BEAM A horizontal or inclined structural member that is designed to resist forces acting to its axis is called a beam

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1 BEM horizontal or inclined structural member that is designed to resist forces acting to its axis is called a beam INTERNL FORCES IN BEM Whether or not a beam will break, depend on the internal resistances built up in the beam and on the strength of material that the beam is made of. What are the internal resistences and how to find them? If the beam can still maintain in one piece without breaking under the actions of lateral loads, then the beam as a whole, and any part of it, must be in the condition of equilibrium. By means of equations of equilibrium for the plane structure, we should be able to find the internal resistance in a beam. ΣFx = 0, ΣFy = 0 and ΣM = 0. The internal forces for a beam section will consist of a shear force V, a bending moment M, and an axial force (normal force) N. For beams with no axial loading, the axial force N is zero. Sign Convention for Internal Forces in a Beam Sign Convention (considering a small segment of the member): Shear Force V: Positive shear tends to rotate the segment clockwise. Moment M: Positive moment bends the segment concave upwards. (so as to hold water ) xial Force N: Tension is positive. Page 4-1

2 n important feature of the above sign convention (often called the beam convention) is that it gives the same (positive or negative) results regardless of which side of the section is used in computing the internal forces. V M N N M V N xial Force N POSITIVE SIGN CONVENTION V Shear Force M Bending Moment M V PROCEDURES FOR FINDING V, M ND N T BEM SECTION 1. Identify whether the beam is a determinate or an indeterminate structure. (This chapter focuses on the analysis of determinate beam only. Indeterminate structure requires the consideration of compatibility condition, i.e. the deformation of the structure). 2. Compute the Support Reactions Make use of the equilibrium equations and the equations of condition if any. 3. Draw a Free-Body Diagram of the Beam Segment Keep all external loading on the member in their exact location. Draw a free-body diagram of the beam segment to the left or right of the beam. (lthough the left or right segment could equally be used, we should select the segment that requires the least computation). Indicate at the section the unknowns V, M and N. The directions of these unknowns may be assumed to be the same as their positive directions. Page 4-2

3 4. Use the Equations of Equilibrium to Determine V, M & N. If the solution gives a negative value for V, M or N, this does not mean the force itself is NEGTIVE. It tells that actual force or moment acts in the reversed direction only. 5. Check the Calculations using the Opposite Beam Segment if necessary. Page 4-3

4 Example 1 Find the shear and moment on sections at 2 m and 4 m from the left end for the following beam. Solution To analyze internal forces in the beam, it will first be necessary to calculate the reaction at one support. ΣM D = 0, V *7 = 40*4 + 20*2 V = kn ΣF x = 0, H = 0 ΣF y = 0, V + V D = V D = KN Page 4-4

5 free-body diagram of the part of the beam to the left of the section at 2 m is drawn to analyze the shear and moment on the section. t x = 2m, ΣF y = 0, V= kn ΣM 0 = 0, M = *2 M = knm (Sagging) new free-body diagram is drawn to shown V and M on the section 4 m from right of the left end. gain, both V and M are drawn in the positive direction for internal shear and moment. Page 4-5

6 t x = 4m, ΣF y = 0, = V V = kn (Negative sign indicates that the shear force acts upwards direction.) ΣM 0 = 0, *4 + 40*1 + M = 0 M = knm (Sagging) Page 4-6

7 Example 2 Determine the shear force V and the bending moment M at the section P of the overhanging beam shown. H B 10 kn 15 kn 4 kn/m B C D P V B VC 2m 4m 10m 3m Solution: No. of reactions = no. of equations of equilibrium This is a determinate beam. Determine reactions H B, V B and V C. ΣX = 0, H B = 0 Take moment about B, ΣM = 0, 4*10*(10/2) + 15*13 10*2 V C *10 = 0 V C = 37.5 kn ΣY = 0, V B + V C = * V B = 27.5 kn Page 4-7

8 Determine V and M at P (using left free-body) 10 kn H B 2m V B 4 kn/m M B P H VP 4m P P X = 0, since H B = 0, H P = 0 Y = 0, V B + V P = * V P = 26 V P = -1.5 kn (This implies that V p acts in downwards direction ) Take moment about P, 10*6 + 4*4*4/2 + M P = V B *4 M P = 27.5* = 18 knm Determine V and M at P (using right free-body) M P H P V P P 4 kn/m 6m V C 3m 15 kn Page 4-8

9 X = 0, H P = 0, Y = 0, V C + V P = * V P = 39 V P = +1.5 kn (This implies that V p acts in upwards direction as assumed) Take moment about P, 15*9 + 4*6*6/2 + M P = V C *6 M P = 37.5* = 18 knm (This implies that M P acts in the direction as indicated in the free-body diagram.) Page 4-9

10 Example 3 Determine the shear force V and the bending moment M at section P of the cantilever beam. 40 kn M H V 5 kn/m 4m 3m B P C 3m Solution: Determine the support reactions X = 0, H = 0, Y = 0, V = 5* = 70 kn Take moment about, 40 * 3 + 5*6*6/2 M =0 M = 210 knm Determine V and M at P (using left free-body) 40 kn 210 knm 5 kn/m P M P 70 kn 4m B V P H P X = 0, H P = 0, Y = 0, V P + 70 = *4 V P = -10 kn Page 4-10

11 Take moment about P, 40*1 + 5*4*4/ *4 - M p = 0 M P = 10 knm Determine V and M at P (using right free-body) M P P H P VP 5 kn/m 2m C X = 0, H P = 0, Y = 0, V P = 5*2 = 10 kn Take moment about P, 5*2*1 M p = 0 M P = 10 knm *Both the left free-body and the right free-body can be used to obtain the results. However, it is noted that by using the right free-body will greatly simplified the calculations. This shows importance of choosing the appropriate free-body. Page 4-11

12 SHER FORCE ND BENDING MOMENT DIGRMS By the methods discussed before, i.e. by using free-body diagrams, the magnitude and sign of the shear forces and bending moments may be obtained at many sections of a beam. When these values are plotted on a base line representing the length of a beam, the resulting diagrams are called, respectively, the shear force diagram and the bending moment diagram. Shear force and bending moment diagrams are very useful to a designer, as they allow him to see at a glance the critical sections of the beam and the forces to design for. Draftsman like precision in drawing the shear force and bending moment diagrams is usually not necessary, as long as the significant numerical values are clearly marked on the diagram. The most fundamental approach in constructing the shear force and bending moment diagrams for a beam is to use the procedure of sectioning. With some experience, it is not difficult to identify the sections at which the shear force and bending moment diagrams between these sections are readily identified after some experience and can be sketched in. Page 4-12

13 Example 4 Draw the shear force and the bending moment diagrams for the beam shown. 8 kn 2 kn/m H V V C B C 5m 3m Solution: X = 0, H = 0 Take moment about, 2*8*8/2 + 8*5 V C *8 = 0 V C = 13 kn Y = 0, V + V C = 2*8 + 8 V + 13 = 2*8 + 8 V = 11 kn 2 kn/m H X M X 11 kn x V X For 0 x 5m X = 0, H X = 0 Y = 0, V X + 2x = 11, V X =11-2x Page 4-13

14 Take moment about the cut, 11x 2x(x)/2 M x = 0, M x = 11x x 2 8 kn 2 kn/m H X 11 kn 5m B V X M X x For 5 x 8m X = 0, H X = 0 Y = 0, V X + 11 = 2x + 8, V X = 2x - 3 Take moment about X, 11x 2x(x)/2 8*(x-5) - M x = 0, M x = 11x x 2 8x + 40 M x = 3x x X (m) V (kn) M (knm) Page 4-14

15 8 kn 2 kn/m C B 11 kn 13 kn 5m 3m +ve B Shear Force (kn) C +ve B C Bending Moment (knm) Page 4-15

16 Example 5 Draw the shear force and bending moment diagrams for a cantilever beam carrying a distributed load with intensity varies linearly from w per unit length at the fixed end to zero at free end. M H V w kn/m wx l X B Solution: t any section distance x from the free end B, wx x wx V x = = l 2 2l 2 wx x wx M x = = 2l 3 6l 2 3 Page 4-16

17 M H V w kn/m M x H x V wx l wx l x X x B wl/2 2 V x = wx /2l Shear Force Diagram -wl /6 M = -wx 3 Bending Moment Diagram x /6l Page 4-17

18 Example 6 Draw the shear force and bending moment diagrams for the beam subject to a concentrated moment M* at point C. C M * B V a L b H B V B Solution: X = 0, H B = 0 Take moment about, V B * L = M *, V B = M * /L ( ) Y = 0, V B V = 0, V = M * /L ( ) Take the left free-body (for 0 x < a) M */L x H X M X V X X = 0, H X = 0 Y = 0, V X = M * /L Take moment about the cut section, (M * /L)*(x) = M X (hogging moment) Page 4-18

19 Take the left free-body (for a < x L) C M * M X M */L a Y = 0, V X = M * /L x V X Take moment about the cut section, (M * /L)*(x) + M X = M * M X = M * - (M * /L)*(x) M X = M * (L x)/l Page 4-19

20 C M * B M */L M */L a b L -M* /L Shear Force Diagram -M* a/l B C Bending Moment Diagram M * b/l Page 4-20

21 RELTIONSHIPS BETWEEN LOD, SHER FORCE ND BENDING MOMENT Consider a beam element subject to distributed load as shown below. q M M+dM V dx V+dV Y = 0, V = q(dx) + (V+dV) B dv q dv = dx V B V = = B qdx B qdx = - (area of load-intensity diagram between points and B) M = 0, -M + q(dx)(dx)/2 + (M+dM) - V(dx) = 0 Ignore the higher order terms, we get dm = V B B dm = Vdx dx M B M = Vdx B = area of shear force diagram between points and B. Page 4-21

22 SUMMRY OF THE RELTONSHIPS BETWEEN LODS, SHER FORCE ND BENDING MOMENTS Slope of shear force diagram at a point = Intensity of distributed load at that point Change in shear between points and B = rea under the distributed load diagram between points and B. Slope of bending moment diagram at a point. = Shear at that point Change in bending moment between points and B. = rea under the shear force diagram between points and B. Concentrated Loads Change in shear at the point of application of a concentrated load. = Magnitude of the load. Couples or Concentrated Moments Change in bending moment at the point of application of a couple. = Magnitude of the moment of the couple. Page 4-22

23 SHPES OF SHER FORCE ND BENDING MOMENT DIGRMS. Beams under Point Loads only 1. Shears are constant along sections between point loads. 2. The shear force diagram consists of a series of horizontal lines. 3. The bending moment varies linearly between point loads. 4. The bending moment diagram is composed of sloped lines. B. Beams under Uniformly Distributed Loads (UDL) only 1. Uniformly Distributed Load produces linearly varying shear forces. 2. The shear force diagram consists of a sloped line or a series of sloped lines. 3. UDL produces parabolically varying moment. 4. The bending moment diagram is composed of 2 nd -order parabolic curves. C. Beams under General Loading 1. Section with No Load: Shear force diagram is a Horizontal Straight Line. Moment Diagram is a Sloping Straight Line. 2. t a Point Load: There is a Jump in the Shear Force Diagram. 3. t a Point Moment: There is a Jump in the Bending Moment Diagram. 4. Section under UDL: Shear Force Diagram is a sloping straight line (1 st order) Bending Moment Diagram is a Curve (2 nd order parabolic) 5. Section under Linearly Varying Load Shear Force Diagram is a Curve (2 nd order) Bending Moment Diagram is a Curve (3 rd order) Page 4-23

24 6. The Curve of the Bending Moment Diagram is 1 order above the Curve of the Shear Force Diagram. 7. Maximum and Minimum Bending Moments occur where the Shear Force Diagram passes through the X-axis (i.e. at points of zero shear) (This characteristics is very useful in finding Max. and Min. bending moment.) Page 4-24

25 Example 7 Draw the shear force and bending moment diagrams for the beam shown. 20 kn 40 kn 4 kn/m B C D E F H V V F 1.5m 1.5m 3m 1.5m 1.5m Solution: X = 0, H = 0 kn Take moment about, 20* *3*(3 + 3/2) + 40*7.5 V F *9 = 0 V F = 42.7 kn Y = 0, V + V F = *3 V = 29.3 kn Page 4-25

26 Shear Force and Bending Moment 20 kn 40 kn 4 kn/m B C D E F m B C Shear Force (kn) D E F B C D E F Bending Moment (knm) Page 4-26

27 Example 8 Draw the shear force and bending moment diagrams for the beam shown. 10 kn 20 kn 4 kn/m B C D E H B VB V D 2m 2m 4m 2m Solution: X = 0, H B = 0 kn Take moment about B, 20*2 + 4*8*4 10*2 V D *6 = 0 V D = 24.7 kn Y = 0, V B + V D = *8 V B = 37.3 kn Page 4-27

28 Shear Force and Bending Moment 10 kn 20 kn 4 kn/m B C D E B C D Shear Force (kn) E B C -8 D 0 E Bending Moment (knm) Page 4-28

29 Deflected Shape of a Beam The qualitative deflected shape (also called elastic curve ) of a beam is simply an approximate and exaggerated sketch of the deformed beam due to the given loading. The deflected shape is useful in understanding the structural behaviour. Sketching the Deflected Shape of a Beam 1. The deflected shape must be consistent with the support conditions: (a) t a Roller Support, the vertical deflection is zero but the beam may rotate freely. (b) t a Pin Support, the vertical and horizontal deflections are zero but the beam may rotate freely. (c) t a Fixed Support, the vertical and horizontal deflections are zero and there is no rotation. 2. The deflected shape must be consistent with the Bending Moment Diagram. (a) Where the moment is positive, the deflected shape is concave upwards ( ). (b) Where the moment is negative, the deflected shape is concave downwards ( ). 3. The transition points between positive and negative moment regions are points of zero moment. These points are called point of inflection or point of contraflexure. 4. The deflected shape must be a smooth curve except at internal hinges. Normally, the vertical deflection at an internal hinge is not zero. 5. Quite often it is possible to sketch the deflected shape of a structure first and then to infer the shape of the bending moment diagram from the sketch. This is useful for checking whether a bending moment diagram obtained through calculations is correct. Page 4-29

30 Deflected Shape B C Bending Moment +ve moment -ve moment Deflected Shape B C point of inflection Page 4-30

31 Examples of Beam Deflected Shape Page 4-31

32 Examples of Beam Deflected Shape Page 4-32

33 Examples of Beam Deflected Shape Page 4-33

34 Examples of Beam Deflected Shape Page 4-34

35 Example 9 Construct the complete shear force and bending moment diagrams, and sketch the deflected shape for the beam shown. 10 kn 20 kn 30 kn/m B C D 2m 5m 2m Solution: 10 kn 20 kn 30 kn/m B C D H B V B V C 2m 5m 2m X = 0, H B = 0 kn Take moment about B, 20*7 + 30*7*(3.5 2) + (30*2/2)*(5 + 2/3) - 10*2 V C *5 = 0 V C = 121 kn Y = 0, V B + V C = *7 + 30*2/2 V B = 149 kn Page 4-35

36 B C D m -71 Shear Force (kn) B C D m Bending Moment (knm) 10 kn 20 kn 30 kn/m B C D P.I. P.I. Deflected Shape Page 4-36

37 PRINCIPLE OF SUPERPOSITION The principle of superposition states that on a linear elastic structure, the combined effect (e.g., support reactions, internal forces and deformation) of several loads acting simultaneously is equal to the algebraic sum of the effects of each load acting individually. There are two conditions for which superposition is NOT valid. 1. When the structural material does not behave according to Hooke s law; that is, when the stress is not proportional to the strain. 2. When the deflections of the structure are so large that computations cannot be based on the original geometry of the structure. Page 4-37

38 Principle of Superposition M H V w kn/m P2 P1 L L 2P L 1 (a) m1 B.M. due to P1 + (b) + P 2L 2wL2 m2 B.M. due to P2 (c) (d) m3 2P 1 L + P 2L + 2wL 2 m1 + m2 + m3 B.M. due to w Complete bending moment diagram Page 4-38

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40 SUPERPOSITION kn/m 40 knm 5 kn m 2.5m kn/m kn knm Superposition of loading Superposition of bending moment (knm) Page 4-40

41 Bending Moment Diagrams by Parts 10 kn -20 knm 5 kn/m 5 kn 25 kn 10 knm -20 knm 4m 2m 2.5 knm 5 kn/m 10 kn 10 kn 10 knm + 10 kn knm 5 kn 15 kn Bending Moment Page 4-41

42 Beam Deflection Beam deflection can be determined by using the following beam deflection table. The deflection of the beam is inversely proportional to the quantity EI, which is the flexural rigidity of the beam. Page 4-42

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45 Example 10 Use the methods of superposition to find the deflection at the free end of the following cantilever beam. EI of the beam is constant. Page 4-45

46 Solution Page 4-46

47 where W T = W 1 + W 2 From the deflection tables, W T = PL 3 /3EI + wl 4 /8EI W T = 2(9) 3 /(3EI) + 0.5(9) 4 /(8EI) W T = 896 /(EI) kn-m 3 Example 11 Use the method of superposition to find the deflection at the middle of the following simply supported beam. EI of the beam is constant. Page 4-47

48 Solution 6 m 3 m 10 kn B y kn/m B y 2 Total deflection at the mid-span of the beam, y M y M = y 1 + y 2 From the beam deflection tables, y M = Pa(3L 2-4a 2 ) / 48EI + 5wL 4 / 384 EI = 10(3)[3(12) 2-4(3) 2 ] / 48EI + 5(1.5)(12) 4 / 384EI = 652/EI kn-m 3 Page 4-48

49 Example 12 Determine the displacement at point C and the slope at the support of the beam. EI of the beam is constant. Solution Page 4-49

50 Total deflection at the point C of the beam, y c y c = y 1 + y 2 y c = PL 3 / 48EI + 5wL 4 / 768EI y c = 8(8) 3 / 48EI + 5(2)(8) 4 / 768EI y c = 139/EI kn-m 3 Total slope at the support of the beam, θ c θ c = θ 1 + θ 2 = PL 2 / 16EI + 3wL 3 / 128EI = 8(8) 2 / 16EI + 3(2)(8) 3 / 128EI = 56/EI (Clockwise direction) Page 4-50

51 Example 13 Use the method of superposition to find the deflection at the free end of the following cantilever beam. Page 4-51

52 Solution The uniform loading produces a deflection at B and because the beam is continuous and has a slope at B, an additional deflection occurs at C equal to (L 1 -L 2 )θ B as shown in the following figures. 0.5 kn/m B 0 L 1- L2 C y 1 y 2 2 kn B C y 3 Total deflection at C = y 1 + y 2 + y = wl 2 / 8EI + (L 1 -L 2 )*wl 2 /6EI + PL 1 / 3EI = 0.5(6) 4 / 8EI + 3*0.5*(6) 3 /6EI + 2*(9) 3 / 3EI = 624/EI kn-m 3 Page 4-52

53 Example 14 cantilever beam is subject to a uniformly distributed load of 0.5 kn/m over the length BC as shown in the following figure. If the flexural rigidity (EI) of the beam is 6000 kn-m 2, calculate the deflection at the free end C of the beam. Page 4-53

54 Solution 0.5 kn/m B C B 0.5 kn/m C y kn/m B 0 C y 2 y 3 Total deflection at C = y 1 - y 2 - y 3 = 0.5(7) 4 / 8EI - 0.5(2) 4 / 8EI -5*0.5*(2) 3 /6EI = /EI kn-m 3 = /6000 m = m Page 4-54

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Shear Force V: Positive shear tends to rotate the segment clockwise.

Shear Force V: Positive shear tends to rotate the segment clockwise. INTERNL FORCES IN EM efore a structural element can be designed, it is necessary to determine the internal forces that act within the element. The internal forces for a beam section will consist of a shear