Shear Force V: Positive shear tends to rotate the segment clockwise.
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1 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 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 (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. 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 ending Moment M V HD in Civil Engineering Page 2-1
2 PROCEDURES FOR FINDING V, M ND N T EM 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-ody Diagram of the eam 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. 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 eam Segment if necessary. HD in Civil Engineering Page 2-2
3 Example 1 Determine the shear force V and the bending moment M at the section P of the overhanging beam shown. H 10 kn 15 kn 4 kn/m C D P V VC 2m 4m 10m 3m Solution: No. of reactions = no. of equations of equilibrium This is a determinate beam. Determine reactions H, V and V C. ΣX = 0, H = 0 Take moment about, ΣM = 0, 4*10*(10/2) + 15*13 10*2 V C *10 = 0 V C = 37.5 kn ΣY = 0, V + V C = * V = 27.5 kn HD in Civil Engineering Page 2-3
4 Determine V and M at P (using left free-body) 10 kn 4 kn/m MP P H H V VP 2m 4m P X = 0, since H = 0, H P = 0 Y = 0, V + 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 *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 HD in Civil Engineering Page 2-4
5 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.) HD in Civil Engineering Page 2-5
6 Example 2 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 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 V P H P X = 0, H P = 0, Y = 0, V P + 70 = *4 V P = -10 kn HD in Civil Engineering Page 2-6
7 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 *oth 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. HD in Civil Engineering Page 2-7
8 SHER FORCE ND ENDING MOMENT DIGRMS y 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. Draftmanlike 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. HD in Civil Engineering Page 2-8
9 Example 3 Draw the shear force and the bending moment diagrams for the beam shown. 8 kn 2 kn/m H V V C 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 HD in Civil Engineering Page 2-9
10 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 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) HD in Civil Engineering Page 2-10
11 8 kn 2 kn/m C 11 kn 13 kn 5m 3m +ve Shear Force (kn) C +ve C ending Moment (knm) HD in Civil Engineering Page 2-11
12 Example 4 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 Solution: t any section distance x from the free end, wx x wx V x = = l 2 2l 2 wx x wx M x = = 2l 3 6l 2 3 HD in Civil Engineering Page 2-12
13 M H V w kn/m wx l X M x H x V wx l x x wl/2 V = wx 2 x /2l Shear Force Diagram -wl /6 M = -wx 3 ending Moment Diagram x /6l HD in Civil Engineering Page 2-13
14 Example 5 Draw the shear force and bending moment diagrams for the beam subjected to a concentrated moment M* at point C. C M * V a L b H V Solution: X = 0, H = 0 Take moment about, V * L = M *, V = M * /L ( ) Y = 0, V 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) HD in Civil Engineering Page 2-14
15 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 HD in Civil Engineering Page 2-15
16 C M * M */L M */L a b L -M* /L Shear Force Diagram -M* a/l C ending Moment Diagram M * b/l HD in Civil Engineering Page 2-16
17 Example 6 Draw the shear force and bending moment diagrams for the beam shown. C is an internal hinge of the beam. M W W H V C D L L L L E V E Solution: This beam has four support reactions. We can use the equations of equilibrium (3 nos.) together with the equation of condition (1 no. internal hinge at C) to find the support reactions. Therefore this is a determinate beam. Cut the beam into the left and right free-body diagrams. M H V W W V C H C HC C D V L L L L C E V E Remember: 1. The internal forces at the hinge of the left and right free-body diagrams should be equal but opposite in direction. 2. There is no bending moment at the internal hinge. HD in Civil Engineering Page 2-17
18 Consider the right free-body, X = 0, H C = 0 Take moment about C, V E *(2L) WL = 0, V E = W/2 Y = 0, V E + V C = W, V C = W/2 Consider the left free-body, X = 0, H C = H = 0 Y = 0, V = V C + W, V = 3W/2 Take moment about, M = W*(L) + V C *(2L), M = 2WL HD in Civil Engineering Page 2-18
19 W W 2WL C D 3W/2 L L L L E W/2 3W/2 3W/2 W/2 W/2 D C -W/2 E Shear Force Diagram -2WL -WL/2 C D E ending Moment Diagram WL/2 HD in Civil Engineering Page 2-19
20 RELTIONSHIPS ETWEEN LOD, SHER FORCE ND ENDING MOMENT Consider a beam element subjected to distributed load as shown below. q M M+dM V dx V+dV Y = 0, V = q(dx) + (V+dV) dv q dv = dx V V = = qdx qdx = - (area of load-intensity diagram between points and ) M = 0, -M + q(dx)(dx)/2 + (M+dM) - V(dx) = 0 Ignore the higher order terms, we get dm = V dm = Vdx dx M M = Vdx = area of shear force diagram between points and. HD in Civil Engineering Page 2-20
21 SUMMRY OF THE RELTONSHIPS ETWEEN LODS, SHER FORCE ND ENDING MOMENTS Slope of shear force diagram at a point = Intensity of distributed load at that point Change in shear between points and = rea under the distributed load diagram between points and. Slope of bending moment diagram at a point. = Shear at that point Change in bending moment between points and. = rea under the shear force diagram between points and. 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. HD in Civil Engineering Page 2-21
22 SHPES OF SHER FORCE ND ENDING MOMENT DIGRMS. eams 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.. eams 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. eams 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 ending Moment Diagram. 4. Section under UDL: Shear Force Diagram is a sloping straight line (1 st order) ending Moment Diagram is a Curve (2 nd order parabolic) 5. Section under Linearly Varying Load Shear Force Diagram is a Curve (2 nd order) ending Moment Diagram is a Curve (3 rd order) HD in Civil Engineering Page 2-22
23 6. The Curve of the ending Moment Diagram is 1 order above the Curve of the Shear Force Diagram. 7. Maximum and Minimum ending 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.) HD in Civil Engineering Page 2-23
24 Example 7 Draw the shear force and bending moment diagrams for the beam shown. 20 kn 40 kn 4 kn/m 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 HD in Civil Engineering Page 2-24
25 Shear Force and ending Moment 20 kn 40 kn 4 kn/m C D E F m C Shear Force (kn) D E F C D E F ending Moment (knm) HD in Civil Engineering Page 2-25
26 Example 8 Draw the shear force and bending moment diagrams for the beam shown. 10 kn 20 kn 4 kn/m C D E H V V D 2m 2m 4m 2m Solution: X = 0, H = 0 kn Take moment about, 20*2 + 4*8*4 10*2 V D *6 = 0 V D = 24.7 kn Y = 0, V + V D = *8 V = 37.3 kn HD in Civil Engineering Page 2-26
27 Shear Force and ending Moment 10 kn 20 kn 4 kn/m C D E C D Shear Force (kn) E C -8 D 0 E ending Moment (knm) HD in Civil Engineering Page 2-27
28 Deflected Shape of a eam 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 eam 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 ending 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. HD in Civil Engineering Page 2-28
29 Deflected Shape C ending Moment +ve moment -ve moment Deflected Shape C point of inflection HD in Civil Engineering Page 2-29
30 Examples of eam Deflected Shape HD in Civil Engineering Page 2-30
31 Examples of eam Deflected Shape HD in Civil Engineering Page 2-31
32 Examples of eam Deflected Shape HD in Civil Engineering Page 2-32
33 Examples of eam Deflected Shape HD in Civil Engineering Page 2-33
34 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 C D 2m 5m 2m Solution: 10 kn 20 kn 30 kn/m C D H V V C 2m 5m 2m X = 0, H = 0 kn Take moment about, 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 + V C = *7 + 30*2/2 V = 149 kn HD in Civil Engineering Page 2-34
35 C D m -71 Shear Force (kn) C D m ending Moment (knm) 10 kn 20 kn 30 kn/m C D P.I. P.I. Deflected Shape HD in Civil Engineering Page 2-35
36 Example 10 Construct the complete shear force and bending moment diagrams, and sketch the deflected shape for the beam shown. is an internal hinge of the beam. 20 kn/m 50 kn C D 6m 6m 2m Solution: M H V 20 kn/m 50 kn C D V C 6m 6m 2m Consider the free-body diagram CD, HD in Civil Engineering Page 2-36
37 M H V H V 50 kn 20 kn/m C D V C V H Free-body Diagram X = 0, H = 0 kn Take moment about, 20*6*3 + 50*8 - V C *6 = 0 V C = kn Y = 0, V + V C = 20* V = 43.3 kn Consider the free-body diagram of X = 0, H = H = 0 kn Take moment about, 43.3*6 M = 0 M = knm Y = 0, V 43.3 = 0 V = 43.3 kn HD in Civil Engineering Page 2-37
38 C D m Shear Force (kn) HD in Civil Engineering Page 2-38
39 C D m ending Moment (knm) 20 kn/m 50 kn C D Deflected Shape pt. of inflection HD in Civil Engineering Page 2-39
40 Example 11 Construct the complete shear force and bending moment diagrams, and sketch the deflected shape for the beam shown. Joint C is an internal hinge. 30 kn/m 100 kn C D E 10 m 5m 3m 3m Solution: 30 kn/m 100 kn C D E H V V V E 10 m 5m 3m 3m HD in Civil Engineering Page 2-40
41 Consider the free-body diagram of CDE 100 kn 30 kn/m V c H C H c E c C D H V V V c V E y symmetry, V C = V E = 100/2 = 50 kn X = 0, H C = 0 kn Consider the free-body diagram of C, X = 0, H = H C = 0 kn Take moment about, 50* *15/2*5 V *10 = 0 V = kn Y = 0, V + V = *15/2 V = 87.5 kn HD in Civil Engineering Page 2-41
42 Determination of the position of zero shear force 30 kn/m 2X kn/m 50 kn 87.5 P M M P C X Let P be the position where the shear force equals to zero. Consider the vertical equilibrium of the right hand side diagram, 2X (X/2) = 0 X = m Determination of the value of maximum bending moment Consider the right hand side free body, Take moment about P, HD in Civil Engineering Page 2-42
43 2X (X/2)*(X/3) + 50X (X 5) + M =0 X 3 / X M = 0 M = / * M = knm. Determination of the position of zero bending moment Consider the free-body diagram below, 2X kn/m 50 kn Q C X Take moment about Q, 2X (X/2)*(X/3) + 50X (X 5) =0 X 3 / X = 0 X = 8.11 m HD in Civil Engineering Page 2-43
44 C D E m Shear Force (kn) C D E m m ending Moment (knm) kn/m 100 kn C D E P.I. Deflected Shape HD in Civil Engineering Page 2-44
45 Example 12 Construct the complete shear force and bending moment diagrams, and sketch the deflected shape for the beam shown. Joints D and F are internal hinges. Solution: 120 kn C 150 kn D E F G 10 kn/m 5m 5m 2m 3m 3m 2m 6m H Resolve the inclined external load into vertical and horizontal components. 120 kn 120 kn 90 kn 10 kn/m C H H D E F G V V C V G V H 5m 5m 2m 3m 3m 2m 6m HD in Civil Engineering Page 2-45
46 reak the beam into three free-body diagrams, namely CD, DEF and FGH. 120 kn 120 kn C H D V D V D 90 kn D E F H V F F H V 10 kn/m D F H F D F G H V V C V G V H H Consider the free-body FGH first, X = 0, H F = 0 Consider the free-body DEF, y symmetry, V D = V F = 120/2 = 60 kn X = 0, H D = H F +90 = = 90 kn. Consider the free-body CD, X = 0, H = H D, H = 90 kn. Take moment about, 120*5 + 60*12 = V C *10, V C = 132 kn. Y = 0, V + V C = , V = 48 kn. Consider the free-body FGH, Take moment about H, 60*8 + 10*8*4 = V G *6, V G = kn. Y = 0, V G + V H = 10*8 + 60, V H = 6.7 kn. HD in Civil Engineering Page 2-46
47 C D E F G m -6.7 H Shear Force (kn) C D E F G 2.2 H ending Moment (knm) 5.33 m 120 kn C D E 150 kn F G 10 kn/m H Deflected Shape HD in Civil Engineering Page 2-47
48 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. HD in Civil Engineering Page 2-48
49 Principle of Superposition M H V w kn/m P2 P1 L L 2P L 1 (a) m1.m. due to P1 + (b) + P 2L 2wL2 m2.m. due to P2 (c) (d) m3 2P 1 L + P 2L + 2wL 2 m1 + m2 + m3.m. due to w Complete bending moment diagram HD in Civil Engineering Page 2-49
50 SUPERPOSITION kn/m 40 knm 5 kn m 2.5m kn/m kn knm Superposition of loading Superposition of bending moment (knm) HD in Civil Engineering Page 2-50
51 ending 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 ending Moment HD in Civil Engineering Page 2-51
52 Tutorial 2 (nalysis of eams) Find the support reactions, draw the shear force and bending moment diagrams, and sketch the deflected shapes of the beams shown below. Q1. 15 kn/m C D (internal hinge) 8m 8m 4m Q kn 40 kn/m C D 5m 5m 12m C is an internal hinge Q kn 150 kn 25 kn/m C D E F 3m 3m 4m 4m 8m C is an internal hinge HD in Civil Engineering Page 2-52
53 Q kn 15 kn/m 10 kn/m C D E F G 6m 3m 3m 3m 3m 6m C & E are internal hinges Q kn 60 kn 80 kn 30 kn C D E F G 4m 4m 4m 4m 4m 4m E is an internal hinge Q6. 20 kn C D 3 kn/m E 2.5m 2.5m 2m 4m D is an internal hinge HD in Civil Engineering Page 2-53
54 Q kn 50 kn 50 kn 20 kn C D E F G 3m 3m 6m 3m 3m 6m D is an internal hinge Q8. 5 kn/m 5 kn 4 kn 3 kn C D E 3m 2m 2m 2m is an internal hinge Q9. 6 kn/m 3 kn/m C D E 5m 2m 5m 2m C is an internal hinge HD in Civil Engineering Page 2-54
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