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INDEX 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Types of beams Types of loads Types of support Shear force Bending Moment Sign conventions. Point of contraflexure Example on Simply supported beam Example on cantilever beam Example on overhanging beam Example on overhanging beam with point of contra flexure. BEAMS Members that are slender and support loads applied perpendicular to their longitudinal axis. Distributed Load, w(x) Concentrated Load, P Longitudinal Axis Span, L TYPES OF BEAMS Depends on the support configuration FH Pin FH FV FV Fixed Roller M Fv Roller FV Pin FV FH STATICALLY INDETERMINATE BEAMS Continuous Beam Propped Cantilever Beam Can you guess how we find the “extra” reactions? INTERNAL REACTIONS IN BEAMS At any cut in a beam, there are 3 possible internal reactions required for equilibrium: normal force, shear force, bending moment. P a b L INTERNAL REACTIONS IN BEAMS At any cut in a beam, there are 3 possible internal reactions required for equilibrium: normal force, shear force, bending moment. M Left Side of Cut N V Pb/L x Positive Directions Shown!!! INTERNAL REACTIONS IN BEAMS At any cut in a beam, there are 3 possible internal reactions required for equilibrium: normal force, shear force, bending moment. M V Positive Directions Shown!!! Right Side of Cut N Pa/L L-x FINDING INTERNAL REACTIONS Pick left side of the cut: Find the sum of all the vertical forces to the left of the cut, including V. Solve for shear, V. Find the sum of all the horizontal forces to the left of the cut, including N. Solve for axial force, N. It’s usually, but not always, 0. Sum the moments of all the forces to the left of the cut about the point of the cut. Include M. Solve for bending moment, M Pick the right side of the cut: Same as above, except to the right of the cut. EXAMPLE: FIND THE INTERNAL REACTIONS AT POINTS INDICATED. ALL AXIAL FORCE REACTIONS ARE ZERO. POINTS ARE 2-FT APART. P = 20 kips 1 2 3 4 5 8 9 10 6 7 8 kips 12 kips 12 ft 20 ft Point 6 is just left of P and Point 7 is just right of P. P = 20 kips 1 2 3 4 5 8 9 10 6 7 8 kips 12 kips 12 ft 20 ft 8 kips V x (kips) -12 kips M (ft-kips) 16 32 48 64 80 96 72 48 24 x V & M Diagrams 8 kips P = 20 kips 12 kips 12 ft 20 ft 8 kips V x (kips) What is the slope of this line? 96 ft-kips/12’ = 8 kips M (ft-kips) -12 kips 96 ft-kips b What is the slope of this line? -12 kips a c x V & M Diagrams 8 kips P = 20 kips 12 kips 12 ft 20 ft 8 kips V x (kips) What is the area of the blue rectangle? 96 ft-kips -12 kips 96 ft-kips b What is the area of the green rectangle? -96 ft-kips M (ft-kips) a c x DRAW SOME CONCLUSIONS The magnitude of the shear at a point equals the slope of the moment diagram at that point. The area under the shear diagram between two points equals the change in moments between those two points. At points where the shear is zero, the moment is a local maximum or minimum. The Relationship Between Load, Shear and Bending Moment w( x ) the load function M( x ) V( x )dx V( x ) w( x )dx Common Relationships 0 Constant Linear Constant Linear Parabolic Linear Parabolic Cubic Load Shear Moment Common Relationships Load 0 0 Constant Constant Constant Linear Linear Linear Parabolic M Shear Moment 12 kN 8 kN A C D B 1m RA = 7 kN 3m 1m RC = 13 kN 12 kN 8 kN A C D B 1m 3m 8 7 V 1m 8 7 -15 (kN) -5 7 M (kN-m) 2.4 m -8 TYPES OF BEAMS 1.Simply supported beam:A beam supported freely on the walls or columns at its both ends is known as simply supported beam. 2.Over hanging beam:A Beam is freely supported on two supports. But its one end or both the ends are projected beyond the support. TYPES OF BEAMS 3.Cantilever Beam: A Beam Fixed at one end and free at the other end is known as cantilever Beam. 4.Fixed Beam:A beam whose both ends are rigidly fixed is known as cantilever beam. TYPES OF BEAMS 5.Continuous beam:A Beam Supported on more then two support is known as continuous beam. TYPE OF LOADS 1. Point load or concentrated load:When a load is acting on a relatively small area it is considered as point load or concentrated load. W = Point Load It is given in N or KN. TYPES OF LOADS 2. Uniformly Distributed Load:A Load which is spread over a beam in such a manner that each unit length of beam is loaded to the same intensity is known as uniformly distributed load. W = U.D.L It is given in N/m or KN/m. TYPES OF LOADS 3.Uniformly Varying Load:A Load which is spread over a beam in such a manner that its intensity varies uniformly on each unit is called uniformly varying load. When load is zero at one end and increases uniformly to the other end it is known as triangular load. TYPES OF SUPPORT 1. Simple Support:Beam is freely supported on the support. There is no monolithic construction between the beam and the support. Only vertical reaction can devlope at the support. TYPES OF SUPPORT 2. Hinge Support: Beam is hinged to the support at end Beam can rotate about the hinge Vertical reaction (V) and horizontal reaction (H) can developed at the support. TYPES OF SUPPORT 3. Fixed Support :End of beam is rigidly fixed or built in wall. Beam can not rotate at the end. Vertical reaction (V), Horizontal Reaction (H), Moment (M) can devlope at support. TYPES OF SUPPORT 4. Roller Support :Beam is hinged to the support at the end roller are provided below the support. End of beam can move on rollers. Only vertical reaction (V) normal to the plane of roller can devlope. SHEAR FORCE AND BENDING MOMENT Shear Force:It is the algebraic sum of the vertical forces acting to the left or right of a cut section along the span of the beam Unit of S.F is N or kN Bending Moment:It is the algebraic sum of the moment of the forces to the left or to the right of the section taken about the section Unit of B.M is N.m or kN.m SIGN CONVENTION • The Shear Force is positive if it tends to rotate the beam section clockwise with respect to a point inside the beam section. • The Bending Moment is positive if it tends to bend the beam section concave facing upward. (Or if it tends to put the top of the beam into compression and the bottom of the beam into tension.) POINT OF CONTRAFLEXURE In B.M diagram, The point at which B.M change its sign from positive to negative or negative to positive is called point of contraflexure. It is a point where the beam tends to bend in opposite direction. It is the point at which curvature of beam changes. EXAMPLE ON SIMPLY SUPPORTED BEAM Problem 1 :Calculate the value and draw a bending moment and shear force diagram for following beam shown in fig. SOLUTION PROBLEM 1 Solution:- The beam has two unknown reaction Ay and By. Ay + By = 90……… (1) Taking moment from Ay By10 = (20 42)+(10 8) By = 240/10 By = 24kN Ay = 90-24 = 66kN S.F Calculation FA left = 0 FA Right = 66kN FC = 66-(204) = -14kN FD left= -14kN FD Right=(-14)-10=-24kN FB Left = -24 kN FB Right= -24 + 24 = 0kN B.M Calculation MA = 0 MC = (6640) – (2042) = (264-160) = 104kN.m MD =(66 8)-(20 4 6) =(528-140) =48kN.m MB = 0kN.m EXAMPLE ON CANTILEVER BEAM Problem 2 :- Calculate the value and draw a bending moment and shear force diagram for following cantilever beam shown in fig. SOLUTION PROBLEM 2 The support reaction for given beam can be easily determined by following method. Dy = (42 1/2) + (2 4) Dy =10 kN S.F Calculation FA Left = 0kN FA Right= -2 kN FB = -2-0 = -2kN FC = -2-8 = -10 FD = -10 B.M Calculation MA = 0 MB = -2 1 = -2kN.m MC = (-2 3) – (4 2 1) = -14kN.m MD = -14-10 = -24kN EXAMPLE ON OVERHANGING BEAM Problem 3 :- Calculate the value and draw a bending moment and shear force diagram for following Overhanging beam shown in fig. SOLUTION PROBLEM 3 Applying equation of static equilibrium. Ay - By = 45kN Taking Moment Of A By6 =(2510)+(5 4 4) By = (250+80)/6 = 55kN Ay = -10Kn S.F Calculation FA Left = 0kN FA Right = -10kN FC = -10kN FB Left = -10-35 = 25kN FD Right = 25kN FD Right = 25-25 = 0kN B.M Calculation MA = 0kN.m MC =(-102) =-20kN.m MB = -60-40= -100kN.m MD = 0kN.m EXAMPLE ON OVERHANGING BEAM WITH POINT OF CONTRAFLEXURE Problem 4 :- Calculate the value and draw a bending moment and shear force diagram for following beam shown in fig also find the point of contraflexure. SOLUTION PROBLEM 4 The support reaction can be find as following Ay+Cy=(104)+20 =60kN Taking moment from A Cy8=(1042)+(20 10) Cy=280/8=35kN Ay=60-35 = 25kN S.F Calculation FA left=0 FA right=25kN FB=25-(10 4)= -15 kN FC left= -15 kN FC right=25-5=20kN FD left=20kN FD right=20-20=0kN B.M Calculation MA = 0 MB=(254)-(10 4 2)= =20kN.m MC=(25 8)-(10 4 6) = -40kN.m Point of contra-flexure can be determined as writing the equation for BM and put part BC=0 Mx = 25x – 10*4(x-2) = 5.33 so X = 5.33 from A THANK U