Beam Structures and Internal Forces

Transcription

1 RCH 331 Note Set 7 Su2011abn Beam Structures and Internal Forces Notation: a = algebraic quantity, as is b, c, d = name for area b = intercept of a straight line d = calculus symbol for differentiation (C) = shorthand for compression F = name for force vectors, as is P, F, P = internal aial force F = force component in the direction F y = force component in the y direction FBD = free body diagram L = beam span length m = slope of a straight line M = internal bending moment M() = internal bending moment as a function of distance R = name for reaction force vector (T) = shorthand for tension = internal shear force () = internal shear force as a function of distance w = name for distributed load W = name for total force due to distributed load = horizontal distance y = vertical distance º = symbol for order of curve = symbol for integration = calculus symbol for small quantity = summation symbol BEMS - Important type of structural members (floors, bridges, roofs) - Usually long, straight and rectangular - Have loads that are usually perpendicular applied at points along the length Internal Forces 2 Internal forces are those that hold the parts of the member together for equilibrium - Truss members: F B F F F F B F - For any member: T F = internal aial force (perpendicular to cut across section) = internal shear force (parallel to cut across section) T T M = internal bending moment 1

2 RCH 331 Note Set 7 Su2011abn Support Conditions & Loading Most often loads are perpendicular to the beam and cause only internal shear forces and bending moments M Knowing the internal forces and moments is necessary when designing beam size & shape to resist those loads Types of loads R - Concentrated single load, single moment - Distributed loading spread over a distance, uniform or non-uniform. Types of supports - Statically determinate: simply supported, cantilever, overhang (number of unknowns < number of equilibrium equations) - Statically indeterminate: continuous, fied-roller, fied-fied (number of unknowns < number of equilibrium equations) L Propped Sign Conventions for Internal Shear and Bending Moment (different from statics and truss members!) L Restrained When F y **ecluding ** on the left hand side (LHS) section is positive, will direct down and is considered POSITIE. M When M **ecluding M** about the cut on the left hand side (LHS) section causes a smile which could hold water (curl upward), M will be counter clockwise (+) and is considered POSITIE. On the deflected shape of a beam, the point where the shape changes from smile up to frown is called the inflection point. The bending moment value at this point is zero. 2

3 RCH 331 Note Set 7 Su2011abn Shear nd Bending Moment Diagrams The plot of shear and bending moment as they vary across a beam length are etremely important design tools: () is plotted on the y ais of the shear diagram, M() is plotted on the y ais of the moment diagram. The load diagram is essentially the free body diagram of the beam with the actual loading (not the equivalent of distributed loads.) Maimum Shear and Bending The maimum value, regardless of sign, is important for design. Method 1: The Equilibrium Method Isolate FDB sections at significant points along the beam and determine and M at the cut section. The values for and M can also be written in equation format as functions of the distance to the cut section. Important Places for FBD cuts - at supports - at concentrated loads - at start and end of distributed loads - at concentrated moments Method 2: The Semigraphical Method Relationships eist between the loading and shear diagrams, and between the shear and bending diagrams. Knowing the area of the loading gives the change in shear (). Knowing the area of the shear gives the change in bending moment (M). Concentrated loads and moments cause a vertical jump in the diagram. Δ Δ lim D 0 d d C w D (the negative shows it is down because we give w a positive value) wd the area under the load curve between C & D C *These shear formulas are NOT LID at discontinuities like concentrated loads 3

4 RCH 331 Note Set 7 Su2011abn ΔM Δ lim 0 dm d D M M d the area under the shear curve between C & D D C C * These moment formulas RE LID even with concentrated loads. *These moment formulas are NOT LID at discontinuities like applied moments. The MXIMUM BENDING MOMENT from a curve that is continuous can be found dm when the slope is zero 0, which is when the value of the shear is 0. d Basic Curve Relationships (from calculus) for y() Horizontal Line: y =b (constant) and the area (change in shear) = b, resulting in a: Sloped Line: y = m+b y and the area (change in shear) =, resulting in 2 a: Parabolic Curve: y = a 2 + b y and the area (change in shear) =, 3 resulting in a: 3 rd Degree Curve: y = a 3 + b 2 +c + d Free Software Site: 4

5 RCH 331 Note Set 7 Su2011abn BSIC PROCEDURE: 1. Find all support forces. diagram: 2. t free ends and at simply supported ends, the shear will have a zero value. 3. t the left support, the shear will equal the reaction force. load height = shear w (force/length) width = w w 4. The shear will not change in until there is another load, where the shear is reduced if the load is negative. If there is a distributed load, the change in shear is the area under the loading. 5. t the right support, the reaction is treated just like the loads of step t the free end, the shear should go to zero. M diagram: 7. t free ends and at simply supported ends, the moment will have a zero value. 8. t the left support, the moment will equal the reaction moment (if there is one). 9. The moment will not change in until there is another load or applied moment, where the moment is reduced if the applied moment is negative. If there is a value for shear on the diagram, the change in moment is the area under the shear diagram. For a triangle in the shear diagram, the width will equal the height w! 10. t the right support, the moment reaction is treated just like the moments of step t the free end, the moment should go to zero. 5

6 RCH 331 Note Set 7 Su2011abn Parabolic Curve Shapes Based on Triangle Orientation In order to tell if a parabola curves up or down from a triangular area in the preceding diagram, the orientation of the triangle is used as a reference. Geometry of Right Triangles Similar triangles show that four triangles, each with ¼ the area of the large triangle, fit within the large triangle. This means that ¾ of the area is on one side of the triangle, if a line is drawn though the middle of the base, and ¼ of the area is on the other side By how a triangle is oriented, we can determine the curve shape in the net diagram. CSE 1: Positive triangle with fat side to the left. + 75% CSE 2: Positive triangle with fat side to the right. + 25% CSE 3: Negative triangle with fat side to the left. - 75% CSE 4: Negative triangle with fat side to the right. - 25% 6

7 RCH 331 Note Set 7 Su2011abn Eample 1 (pg 273) erify the general equation from Beam Diagrams & Formulas. 7

8 RCH 331 Note Set 7 Su2011abn Eample 2 (pg 275) 10 k 20 k R B R D R D(y) 10 k w=2 k/ft B C D 20 k 10 k + M+ 8

9 RCH 331 Note Set 7 Su2011abn Eample 3 (pg 283) 9

10 RCH 331 Note Set 7 Su2011abn Eample 4 (pg 285) 20 k R C R w =2 k/ft R C(y) B C 15 k 5 k + M+ 10

11 RCH 331 Note Set 7 Su2011abn Eample 5 (pg 286) M R 4 k R R y 2 k 8 k-ft 18 k-ft 8 k-ft 4 k 4 k 2 k 2 k B C + M+ 11

12 RCH 331 Note Set 7 Su2011abn Eample 6 (changed from pg 284) 0.75 m 0.75 m from the SOLUTION: Determine the reactions: 10 kn 6 kn F R 0 RB = 0 kn B F 10kN (2 kn )(3m) R 0 RBy = 16 kn y m By M ( 10kN )(2.25m) (6kN )(1.5 m) M 0 MRB = kn-m B RB Draw the load diagram with the distributed load as given with the reactions. Shear Diagram: Label the load areas and calculate: rea I = (-2 kn/m )(0.75 m) = -1.5 kn rea II = (-2 kn/m )(2.25 m) = -4.5 kn = 0 C = + rea I = kn = -1.5 kn and C = C + force at C = -1.5 kn -10 kn= kn B = C + rea II = kn -4.5 kn = -16 kn and B = B + force at B = -16 kn +16 kn= 0 kn Bending Moment Diagram: Label the load areas and calculate: rea III = (-1.5 kn)(0.75 m)/2 = kn-m rea I = (-11.5 kn)(2.25m) = kn-m rea = ( kn)(2.25m)/2 = kn-m M = 0 MC = M + rea III = kn-m = kn-m MB = MC + rea I + rea = kn-m kn-m kn-m = = kn-m and MB = MB + moment at B = kn-m kn-m = 0 kn-m R B R By M RB + (kn) *M+ (kn-m) 10 kn I C III w = 2 kn/m II I B *31.5 kn-m 16 kn M RB positive bending moment

13 RCH 331 Note Set 7 Su2011abn Eample 7 (pg 287) SOLUTION: Determine the reactions: 450 N 450 N F R 0 RB = 0 kn F y R B y ( 300 N )(3 ) (300 )(3 ) 1 2 N m m m m 2 R 1 By 0 or by load tracing Ry & RBy = ( wl/2)/2 = (300 N/m )(6 m)/4 = 450 N R B R B(y) M 450N)( 2 3m) (450N)(3 1 3m) R (6m) 0 RBy = 450 N ( 3 3 By R (y) w = 300 N/m Draw the load diagram with the distributed load as given with the reactions. I II Shear Diagram: Label the load areas and calculate: 450 N C B 450 N rea I = (-300 N/m )(3 m)/2 = -450 N rea II = -300 N/m )(3 m)/2 = -450 N = 0 and = + force at = N = 450 N C = + rea I = 450 N -450 N = 0 N B = C + rea II = 0 N 450 N = -450 N and B = B + force at B = -450 N N = 0 N Bending Moment Diagram: Label the load areas and calculate: + (N) 450 III reas III & I happen to be parabolic segments with an area of 2bh/3: rea III = 2(3 m)(450 N)/3 = 900 N-m rea I = -2(3 m)(450 N)/3 = -900 N-m I -450 M = 0 MC = M + rea III = N-m = 900 N-m MB = MC + rea I = 900 N-m N-m = 0 We can prove that the area is a parabolic segment by using the equilibrium method at C: M section cut M C ( 450N)(3m) (450N)( 1 3 3m) 0 M+ (N-m) 900 so Mc = 900 N-m 450 N M C 450 N 13