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- 1. Lecturer; Dr. Dawood S. Atrushi November 2014 Shearing Force and Bending Moment M = RA x RA - P1 P1 (x - a1) RA - P1 )(a2 - a1) that point bending simple beam a < x < a + b V = RA - q (x - a) M = RA x - q (x - a)2 / 2
- 2. Content ¢ Beams ¢ Type of beams, loadings, and reactions ¢ Shear Forces and Bending Moments ¢ Relationships Between Loads, Shear Forces, and Bending Moments ¢ Shear-Force and Bending-Moment Diagrams 2 Shearing Force and Bending Moment - DAT November, 2014
- 3. Beams Members which are subjected to loads that are transverse to the longitudinal axis are termed as the beam. The members are either subjected to the forces or the moment having their vectors perpendicular to the axis of the November, 2014 bar. Shearing Force and 3 Bending Moment - DAT
- 4. both. When the load w per unit of the beam (as between A be uniformly distributed over Beams are classified according supported. Several types of Fig. 5.2. The distance L shown L both. When the load w per unit length of the beam (as between A and B be uniformly distributed over that Beams are classified according supported. Several types of beams Fig. 5.2. The distance L shown in L The transverse loading of a beam may consist of concentrated loads P1, P2, Types . . . , expressed of in Beams newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When ① the Statically load w per Determinate unit length has a beams constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is November, 2014 L (b) Overhanging beam L (c) Cantilever beam (b) Distributed load A B C Fig. 5.1 Transversely loaded beams. (a) Simply supported beam Statically Determinate Beams Statically Indeterminate Beams L1 L2 (d) Continuous beam L (b) Overhanging beam L Beam fixed at one end and simply supported (e) (b) Distributed load A B C Fig. 5.1 Transversely loaded beams. (a) Simply supported beam Determinate Indeterminate L1 L2 (d) Continuous beam L (b) Overhanging beam L Beam fixed at one end and simply supported (e) 4 Shearing Force and Bending Moment - DAT
- 5. Beams are classified according to the way in which they are of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is Beams are classified according to the way in which they are Types of Beams supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is ② Statically Indeterminate beams Fig. 5.2 Common beam support configurations. November, 2014 316 L (a) Simply supported beam Statically Determinate Beams Statically Indeterminate Beams L1 L2 (d) Continuous beam configurations. L (b) Overhanging beam L Beam fixed at one end and simply supported at the other end (e) L (c) Cantilever beam L ( f ) Fixed beam configurations. L (b) Overhanging beam L Beam fixed at one end and simply supported at the other end (e) L (c) Cantilever beam L ( f ) Fixed beam 5 Shearing Force and Bending Moment - DAT
- 6. P1 B C B C (a) Concentrated loads w P2 A D (b) Distributed load A B C Fig. 5.1 Transversely loaded beams. Type of Loads November, 2014 P1 (a) Concentrated loads w P2 A D (b) Distributed load A B C Fig. 5.1 Transversely loaded beams. a) Concentrated load (single force) b) Distributed load (measured by their intensity): i. Uniformly distributed load (uniform load) ii. Linearly varying load c) Couple 6 Shearing Force and Bending Moment - DAT
- 7. (a) Roller Support – resists vertical forces only ¢ Pinned support or hinged support. Hinge support or pin connection – resists horizontal and November, 2014 Type of Reactions (b) Hinge support or pin connection – resists horizontal and vertical forces vertical forces Hinge and roller supports are called as simple supports (c) Fixed support or built-in end Shearing Force and 7 Bending Moment - DAT Hinge and roller supports are called as simple supports
- 8. ¢ Roller support November, 2014 Type of Reactions 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 8 forces Shearing Force and Bending Moment - DAT
- 9. and roller supports are called as simple supports ¢ Fixed support November, 2014 Type of Reactions Hinge and roller supports are called as simple supports (c) Fixed support or built-in end Fixed support or built-in end Shearing Force and 9 Bending Moment - DAT
- 10. simply supported beam (simple beam) Calculate the reaction forces of on of the below members cantilever beam (fixed end beam) with an overhang November, 2014 shear beams Reactions cantilever beam (fixed end beam) beam with an overhang Shearing Force and 10 Bending Moment - DAT
- 11. a) MB = 0 - RA L + (P1 sin ) (L - a) + P2 (L - b) + q c2 / 2 = 0 (P1 sin ) (L - a) P2 (L - b) q c2 RA = CCCCCCC + CCCC + CC L L 2 L (P1 sin ) a P2 b q c2 12 P3 q1 b q1 b RB = CCCCC + CC + CC MA = 0 L MA = CC + L CC (L – 2 L 2b/3) + CC (L – b/3) 13 2 2 November, 2014 for the overhanging beam MB 0 - RA L + P4 (L– a) + M1 = 0 MA = 0 - P4 2 a + RB L + M1 = 0 P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC L L 4.3 Shear Forces and Bending Moments RA = CCCCCCC + CCCC + CC L L 2 L (P1 sin ) a P2 b q c2 RB = CCCCC + CC + CC L L 2 L for the cantilever beam Fx = 0 HA = 5 P3 / 13 12 P3 (q1 + q2) b Fy = 0 RA = CC + CCCCC 13 2 12 P3 q1 b q1 b MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam MB = 0 - RA L + P4 (L– a) + M1 = 0 MA = 0 - P4 a + RB L + M1 = 0 P4 (L– a) + M1 P4 a - M1 2 for the cantilever beam Fx = 0 HA = 5 P3 / 13 12 P3 (q1 + q2) b Fy = 0 RA = CC + CCCCC 13 2 b) c) Shearing Force and 11 Bending Moment - DAT
- 12. Shear Forces and Bending Moments At any point (x) along its length, a beam can transmit a bending moment M(x) and a shear force V(x). November, 2014 Bending-Moment Diagrams support beam AB = P a / L Shearing Force and 12 Bending Moment - DAT
- 13. MA = 0 - P4 a + RB L + M1 = 0 MB = RA L P4 (L– a) + M1 = 0 MA = 0 - P4 a + RB L + M1 = 0 B.F.MA = 0 - P4 a + RB L + M1 = 0 P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC Shear Forces P4 (L– and a) + Bending M1 Moments P4 a - M1 RA = CCCCCC RB = CCCC Consider a cantilever beam with a I II A B A ¢ P4 Consider (L– a) + M1 a cantilever L P4 a - beam with a concentrated load P L M1 applied at the end A, at the cross section mn, the shear force and bending moment are found Look RA at = CCCCCC the L FBD of an elemental RB = CCCC L length dx of the above loaded infinitesimal L length, the distributed I L load can be considered as a Uniformly (UDL) with constant magnitude w(x) over the differential length dx. It is now necessary to equate the equilibrium of the element. Starting
- 14. · concentrated load P applied at the end 4.3 Shear Forces and Bending Moments Consider a cantilever beam with a concentrated load P applied at the end A, at the cross section mn, the shear force and bending moment are found Shear Forces and Bending Moments Consider a cantilever beam with a concentrated load P applied at the end 4.3 Shear Forces and Bending Moments Consider a cantilever beam with a concentrated load P applied at the end A, at the cross section mn, the shear force and bending moment are found at the cross section mn, the shear force and bending moment are found F V x w x dx V x dV x y Dividing by dx in the limit as dxĺ0, n¦ § dx 0 0 ¸¹ the cross section mn, the shear and bending moment are found Fy = 0 V = P M = 0 M = P x M P a o ĺ
- 15. L ¦ § · x M x Fy = 0 V = P M = 0 M = P x Fy = 0 V = P M = 0 M = P x Fy = 0 V = P M = 0 M = P x ¦ M dM M x V x dx w x dx dx § M x R.H .Edge November, 2014 Dividing by dx in the limit as dxĺ0, ¨© dx
- 16. wx
- 17. dx dV x Taking moments about the right hand edge of the element:
- 18. 2 0 ¨©
- 19. dM x + P L a RAY=(1-a/L)P RBY=Pa/L RAY=(F.B.D. (global equilibrium) II Fig. 5.3 FBD of beam cut before force Step A: Cut beam just before the force P (i.e. Section I-I), and including the unknown shear force and bending moment as in Fig. Take moments about the right hand end (O):
- 20. 0 1 0 ¸¹ ¨© L M x P a ¨© §1 To determine the shear force, use Eq. (5.4), giving that:
- 21. P V x dM x ¸¹ a §1 · Shearing Force dx and Bending Moment L - DAT To verify Eq. (5.6), equate vertical equilibrium: ¨© + conventions (deformation sign conventions) 13 sign conventions (deformation sign conventions) conventions (deformation sign conventions) sign conventions (deformation sign conventions) the shear force tends to rotate the
- 22. Sign Convention ¢ The shear force tends to rotate the material clockwise is defined as positive ¢ The bending moment tends to compress the upper part of the and elongate the lower part is defined as positive November, 2014 = 0 M = P x conventions (deformation sign conventions) shear force tends to rotate the clockwise is defined as positive bending moment tends to compress part of the beam and elongate the part is defined as positive 3 M = 0 M = P x conventions (deformation sign conventions) shear force tends to rotate the material clockwise is defined as positive bending moment tends to compress upper part of the beam and elongate the part is defined as positive 3 M = 0 M = P x conventions (deformation sign conventions) shear force tends to rotate the material clockwise is defined as positive bending moment tends to compress upper part of the beam and elongate the part is defined as positive Shearing Force and 14 Bending Moment - DAT
- 23. Example 1 A simple beam AB supports a force P and a couple M0, find the shear V and bending moment M at a) x = L/2 b) x = (L/2)+ November, 2014 supports a force P the shear V and (b) at x = (L/2)+ Shearing Force and 15 Bending Moment - DAT
- 24. Example 2 A cantilever beam AB subjected to a linearly varying distributed load as shown, find the shear force V and the bending moment M November, 2014 = RA (L/2) + P (L/4) = P L / 8 - M0 / 2 = (L/2)+ [similarly as (a)] = - P / 4 - M0 / L = P L / 8 + M0 / 2 cantilever beam AB subjected to a varying distributed load as shown, find V and the bending moment M q0 x / L 0 - V - 2 (q0 x / L) (x) = 0 Shearing Force and 16 Bending Moment - DAT
- 25. M + 2 (q0 x / L) (x) (x / 3) = 0 Example 3 An overhanging beam ABC is supported to an uniform load of intensity q and a concentrated load P, calculate the shear force V and the bending moment M at D - q0 x3 / (6 L) - q0 L2 / 6 overhanging beam ABC is uniform load of intensity concentrated load P, calculate V and the bending D 17 Shearing Force and Bending Moment - DAT November, 2014
- 26. x 5 + 28 x 2 + 6 x 5 x 2.5 + M = 0 = 69 kN-m Relationships Between Loads, Shear Forces, and Bending Moments body diagram of the right-hand part, same results can be a) Consider an element of a beam of length dx subjected to distributed loads q. Between Loads, Shear Forces, and Bending Moments element of a beam of length distributed loads q forces in vertical direction November, 2014 Shearing Force and 18 Bending Moment - DAT 5
- 27. Fy = 0 V - q dx - (V + dV) = 0 Equilibrium of forces in vertical direction Fy = 0 V - q dx - (V + dV) = 0 or dV / dx = - q Fy = 0 V - q dx - (V + dV) = 0 integrate between two points A and B integrate between two points A and B Integrate B between two B points A and B; Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between the area of loading diagram may be positive or negative moment equilibrium of the element = - (area of the loading diagram between A = - (area of the loading diagram between the area of loading diagram may be positive November, 2014 or negative or dV / dx = - q or dV / dx = - q integrate between two points A and B B B Н dV = -Н q dx B B Н A dV = -Н A q dx A A i.e. B i.e. B VB - VA = -Н q dx VB - VA = -Н q dx A A and B) = - (area of the loading diagram between the area of loading diagram may be positive or negative Shearing Force and 19 Bending Moment - DAT
- 28. = - (area of the loading diagram between A the area of loading diagram may be positive or negative moment equilibrium of the element = - (area of the loading diagram between A and B) = - (area of the loading diagram between A the area of loading diagram may be positive or negative moment equilibrium of the element area of loading diagram may be positive or negative Moment equilibrium of the element moment equilibrium of the element M = 0 - M - q dx (dx/2) - (V + dV) dx + M or dM / dx = V maximum (or minimum) bending-moment occurs at i.e. at the point of shear force V = 0 = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0 or dM / dx = V M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM or dM / dx = V maximum (or minimum) bending-moment occurs at dM i.e. at the point of shear force V = 0 maximum (or minimum) bending-moment occurs at dM / dx = Maximum (or minimum) bending-moment point of shear integrate occurs force at between V dM/= dx two = 0 0, points i.e. at the A point and of B integrate between shear two points force V A = and 0. B November, 2014 integrate between two points A and B B B Н dM = Н V dx A A B B Н dM = Н V dx A A B B Н dM = Н V dx A A Shearing Force and 20 Bending Moment - DAT i.e. B
- 29. maximum (or minimum) bending-moment occurs at i.e. at the point of shear force V = 0 integrate between two points A and B Integrate between two points A and B = (area of the shear-force diagram between this equation is valid even when concentrated loads between A and B, but it is not valid if a couple and B November, 2014 or dM / dx = V maximum (or minimum) bending-moment occurs at i.e. at the point of shear force V = 0 integrate between two points A and B B B Н dM = Н V dx A A B B Н dM = Н V dx A A i.e. B i.e. B MB - MA = Н V dx MB - MA = Н A V dx A = (area of the loading = (area diagram of the between shear-force A diagram between this and equation B) is valid even when concentrated loads between A and B, but it is not valid if a couple and 21 B Shearing Force and Bending Moment - DAT
- 30. b) Consider an element of a beam of length dx subjected to concentrated load P. November, 2014 concentrated loads equilibrium of force V - P - (V + V1) = 0 or V1 = - P e. an abrupt change in the shear force occurs any point where a concentrated load acts Shearing Force and 22 Bending Moment - DAT
- 31. concentrated concentrated loads equilibrium Equilibrium of force of force; i.e. an abrupt V - P - (V + V1) = 0 change in the or V1 = - P shear force occurs at any point where a i.e. an abrupt change in the shear force occurs concentrated at any Equilibrium point where of a concentrated moment; load acts equilibrium of moment since the length dx of the element is infinitesimally November, 2014 is also infinitesimally small, thus, the bending moment does - M - P (dx/2) - (V + V1) dx + M + M1 or M1 = P (dx/2) + V dx + V1 dx M 0 loads equilibrium of force V - P - (V + V1) = 0 or V1 = - P i.e. an abrupt change in the shear force occurs at any point where a concentrated load acts equilibrium of moment - M - P (dx/2) - (V + V1) dx + M + M1 = 0 or M1 = P (dx/2) + V dx + V1 dx M 0 load acts. Since the length dx of the element is infinitesimally small, i.e. M1 is also infinitesimally small, thus, the bending moment does not change as we pass through the point of application of a concentrated load. since the length dx of the element is infinitesimally small, i.e. M1 is also infinitesimally small, thus, the bending moment does not change as we pass through the point of application of a concentrated load Shearing Force and 23 Bending Moment - DAT
- 32. c) Consider an element of a beam of length dx subjected to concentrated loads in form of couples, M0. November, 2014 - M - P (dx/2) - (V + V1) dx + M + M1 = 0 M1 = P (dx/2) + V dx + V1 dx M 0 the length dx of the element is infinitesimally small, i.e. M1 infinitesimally small, thus, the bending moment does not change as through the point of application of a concentrated load the form of couples equilibrium of force V1 = 0 Equilibrium of force; V1 = 0 change in shear force at the point of application of a couple equilibrium of moment Shearing Force and 24 Bending Moment - DAT M + M0 - (V + V1) dx + M + M1 = 0
- 33. equilibrium of force V1 = 0 i.e. no change in shear force at the point of ¢ i.e. no change in shear force at the point of application of a couple. November, 2014 application of a couple equilibrium of moment Equilibrium of moment; - M + M0 - (V + V1) dx + M + M1 = 0 or M1 = - M0 the bending moment changes abruptly at a point couple ¢ The bending moment changes abruptly at a point of application of a couple. Shearing Force and 25 Bending Moment - DAT
- 34. Shear-Force and Bending- Moment Diagrams Concentrated loads Consider a simply support beam AB with a concentrated load P 4.5 Shear-Force and Bending-Moment Diagrams concentrated loads consider a simply support beam AB with a concentrated load P RA = P b / L RB = P a / L Shear-Force and Bending-Moment Diagrams concentrated loads consider a simply support beam AB concentrated load P = P b / L RB = P a / L 0 x a V = RA = P b / L M = RA x = P b x / L V = RA = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V November, 2014 Shear-Force and Bending-Diagrams concentrated loads a simply support beam AB concentrated load P for 0 x a P b / L RB = P a / L x a = RA = P b / L Shearing Force and 26 Bending Moment - DAT
- 35. consider a simply support beam AB with a concentrated load P RA = P b / L RB = P a / L with a concentrated load P RA = P b / L RB = P a / L concentrated load P = P b / L RB = P a / L 0 x a V = RA = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V with a concentrated load P RA = P b / L RB = P a / L / L RB = P a / L x a RA = P b / L RA x = P b x / L that dM / dx = P b / L = V = P b / L RB = P a / L RA = P b / L RB = P a / L for 0 x a for 0 x a RA for 0 x a V = RA = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V V = RA = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V for 0 x a for 0 x a V = RA = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V V = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V V = RA = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V for a x L for a x L V = RA - P = - P a / L M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V L RA - P = - P a / L RA x - P (x - a) = P a (L - x) / L x L V = RA - P - P a / L M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V V = RA - P = - P a / L M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V for a x L V = RA - P = - P a / L M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V for a x L V = RA - P = - P a / L M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V V - P = - P a / L M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V V = RA - P = - P a / L M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V November, 2014 = P b / L RB = P a / L RA = P b RB = P a / L for 0 x a for 0 x a V = RA = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V V = RA = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V RA for a x L for a x L for a x L V = RA - P = - P a / L M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V that dM / dx = - P a / L = V with Mmax = P a b / L with Mmax = P a b / L RA = P b / L RB = P a / L for 0 x a V = RA = P b / L M = RA x = P b x / L note that dM / dx = P b / L = V for a x L V = RA - P = - P a / L M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V with Mmax = P a b / L Shearing Force and 27 Bending Moment - DAT with Mmax = P a b / L with Mmax = P a b / L with Mmax = P a b / L RB = P a / L a = P b / L = P b x / L dM / dx = P b / L = V L - P = - P a / L - P (x - a) = P a (L - x) / L dM / dx = - P a / L = V Mmax = P a b / L
- 36. - P = - P a / L With Mmax = Pab/L November, 2014 RA x - P (x - a) = P a (L - x) / L dM / dx = - P a / L = V = P a b / L V - Diagram M - Diagram Shearing Force and 28 Bending Moment - DAT
- 37. Uniform loads Consider a simply support beam AB with a uniformly distributed load of constant intensity q; November, 2014 8 load consider a simple beam AB with a distributed load of constant q RA = RB = q L / 2 V = RA - q x = q L / 2 - q x Shearing Force and 29 Bending Moment - DAT x - q x (x/2) = q L x / 2 - q x2 / 2 M = RA
- 38. - q x = q L / 2 - q x - q x (x/2) = q L x / 2 - q x2 / 2 note that dM / dx = q L / 2 - q x / 2 = V V - Diagram M - Diagram Mmax = q L2 / 8 at x = L / 2 November, 2014 = RB = q L / 2 V = RA - q x = q L / 2 - q x x - q x (x/2) = q L x / 2 - q x2 / 2 RA M = RA note that dM / dx = q L / 2 - q x / 2 = V = RB = q L / 2 Mmax = q L2 / 8 at x = L / 2 RA V = RA - q x = q L / 2 - q x M = RA x - q x (x/2) = q L x / 2 - q x2 / 2 several concentrated loads - q x x2 / 2 x / 2 = V for 0 x a1 V = RA M = RA x M1 = RA a1 Shearing Force and 30 Bending Moment - DAT for a1 x a2 V = RA - P1 RB = q L / 2 RA dM / dx = q L / 2 - q x / 2 = V = q L2 / 8 at x = L / 2 concentrated loads V = RA M = RA x
- 39. = RB = q L / 2 V = RA - q x = q L / 2 - q x - q x = q L / 2 - q x x - q x (x/2) = q L x / 2 - q x2 / 2 V = RA M = RA x - q x (x/2) = q L x / 2 - q x2 / 2 M = RA Several concentrated loads Consider a simply support beam AB with a several concentrated pi; RA x - q x (x/2) = q L x / 2 - q x2 / 2 note that dM / dx = q L / 2 - q x / 2 = V note that dM / dx = q L / 2 - q x / 2 = V Mmax = q L2 / 8 at x = L / 2 Mmax = q L2 / 8 at x = L / 2 for 0 x a1 V = RA M = RA x M1 = RA a1 for a1 x a2 V = RA - P1 M = RA x - P1 (x - a1) - M1 = (RA - P1 )(a2 - a1) November, 2014 x (x/2) = q L x / 2 - q x2 / 2 dM / dx = q L / 2 - q x / 2 = V = q L2 / 8 at x = L / 2 loads several concentrated loads several concentrated loads V = RA M = RA x = RA a1 V = RA - P1 = RA x - P1 (x - a1) M2 for 0 x a1 V = RA M = RA x M1 = RA a1 for a1 x a2 V = RA - P1 M = RA x - P1 (x - a1) M2 M1 = (RA - P1 )(a2 - a1) RA = RB = q L / 2 = RA - q x = q L / 2 - q x M = RA note that dM / dx = q L / 2 - q x / 2 = V Mmax = q L2 / 8 at x = L / 2 concentrated loads x a1 V = RA M = RA x M1 = RA a1 a1 x a2 V = RA - P1 M = RA x - P1 (x - a1) M2 - M1 = (RA - P1 )(a2 - a1) 31 Shearing Force and Bending Moment - DAT
- 40. M1 several concentrated loads several concentrated loads = RA a1 V = RA - P1 = RA x - P1 (x - a1) for 0 x a1 V = RA M = RA x several concentrated loads M1 = RA a1 for 0 x a1 V = RA M = RA x M2 for a1 x a2 V = RA - P1 M1 = RA a1 V - Diagram - M1 = (RA - P1 )(a2 - a1) M = RA x - P1 (x - a1) for a1 x a2 V = RA - P1 others because V = 0 at that point M = RA x - P1 (x - a1) M2 - M1 = (RA - P1 )(a2 - a1) similarly for others M2 = Mmax because V = 0 at that point shear-force and bending diagrams for the simple beam November, 2014 M1 = RA a1 for a1 x a2 V = RA - P1 M = RA x - P1 (x - a1) M2 - M1 = (RA - P1 )(a2 - a1) similarly for others M2 = Mmax because V = 0 at that point Example 4-4 construct the shear-force and bending -moment diagrams for the simple beam AB Example 4-4 for 0 x a1 V = RA M = RA x M1 = RA a1 for a1 x a2 V = RA - P1 M = RA x - P1 (x - a1) M2 - M1 = (RA - P1 )(a2 - a1) similarly for others M2 = Mmax because V = 0 at that point M - Diagram Example 4-4 construct the shear-force and bending -moment diagrams for the simple beam concentrated loads = RA M = RA x M1 = RA a1 V = RA - P1 = RA x - P1 (x - a1) M2 - M1 = (RA - P1 )(a2 - a1) others because V = 0 at that point - M1 = (RA - P1 )(a2 - a1) similarly for others M2 = Mmax because V = 0 at that point Example 4-4 construct the shear-force and bending 32 Shearing Force and Bending Moment - DAT -moment diagrams for the simple beam
- 41. Example 4 Construct the shear force and bending moment diagrams for the simple beam AB. November, 2014 M = RA x - P1 (x - a1) M2 - M1 = (RA - P1 )(a2 - a1) others because V = 0 at that point shear-force and bending diagrams for the simple beam b + 2c) / 2L b + 2a) / 2L Shearing Force and 33 Bending Moment - DAT
- 42. Example 5 Construct the V- and M-diagram for the cantilever beam supported to P1 and P2. November, 2014 = q b (2L - b) / 8 = c = 0 (uniform loading over the entire span) = q L2 / 8 V- and M-dia for the supported to P1 and P2 P2 MB = P1 L + P2 b a P1 M = - P1 x Shearing Force and 34 Bending Moment - DAT
- 43. Example 6 Construct the V- and M-diagram for the cantilever beam supporting to a constant uniform load of intensity q. November, 2014 and M-dia for the to a constant uniform MB Shearing Force and 35 Bending Moment - DAT = q L2 / 2
- 44. Example 7 An overhanging beam is subjected to a uniform load of q = 1 kN/m on AB and a couple M0 = 12 kN-m on midpoint of BC. Construct the V- and M-diagram for the beam. November, 2014 x = V - 0 = V = -Н q dx = - q x 0 x = M - 0 = M = -Н V dx = - q x2 / 2 0 overhanging beam is subjected to a q = 1 kN/m on AB M0 = 12 kN-m on midpoint construct the V- and M-dia for Shearing Force and 36 Bending Moment - DAT
- 45. Thank You 37 Shearing Force and Bending Moment - DAT November, 2014

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