 10.7922 10.792 145 145 4 3 ππ − = + 1148475.884 NmmT′∴ = 4 3 Dept. of CE, GCE Kannur Dr.RajeshKN
- 34. To find residual stresses on an elastic rebound, (i.e., on removal of the T R′ ′ 1148475.884 Nmm 16× torque) Shear stress on the surface 2T R J τ′ = 4 1148475.884 Nmm 16 32 32 π = ⎛ ⎞× ⎜ ⎟ ⎝ ⎠ (considering elastic rebound) 2 178.5N mm= Hence, residual shear stress on the surface 2 178.5 145 33.5 N mm= − = maxτ τ′= − max TR J τ= −Residual shear stress at 10.792 mmρ = 145 MPa Residual stress 178.5 MPa max4 1148475.884 Nmm 10.792 32 τ π × = − ⎛ ⎞× ⎜ ⎟ 16 mm10.792 mm 2 120.4 145 24.6 N mm= − = − 32⎜ ⎟ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN Hence the maximum residual shear stress is on the surface.
- 35. Problem 4: A hollow circular steel shaft 1 25 m long 60 mm outer diameter and 36 mmA hollow circular steel shaft 1.25 m long, 60 mm outer diameter and 36 mm inner diameter has been applied with a torque such that the inner surface first reaches plastic zone, and then torque is removed. Steel is elastoplastic with yield strength in shear of 145MPa and G = 77 Gpa What is the maximumyield strength in shear of 145MPa and G 77 Gpa. What is the maximum residual stress in the shaft and residual (permanent) angle of twist, after removal of torque? 30 Applied torque, To find torque applied. . . A T r dAτ′ = ∫ 2 0 0 . .2 . .2 . R R r r dr r drτ π τ π= =∫ ∫ ( ) 30 2 18 145 .2 .r drπ= ∫ ( )( )3 3 2 30 18 145 3 π − = 145 MPa 6428452.551 NmmT′∴ = ( )4 4 60 36π × − 18 mm 30 mm Dept. of CE, GCE Kannur Dr.RajeshKN 35 ( ) 4 60 36 1107449.11 mm 32 J π × = =
- 36. To find residual stresses on an elastic rebound, (i.e., on removal of the T R J τ ′ ′ = 6428452.551 Nmm 30 1107449 11 × = torque) Shear stress on the surface (considering elastic rebound) J 1107449.11(considering elastic rebound) 2 174.142 N mm= Hence, residual shear stress on the outer surface 2 174.142 145 29.142 N mm= − = maxτ τ′= − τ′ max TR J τ= −Residual shear stress at 18 mmρ = 174 142 MPa= 145 MPa Residual stress τ max 6428452.551 Nmm 18 1107449.11 τ × = − 174.142 MPa 18 mm 30 mm 2 104.49 145 40.51 N mm= − = − Dept. of CE, GCE Kannur Dr.RajeshKN Hence the maximum residual shear stress is on the inner surface.
- 37. To find residual twist on an elastic rebound, Final twist per unit length (after yielding, when T=T’) L rG τ θ′ = L rG θ τ =∵ is valid from r=0 to 18. 3 145 1250 rad 0.131 rad 18 77 10 × = = × × Elastic twist recovered (considering elastic rebound) 8 77 0 3 2 4 6428452.551 Nmm 1250mm 77 10 N mm 1107449 11mm × = × × Elastic twist recovered (considering elastic rebound) T L GJ θ ′ = 0.094 rad= Hence, residual twist 0.131 0.09 0.034 7rad= − = 77 10 N mm 1107449.11mm× ×GJ 0 0.037rad 2.1= = Dept. of CE, GCE Kannur Dr.RajeshKN
- 38. Bending moment and shear force T f l d g Types of loads Point load Distributed loadDistributed load Uniformly distributed load Uniformly varying loady y g Couple T f tTypes of supports Fixed (built-in or encastre) Hinged (pinned)Hinged (pinned) Roller Guided fixed Dept. of CE, GCE Kannur Dr.RajeshKN 38 Elastic (Spring)
- 39. Beams Statically determinate and indeterminate beams Beams y • Simply supportedSimply supported • Cantilever • Propped cantilever • Fixed • Continuous Dept. of CE, GCE Kannur Dr.RajeshKN 39
- 40. Sign conventions Bending Momentg S i H iSagging Hogging Shear Force Clockwise Anticlockwise Dept. of CE, GCE Kannur Dr.RajeshKN 40 Clockwise Anticlockwise
- 41. Shear force and bending moment diagramsShear force and bending moment diagrams 1 Cantilever1. Cantilever • with a single load at the free end • with several point loadswith several point loads • with UD load over full span • with UD load over part spanp p • with a couple • with combination of loads Dept. of CE, GCE Kannur Dr.RajeshKN 41
- 42. B l B x (+)P SFD (-) Pl− Dept. of CE, GCE Kannur Dr.RajeshKN 42 BMD
- 43. l B x wl (+) SFDSFD 2 wl− (-) Dept. of CE, GCE Kannur Dr.RajeshKN 43 BMD
- 44. Si l t d b • with a single point load at the centre Simply supported beams • with a single point load at the centre • with a single eccentric point load • with several point loadsp • with UD load over full span • with UD load over part span h b f l d• with combination of loads • with a couple Dept. of CE, GCE Kannur Dr.RajeshKN 44
- 45. C P ( ) 2 SFD C A B 2 P− (+) (-)SFD 2 4 PL A B Dept. of CE, GCE Kannur Dr.RajeshKN 45 BMD C
- 46. 2 wL (+) 2 wL− SF diagram CA B 2 (+) (-) SF diagram 2 8 wL A B (+) 2 8 wL A B (+) Dept. of CE, GCE Kannur Dr.RajeshKN 46BM diagram C BM diagram C
- 47. P ba x Pb ( )L SFD CA B Pa L − (+) (-) SFD L Pab LA B (+) Dept. of CE, GCE Kannur Dr.RajeshKN BMD C
- 48. 60 kN 50 kN 1 A D C B 1m 3m 6m x Dept. of CE, GCE Kannur Dr.RajeshKN 48
- 49. Simply supported beam with overhangp y pp g • Overhang on one side with point loadOverhang on one side with point load • Overhang on one side with UD load over full span • Overhang on both sides with point loadsg • Overhang on both sides with UD load over full span Dept. of CE, GCE Kannur Dr.RajeshKN 49
- 50. P ba x 60 kN 30 kN/m 5 kN/m25 kNm A DCB E F 1m 2m1m 1.5m 2.5m Dept. of CE, GCE Kannur Dr.RajeshKN 50
- 51. Relation between shear force and bending momentRelation between shear force and bending moment w M M dM+ A C V V dV+ B D 0Y =∑ 0BM =∑ dx 0Y =∑ ( ) . 0V V dV w dx− + − = ( ) ( ) 2 0 2 dx M w V dV dx M dM+ + + − + = .dV w dx− = dV− 0Vdx dM− = dM V = Dept. of CE, GCE Kannur Dr.RajeshKN 51 w dx = V dx =
- 52. Example 1Example 1 2 x Px M = 2 x x dM P V dx = = Shear force 0xdV dx − = Load intensity Dept. of CE, GCE Kannur Dr.RajeshKN
- 53. Example 2Example 2 w kN/m Lx L 2 wL 2 wL 2 2 2 x wLx wx M = − 2 2 x x dM wL wx V dx = − = Shear force xdV w dx − = Load intensity Dept. of CE, GCE Kannur Dr.RajeshKN dx
- 54. Example 3Example 3 l B l x x is taken in the negative direction 2 2 x wx M − = x is taken in the negative direction Hence dx is negative 2 x x x dM dM wx V dx dx − = = = Shear forcedx dx− x xdV dV w dx dx − = = Load intensity Shear force Dept. of CE, GCE Kannur Dr.RajeshKN 54 dx dx− y
- 55. Shear force and bending moment diagrams by integrationShear force and bending moment diagrams by integration dV w − = dM V =( ) 1V wdx C∴ = − +∫ 2M Vdx C∴ = +∫ • Slope of SFD at any point is the load intensity at that cross section dx dx ( ) 1∫ 2∫ Slope of SFD at any point is the load intensity at that cross section • Slope of BMD at any point is the shear force at that cross section • Area of load diagram on the left (or right) of a cross section +g ( g ) reaction = shear force at that cross section • Area of SFD on a segment of a beam = Change in bending moment at that cross section i.e., dM Vdx= • For bending moment M to be a maximum or minimum, i.e.0 0, dM V= = Dept. of CE, GCE Kannur Dr.RajeshKN i.e.0 0, V dx
- 56. Example 1 Total load = W maxw L W= max 2 2 W W w L = ∴ = L L 3 W 2 3 W dV− ( )∫ 2w W To draw SFD dV w dx = ( ) 1V wdx C∴ = − +∫ max 2 2 At , w W x w x x L L = = 2 1 12 2 2 2 2 W W x V xdx C C L L − − ∴ = + = +∫ Dept. of CE, GCE Kannur Dr.RajeshKN
- 57. 22 12 Wx V C L − ∴ = + W W W At 0, 3 W x V= = x maxw W 1 10 3 3 W W C C∴ = + ⇒ = max 2 2 w x x L L = =2 2 3 Wx W V L ∴ − += Parabolic variation W 3 2 3 W− SFD Dept. of CE, GCE Kannur Dr.RajeshKN 3SFD
- 58. To draw BMD dM V dx = 2 22 3 Wx W M dx C L ⎛ ⎞− ∴ = + +⎜ ⎟ ⎝ ⎠ ∫⎝ ⎠ 3 22 i.e., 3 3 Wx Wx M C L − = + + 3 3L At 0 0x M= = 2i.e., 0 C= 3 Wx Wx M − ∴ = +At 0, 0x M 2, 2 3 3 M L ∴ = + Cubic variation i.e.0 0, dM V= = For maximum bending moment, 2 9 3 WL i.e.0 0, V dx 2 0 Wx W L− + ⇒ 9 3 3L Dept. of CE, GCE Kannur Dr.RajeshKN BMD 2 0 3 3 x L + = ⇒ =
- 59. Example 2 d ( ) 1V wdx C= − +∫ PP To draw SFD ∫ F t0 0 L Portion AB L/4 L/4 A B C D From to0 , 0 4 x x w= = = V C∴ = L PP Constant1V C∴ = But at 0,x V P= = 1C P⇒ = from 0 to, L x xV P =∴ == Constant , 1 from 0 to 4 , x xV P∴ 3L L L Portion BC From to 3 , 0 4 4 L L x x w= = = Also, at , 0 4 L x V= = 3L L Dept. of CE, GCE Kannur Dr.RajeshKN 3 from to 4 0, 4 L L xV x= =∴ = Constant
- 60. P P−P− SFD To draw BMD Portion AB 2M Vdx C= +∫ 2M Pdx C∴ = +∫ 2Px C= +From to0 , 4 L x x V P= = = Also, at 0, 0x M= = 2 0C⇒ = , from 0 4 to L M Px x x∴ = = = Dept. of CE, GCE Kannur Dr.RajeshKN Linear variation
- 61. Portion BC 2M Vdx C= +∫ Constant2M C∴ =From to 3 , 0 4 4 L L x x V= = = , 4 4 Also, at L PL x M= = 2 4 PL C⇒ = from 3 , 4 t 4 o 4 PL L L M x x∴ = = = 4 4 4 L L 4 PL 4 PL BMD4 4 Dept. of CE, GCE Kannur Dr.RajeshKN
- 62. SummarySummary Torsion - torsion of circular elastic bars - statically indeterminate problems - torsion of inelastic circular bars Axial force, shear force and bending moment - diagrammatic conventions for supports and loading, axial force, shear force and bending moment diagrams - shear force and bending moments bybending moment diagrams shear force and bending moments by integration and by singularity functions Dept. of CE, GCE Kannur Dr.RajeshKN 62