CHPTER 7 DEFLECTIONS OF EMS OJECTIVES Determine the deflection and slope at specific points on beams and shafts, using various analytical methods including: o o o The integration method The use of discontinuity functions (McCaulay) The virtual unit-load method 1
INTRODUCTION Deflection is a result from the load action to the beam (self weight, service load etc.) If the deflection value is too large, the beam will bend and then fail. Therefore it is vital that deflection must be limited within the allowable values as stipulated in the Standards The theory and background of deflection comes from curvature CURVTURE The deflection diagram of the longitudinal ais that passes through the centroid of each cross-sectional area of the beam is called curvature or elastic curve, which is characterized by the deflection and slope along the curve P P 2
ELSTIC CURVTURE Moment-curvature relationship: o Sign convention: From the figure, if DE = L ; = DE = L L = = Rθ R y θ CURVTURE Displacement, δ = L L = Rθ R y θ = y From strain, ε = δ L = yθ Rθ = y R R (R y) Therefore, curvature: 1 R = ε y In elastic region; E = σ ε ε = σ E D L C y E L
CURVTURE It is known that σ = My I Therefore : 1 R = ε y = σ Ey = My y 1 R = M where: = Stiffness or Fleure Rigidity (The higher the value, the stiffer the material the smaller the curvature) Deflection is influenced by I, E and L (and load) From 1 R = M SLOPE & DEFLECTION Y DOULE INTEGRTION. (1) Kinematic relationship between radius of curvature R and location : 1 R = d 2 y d 2 1 + dy d 2 /2 ut, in the case of elastic curve, the slope (dy/d) is too small ( 0) and can be ignored. Then: 1 = d2 y R d 2. (2) 4
SLOPE & DEFLECTION Y DOULE INTEGRTION Substitute Eq. (1) into Eq. (2): M = d2 y or d2 y = M d 2 d 2 Elastic curve differential equation (Moment Equation) fter integration: dy = θ = Md + C d 1 Double Integration will produce: y = [ Md]d + C 1 + C 2 (Slope Equation) (Deflection Equation) where C1 and C2 are the constant to be determined from the boundary conditions OUNDRY CONDITIONS The integration constants can be determined by imposing the boundary conditions, or continuity condition at specific locations beam cases are considered: I. Simply Supported eam y P L = 0; y = 0 = L; y = 0 5
II. Overhanged eam y OUNDRY CONDITIONS P L L 1 = 0; y = 0 III. Cantilever eam y = L; y = 0 P L = 0; y = 0 = 0; = 0 SIGN CONVENTIONS Deflection: + - Slope: + - ending moment: +ve -ve 6
EXMPLE 1 (Double Integration) Determine the mid-span deflection of beam shown below. Given E = 20 kn/mm 2 and I = 1600 10 6 mm 4. y 18 kn 2 m 4 m C 6 m EXMPLE 1: Solution Determine the Reaction Forces at and C Taking moment at C; M C = 0 V (6) 18(4) = 0 V = 12 kn F y = 0 V + V C = 0 V C = 6 kn 7
EXMPLE 1: Solution Segment (0 2) Segment C (2 6) 18 M M 12 12 M = 12 d2 y = M = 12 d2 dy d = 122 2 + C 1 y = 12 6 + C 1 + C 2 y = 2 + C 1 + C 2.(1) M = 12 18 2 = 6 + 6 d2 y = M = 6 + 6 d2 dy d = 62 + 6 + C 2 y = 6 + 62 6 2 + C + C 4 y C = + 18 2 + C + C 4.(2) oundary Conditions When = 0, y = 0 When = 6, y = 0 EXMPLE 1: Solution. (). () Matching Conditions t = 2; dy = dy d d C and y = y C Substitute () into Eq. (1): (0) = 2(0) + C 1 (0) + C 2 C 2 = 0 Substitute () into Eq. (2): (0) = -(6) + 18(6) + 6C + C 4 C 4 = -42 6C.() 8
From the Matching Conditions: EXMPLE 1: Solution 12(2) 2 + C 2 1 = 6(2)2 + 6(2) + C 2 C 1 = 6 + C.(4) 2(2) +C 1 2 = 2 + 18 2 2 + C 2 + C 4 2C 1 = 48 + 2C + C 4.(5) EXMPLE 1: Solution Substitute Eq. () and Eq. (4) into Eq. (5): 2(6 + C ) = 48 + 2C + ( 42 6C ) C = -76 From Eq. (): C4 = -42 6(-76) = 24 From Eq. (4): C1 = 6 76 = -40 Therefore: y = 1 2 40 (0 2) y = 1 + 18 2 76 + 24 (2 6) 9
To determine the deflection at mid-span, = m: y m = 1 = () + 18() 2 76() + 24 = 69 69 20 10 6 1600 10 EXMPLE 1: Solution 6 = 0.00215 m = 2.15 mm (NS) *Negative deflection value shows downward direction Conclusion: Each different load produces different section and 2 constant unknowns. Say if we have 4 sections (8 unknowns).therefore, this method is not practical. MC CULY METHOD This is a simplified method based on the double integration concept. In this method, only ONE section will be considered which is at the last loading type. 10
USE OF CONTINUOUS FUNCTIONS Mac Caulay functions a n = 0 ( a) n n a for a for a Integration of Mac Caulay functions: a n d = n+1 a + C n + 1 EXMPLE 2: Mac Caulay Method The cantilever beam shown in the figure below is subjected to a vertical load P at its end. Determine the equation of the elastic curve. is constant. y P y L Elastic curve 11
EXMPLE 2: Solution From the free-body diagram, with M acting in the positive direction as shown in figure, we have M = -P Integrating twice yields; d2 y d2 = P. (1) P dy d = P2 2 + C 1. (2) y = P 6 + C 1 + C 2. () V M Using the boundary conditions dy/d = 0 at = L and y = 0 at = L, Eq. (2) and Eq. () becomes; 0 = PL2 2 + C 1 0 = PL 6 + C 1L + C 2 Therefore, C 1 = PL2 and C 2 2 = PL Substituting these results, we get; EXMPLE 2: Solution θ = P 2 L2 2 and y = P 6 + L 2 2L (NS) P V M 12
EXMPLE Repeat Eample 1 with using the Mac Caulay Method. y 18 kn 2 m 4 m C 6 m EXMPLE : Solution Determine the Reaction Forces at and C Taking moment at C; M C = 0 V (6) 18(4) = 0 V = 12 kn F y = 0 V + V C = 0 V C = 6 kn 1
EXMPLE : Solution Consider to make a section after last load, i.e. in region C (section made from left to right). Therefore, the moment equation: M = 12 18( 2) 2 m 18 12 M Moment: M = 12 18 2 EXMPLE : Solution d2 y d2 = 12 18 2. (1) Slope: dy = 122 18 2 2 + C d 2 2 1 = 6 2 9 2 2 + C 1. (2) Deflection: 18 2 y = 12 +C 6 6 1 + C 2 = 2 2 + +C 1 + C 2. () 14
EXMPLE : Solution oundary Conditions t = 0, y = 0 and from Eq. (): (0) = 2 0 0 2 + +C 1 (0) + C 2 C 2 = 0 t = 6, y = 0 and from Eq. (): (6) = 2 6 6 2 + +C 1 (6) + 0 C 1 = 40 Therefore, Eq. (2) becomes: dy d = 62 9 2 2 40. (4) and Eq. () becomes: y = 2 2 40. (5) t mid-span, = m: y m = 2() 2 40 = 69 y m = 69 20 10 6 1600 10 EXMPLE : Solution 6 = 0.00215 m = 2.15 m (NS) dy d m = 6() 2 9 2 2 40 = 5 dy d m = 5 20 10 6 1600 10 6 = 1. 56 10 4 rad (NS) 15
EXMPLE 4 Determine the maimum deflection of the beam shown in the figure below. is constant. y C 8 kn D y D 120 knm C 10 m 20 m The beam deflects as shown in the figure. The boundary conditions require zero displacement at and (y = y = 0). 8 kn 10 m V = 6 kn EXMPLE 4: Solution 0 m The moment equation section at - is: M = 8 + 6 10 = 8 + 6 10 knm 120 knm V = 2 kn 16
Integrating twice yields: EXMPLE 4: Solution d2 y + 6 10 1 d 2. (1) dy d 42 + 10 2 + C 1. (2) y = 4 + 10 + C 1 + C 2. () oundary Conditions y = 0 at = 10 m and from Eq. (): 0 = 1 + 10 10 + C 1 10 + C 2 y = 0 at = 0 m and from Eq. (): 0 = 6000 + 0 10 + C 1 0 + C 2 C 1 = 1 and C 2 = 12000 From Eq. (2): EXMPLE 4: Solution dy d = 42 + 10 2 + 1. (4) From Eq. (): y = 4 + 10 + 1 12000. (5) To obtain the deflection at C, = 0. Therefore, from Eq. (5): y C = 12000 knm (NS) *The negative sign indicates that deflection is downward 17
EXMPLE 4: Solution To determine the length at point D, use Eq. (4) with 10 and dy d = 0 0 = 4 D 2 + D 10 2 + 1 4 D 2 + 60 D 16 = 0 Solving for the positive root, D = 20. m Hence from Eq. (5): EXMPLE 4: Solution y D = 4 20. + 20. 10 + 1(20.) 12000 y D = 5006 knm (NS) *The positive sign indicates that deflection is upward Comparing y D with y C, y ma = y C (NS) 18
EXMPLE 5 Determine the slope and deflection at = m. lso, determine the location and magnitude of the maimum deflection. Given = 4000 knm 2. 5 kn 10 knm 2 kn/m 2 m 2 m 2 m 2 m 2 m EXMPLE 5: Solution Determine the reaction forces at and Taking moment at, M = 0 10 + (2)(2)(5) + (5)(8) V (10) = 0 V = 7 kn F y = 0 V 2(2) 5 + V = 0 V = 2 kn 19
EXMPLE 5: Solution Cut at section - as shown in the figure. Therefore, the moment equation is given as: M = 2 1 + 10 2 0 2 4 2 + 2 6 2 5 8 1 2 2 5 kn 10 knm 2 kn/m 2 kn 2 m 2 m 2 m 2 m d2 y d 2 = M = 2 1 + 10 2 0 2 4 2 2 dy 2 2 10 2 1 = θ = + 2 4 d 2 1 6 y = 2 10 2 2 + 2 4 4 + 2 6 4 6 2 24 24 oundary Conditions t = 0 m, y = 0: C 2 = 0 t = 10 m, y = 0: 0 = 2 10 + 10 8 2 6 2 C 1 = 56 EXMPLE 5: Solution + 2 6 2 5 8 1 2 + 2 6 6 2 6 4 24 + 2 4 4 24 5 2 5 8 2 + C 2 1 5 8 + C 6 1 + C 2 6 + 10C 1 20
EXMPLE 5: Solution Slope Equation: θ = 2 + 10 2 1 4 Deflection Equation: y = + 5 2 2 4 4 12 + 6 + 6 4 12 5 8 2 56 2 5 8 56 6 Now, we can find the slope and deflection at = m. Given = 4000 knm 2 θ = 1 2 + 10 2 1 56 = 0.00925 rad 4000 y = 1 = + 5 2 2 56() = 0.0085 m 4000 EXMPLE 5: Solution To determine the position and magnitude of the maimum dy deflection, y ma when = 0: d (m) dy d 5-1. 5.5 +8.1 5.2 +2.5 From interpolation; ma = 5.1 m Therefore, at ma = 5.1 m; y ma = 48. mm (NS) 21
EXMPLE 6 Find the deflection equation for the given beam. Then, determine the maimum deflection at mid-span along span. Given E = 200 kn/mm 2 and I = 10 10 6 mm 4. 16 kn/m 20 kn C 4 m 2 m Determine the reactions forces at and Taking moment at, M = 0; 20(2) (16)(4)(2)+ V (4) = 0 V = 22 kn F y = 0; V + V 16(4) 20 = 0 V = 62 kn EXMPLE 6: Solution Cut at section - as shown in the figure. 16 kn/m 4 m 22 62 22
EXMPLE 6: Solution Therefore, the moment equation is given as: M = 22 1 + 62 4 1 16 4 + 16 4 2 2 d2 y d 2 = M = 22 1 + 62 4 1 8 2 + 8 4 2 dy 2 22 = θ = d 2 22 y = + 6 6 11 1 4 = + 2 62 4 + 8 2 + 8 4 + C 1 62 4 8 4 12 + 8 4 4 + C 12 1 + C 2 + C 1 + C 2 2 4 + 2 4 4 oundary Condition at t = 0 m, y = 0: C 2 = 0 oundary condition at t = 4 m, y = 0: (0) = 11 4 + 1 4 4 C 1 = 16 EXMPLE 6: Solution 2 4 4 + 2 4 4 4 + 4C 1 Therefore, the slope and deflection equations are: dy d = 11 2 + 1 4 2 8 + 8 4 16 11 1 4 y = + 2 4 + 2 4 4 16 2
To determine the maimum deflection and where it occurred, dy d = 0. Therefore: (0) = 11 2 + 1 4 2 8 EXMPLE 6: Solution + 8 4 16 Consider the maimum deflection occurs along span (0 4) (0) = 11 2 + 1 0 2 8 + 8 0 16 0 = 11 2 8 16 or 8 2 + 48 = 0 y try and error: ma = 1.52 m from. Therefore, the maimum deflection occurs when = 1.52 m from. To calculate the maimum deflection: 11 1.52 y ma = y ma = 15 y ma = 15 (200 10 6 )(10 10 6 ) EXMPLE 6: Solution 1 1.52 4 + 2 1.52 4 + 2 1.52 4 4 16(1.52) = 0.0075 m = 7.5 mm (downward) 24
MOMENT RE METHOD ased on the properties of elastic curve and bending moment diagram. Suitable to use for determining deflection and slope at a particular point. lso suitable for beam with different cross-section. There are two important theorems used in this method. Theorem 1 The angle between the tangents at any two points on the elastic curve equals the area under the M diagram between these two points. d2 y d 2 = d dy d d = M dθ = M d MOMENT RE METHOD Since θ dy, so dθ = M d d Therefore, θ = M d = 1 M rea 25
MOMENT RE METHOD Theorem 1 (Continued) This equation forms the basis for the first moment-area theorem: θ = θ / = M d Theorem 2 The vertical deviation of the tangent at point on the elastic curve with respect to the tangent etended from another point equals the moment of the area under the M MOMENT RE METHOD diagram between these two points ( and ). This moment is computed about point, where the vertical deviation t / is to be determined. 26
Theorem 2 (Continued) The vertical deviation of the tangent at with respect to the tangent at is given as: t / = M d MOMENT RE METHOD Then: t / = M d where is the location of the centroid of the shaded area M d between and. Theorem 2 (Continued) Centroid and rea MOMENT RE METHOD rea, = Centroid, = bh (n + 1) b (n + 2) b h *Notes: o Draw MD separately for each load, with one reference point o Write the bending moment equation in f() terms o Upward force produces positive bending moment and vice versa 27
Procedures: MOMENT RE METHOD EXMPLE 7 The beam is subjected to the concentrated force shown in the figure. Determine the reactions at the supports. is constant. P L L 28
EXMPLE 7: Solution The free-body diagram is shown in Fig. (b). P M V H L V L (b) EXMPLE 7: Solution Using the method of superposition, the separate M diagrams for the redundant reaction V and the load P are shown in Fig. (c). 2PL PL V L C + L 2L (c) 29
EXMPLE 7: Solution The elastic curve for the beam is shown in Fig. (d). t / = 0 tan tan (d) EXMPLE 7: Solution pplying Theorem 2, we have: t / = 2 L 1 2 V L V = 2.5P (NS) L + L 2 PL L + 2 L 1 2 PL L = 0 Using this result, the reactions at on the free-body diagram, of Fig. (b), are: + F = 0; H = 0 (NS) + F y = 0; V + 2.5P P = 0 V = 1.5P (NS) + M = 0; M + 2.5P(L) P(2L) = 0 M = 0.5PL (NS) 0
EXMPLE 8 The beam shown in the figure is pin supported at and roller at. point load of 60 kn is applied at 6 m from. Determine: i) Deflection at ii) Slope at iii) Maimum deflection m m m 60 kn C EXMPLE 8: Solution The deflection and slope diagram of the beam. Reference point is taken from. m m m 60 kn C V = 20 kn y t / V = 40 kn t C/ From the figure, y = t /. (1) where C/ 9 = t C/. (2) 1
EXMPLE 8: Solution The MD for each loading i.e. V = 20 kn and P = 60 kn 180 knm m 60 knm + C 6 m m 180 knm t / = + 1 1 2 60 1 = + 90 t C/ = + 1 1 2 9 180 1 9 1 2 180 1 = + 2160 EXMPLE 8: Solution = 1 2160 = 720 From Eq. (1); y = 720 90 = 60 Let say = 0,000 knm 2 y = 21 mm (NS) = + = θ tanθ = t C/ 9 = 2160 9 1 θ = + 1 2 θ = 2160 90 = 150 9 60 = + 90 60 kn C Let say = 0,000 knm 2 = 0.005 rad (NS) t C/ 2
Say y ma between 0 6 as shown in the figure: EXMPLE 8: Solution M 60 kn C y ma y ma = MM t M/ where MM = t C/ MM = t 9 C/ t M/ = + 1 1 2 20 1 10 = M t M/ t C/ y ma = 9 2160 = 1 10 240 10. () EXMPLE 8: Solution Maimum deflection occurs when dy d = 0 From Eq. (): dy ma = 1 240 10 = 0 d = 24 = 4.9 m < 6 m OK as assumed y ma = 1 240(4.9) 10(4.9) = 785 Let say = 0,000 knm 2 y ma = 785 0,000 = 26 mm (NS)