Simply Supported And Cantilever Beams

Simply support And Cantilever dispersesA s repeal is a structural member which safely carries rafts i.e. without failing due to the applied loads. We will be restricted to beams of uniform cross- varianceal area.Simply Supported radiateA beam that rests on both supports only along the length of the beam and is allowed to deflect freely when loads are applied. bloodline see fraction A of unit.Cantilever BeamA beam that is supported at one finish only. The leftover could be built into a wall, bolted or welded to another structure for means of support. peak or Concentrated effectA load which acts at a particular dose along the length of the beam. This load is commonly called a pull in (F) and is verbalise in Newtons (N). A mass may be converted into a push back by multiplying by gravity whose revalue is constant at 9.81 m/s2.Uniformly Distributed Load (UDL)A load which is spread evenly over a given length of the beam. This may be the weight of the beam itself. The UDL i s quoted as Newtons per metre (N/m).Beam FailureIf excessive loads are engaged and the beam does not have the necessary material properties of strength then visitation will occur. Failure may occur in two ways-Calculating pluck Forces (we must use the shear pass rule).When tone cover of a branch down forces are affirmatory and upwards forces are prejudicial.When flavour left of a section downward(prenominal)lys forces are nix and upward forces are positive. scratch at point A and facial expression left(note the negative house (-) means just to the left of the position and the positive sign (+) means just to the flop of the position.)SFA = 0 kNSFA + = 6 kNAn alternative method of drawing the shear force diagram is to follow the directions of each force on the line diagram.SFB = 6 kNSFB + = 6 kNSFC = 6 kNSFC + = 6 kNSFD = 6 kNSFD + = 6 12 = -6 kNSFE = 6 12 = -6 kNSFE + = 6 12 = -6 kNSFF = 6 12 = -6 kNSFF + = 6 12 = -6 kNSFG = 6 12 = -6 kNSFG + = 6 12 + 6 = 0 kN bankers bill the shear force at either end of a apparently supported beam must equate to nothing.Calculating Bending Moments (we must use the diversion blink of an eye rule).When looking refine of a section downward forces are negative and upward forces are positive.When looking left of a section downward forces are negative and upward forces are positive.sectionF F sectionF +F +Hogging BeamSagging Beam starting at point A and looking leftBMA = 0 kNmBMB = (6 x 1) = 6 kNmBMC = (6 x 2) = 12 kNmBMD = (6 x 3) = 18 kNmBME = (6 x 4) + ( -12 x 1) = 12 kNmBMF = (6 x 5) + ( -12 x 2) = 6 kNmBMG = (6 x 6) + ( -12 x 3) = 0 kNm pit the deflexion moment at either end of a exclusively supported beam must equate to zero.The adjacent page shows the line, shear force and bending moment diagrams for this beam.Simply Supported Beam with Point Load6 mFEDCGBA6 kN6 kNF =12 kNShear Force Diagram (kN)00-660Line Diagram121218606Bending Moment Diagram (kNm) scoopful Tensile Stress swag (+ve bending) liquid ecstasy Compressive StressFFA maximum bending moment of 18 kNm occurs at position D. bank note the shear force is zero at this point.Simply Supported Beam with Distributed LoadUDL = 2 kN/mFEDCGBA6 mRAThe force from a UDL is considered to act at the UDL mid-point.e.g. if we take moments about D then the total force from the UDL (looking to the left) would be (2 x 3) = 6 kN. This force must be multiplied by the distance from point D to the UDL mid point as shown below.e.g. Take moments about D, then the moment would be (-6 x 1.5) = -9 kNm1.5mUDL = 2 kN/mDCBA3 mTaking moments about point D (looking left)We must first code the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to vex RATake moments about RGClockwise moments (CM) = Anti-clockwise moments (ACM)RA x 6 = 2 x 6 x 3RA = 6 kN now, upwardly Forces = Downward ForcesRA + RG = 2 x 66 + RG = 12RG = 6 kNsectionF +F F F +Calculating Shear Forces (we must use the shear force rule).When looking right of a section downward forces are positive and upward forces are negative.When looking left of a section downward forces are negative and upward forces are positive.Starting at point A and looking left(note the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.)SFA = 0 kNSFA + = 6 kNSFB = 6 (21) = 4 kNSFB + = 6 (21) = 4 kNSFC = 6 (22) = 2 kNSFC + = 6 (22) = 2 kNSFD = 6 (23) = 0 kNSFD + = 6 (23) = 0 kNSFE = 6 (24) = -2 kNSFE + = 6 (24) = -2 kNSFF = 6 (25) = -4 kNSFF + = 6 (25) = -4 kNSFG = 6 (26) = -6 kNSFG + = 6 (26) + 6 = 0 kNNote the shear force at either end of a simply supported beam must equate to zero.Calculating Bending Moments (we must use the bending moment rule).When looking right of a section downward forces are negative and upward forces are positive.When looking left of a section downward forces are negative and upward forces are positive.sectionF F sectionF +F +Hogging BeamSagging BeamStarting at point A and looking leftBMA = 0 kNmBMB = (6 x 1) + (-2 x 1 x 0.5) = 5 kNmBMC = (6 x 2) + (-2 x 2 x 1) = 8 kNmBMD = (6 x 3) + (-2 x 3 x 1.5) = 9 kNmBME = (6 x 4) + (-2 x 4 x 2) = 8 kNmBMF = (6 x 5) + + (-2 x 5 x 2.5 = 5 kNmBMG = (6 x 6) + + (-2 x 6 x 3) = 0 kNmNote the bending moment at either end of a simply supported beam must equate to zero.The following page shows the line, shear force and bending moment diagrams for this beam.Simply Supported Beam with Distributed Load420-2-4UDL = 2 kN/m6 mFEDCGBAShear Force Diagram (kN)00-660Line Diagram88950Bending Moment Diagram (kNm)56 kN6 kNMax Tensile StressSAGGING (+ve bending)Max Compressive StressFFA maximum bending moment of 9 kNm occurs at position D. Note the shear force is zero at this point.Simply Supported Beam with Point oodles6 mFEDCGBARARGF = 15 kNF = 30 kNWe must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, theref ore to find RATake moments about RGClockwise moments (CM) = Anti-clockwise moments (ACM)RA x 6 = (15 x 4) + (30 x 2)RA = 20 kN now,Upward Forces = Downward ForcesRA + RG = 15 + 3020 + RG = 45RG = 25 kNsectionF +F F F +Calculating Shear Forces (we must use the shear force rule).When looking right of a section downward forces are positive and upward forces are negative.When looking left of a section downward forces are negative and upward forces are positive.Starting at point A and looking left(note the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.)SFA = 0 kNSFA + = 20 kNSFB = 20 kNSFB + = 20 kNSFC = 20 kNSFC + = 20 -15 = 5 kNSFD = 20 -15 = 5 kNSFD + = 20 -15 = 5 kNSFE = 20 -15 = 5 kNSFE + = 20 -15 30 = -25 kNSFF = 20 -15 30 = -25 kNSFF + = 20 -15 30 = -25 kNSFG = 20 -15 30 = -25 kNSFG + = 20 -15 30 + 25 = 0 kNNote the shear force at either end of a simply supported beam must equate to zero.Calc ulating Bending Moments (we must use the bending moment rule).When looking right of a section downward forces are negative and upward forces are positive.When looking left of a section downward forces are negative and upward forces are positive.sectionF F sectionF +F +Hogging BeamSagging BeamStarting at point A and looking leftBMA = 0 kNmBMB = (20 x 1) = 20 kNmBMC = (20 x 2) = 40 kNmBMD = (20 x 3) + (-15 x 1) = 45 kNmBME = (20 x 4) + (-15 x 2) = 50 kNmBMF = (20 x 5) + (-15 x 3) + (-30 x 1) = 25 kNmBMG = (20 x 6) + (-15 x 4) + (-30 x 2) = 0 kNmNote the bending moment at either end of a simply supported beam must equate to zero.The following page shows the line, shear force and bending moment diagrams for this beam.020-250Shear Force Diagram (kN)5Simply Supported Beam with Point haemorrhoid6 mFEDCGBA20 kN25 kNF = 15 kNF = 30 kNBending Moment Diagram (kNm)004540205025Max Tensile StressSAGGING (+ve bending)Max Compressive StressFFA maximum bending moment of 50 kNm occurs at position E . Note the shear force is zero at this point.Simply Supported Beam with Point and Distributed Loads (1)6 mFEDCGBARARG15 kN30 kNUDL = 10 kN/mWe must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RATake moments about RGClockwise moments (CM) = Anti-clockwise moments (ACM)RA x 6 = (15 x 4) + (10 x 2 x 3) + (30 x 2)RA = 30 kN now,Upward Forces = Downward ForcesRA + RG = 15 + (10 x 2) + 3030 + RG = 65RG = 35 kNsectionF +F F F +Calculating Shear Forces (we must use the shear force rule).When looking right of a section downward forces are positive and upward forces are negative.When looking left of a section downward forces are negative and upward forces are positive.Starting at point A and looking left(note the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.)SFA = 0 kNSFA + = 30 kNSFB = 30 kNSFB + = 30 kNSFC = 30 kNSFC + = 30 15 = 15 kNSFD = 30 15 (10 x 1) = 5 kNSFD + = 30 15 (10 x 1) = 5 kNSFE = 30 15 (10 x 2) = -5 kNSFE + = 30 15 (10 x 2) 30 = -35 kNSFF = 30 15 (10 x 2) 30 = -35 kNSFF + = 30 15 (10 x 2) 30 = -35 kNSFG = 30 15 (10 x 2) 30 = -35 kNSFG + = 30 15 (10 x 2) 30 + 35 = 35 kNNote the shear force at either end of a simply supported beam must equate to zero.Calculating Bending Moments (we must use the bending moment rule).When looking right of a section downward forces are negative and upward forces are positive.When looking left of a section downward forces are negative and upward forces are positive.sectionF F sectionF +F +Hogging BeamSagging BeamStarting at point A and looking leftBMA = 0 kNmBMB = (30 x 1) = 30 kNmBMC = (30 x 2) = 60 kNmBMD = (30 x 3) + (-15 x 1) + (-10 x 1 x 0.5) = 70 kNmBME = (30 x 4) + (-15 x 2) + (-10 x 2 x 1) = 70 kNmBMF = (30 x 5) + (-15 x 3) + (-10 x 2 x 2) + (-30 x 1) = 35 kNmBMG = (30 x 6) + (-15 x 4) + (-10 x 2 x 3) + (-30 x 2) = 0 kNmNotesthe b ending moment at either end of a simply supported beam must equate to zero.The value of the maximum bending moment occurs where the shear force is zero and is therefore still mystic (see Shear Force diagram). The distance from point A to this zero SF point must be determined as follows-x = 215 20x = 1.5 m Total distance from point A = 2 + 1.5 = 3.5 mtherefore,BM max = (30 x 3.5) + (-15 x 1.5) + (-10 X 1.5 x 0.75) = 71.25 kNmThe following page shows the line, shear force and bending moment diagrams for this beam.7071.253530607000Simply Supported Beam with Point and Distributed Loads (1)2 mx30-5Shear Force Diagram (kN)0-351506 mFEDCGBA30 kN35 kN15 kN30 kNUDL = 10 kN/m20 kNBending Moment Diagram (kNm)Max Tensile StressSAGGING (+ve bending)Max Compressive StressFFA maximum bending moment of 71.25 kNm occurs at a distance 3.5 m from position A.Simply Supported Beam with Point and Distributed Loads (2)1 mRB12 mEDCFBA8 kNREUDL = 6 kN/mUDL = 4 kN/m12 kNWe must first calculate the reactions RB and RE. We take moments about one of the reactions to calculate the other, therefore to find RB.Take moments about REClockwise moments (CM) = Anti-clockwise moments (ACM)(RBx10)+(6x1x0.5) = (4 x 4 x 9) + (8 x 7) + (12 x 3) + (6 x 3 x 1.5)RB = 26 kNnow,Upward Forces = Downward ForcesRB + RE = (4 x 4) + 8 + 12 + (6 x 4)26 + RE = 60RE = 34 kNCalculating Shear ForcesStarting at point A and looking leftSFA = 0 kNSFA + = 0 kNSFB = -4 x 1 = -4 kNSFB + = (-4 x 1) + 26 = 22 kNSFC = (-4 x 4) + 26= 10 kNSFC + = (-4 x 4) + 26 8 = 2 kNSFD = (-4 x 4) + 26 8 = 2 kNSFD + = (-4 x 4) + 26 8 12 = -10 kNSFE = (-4 x 4) + 26 8 12 (6 x 3) = -28 kNSFE + = (-4 x 4) + 26 8 12 (6 x 3) + 34 = 6 kNSFF = (-4 x 4) + 26 8 12 (6 x 4) + 34 = 0 kNSFF + = (-4 x 4) + 26 8 12 (6 x 4) + 34 = 0 kNCalculating Bending MomentsStarting at point A and looking leftBMA = 0 kNmBMB = (-4 x 1 x 0.5) = -2 kNmBM 2m from A = (-4 x 2 x 1) + (26 x 1) = 18 kNmBM 3m from A = (-4 x 3 x 1.5) + (26 x 2) = 34 kNmBMC = (-4 x 4 x 2) + (26 x 3) = 46 kNmBMD = (-4 x 4 x 6) + (26 x 7) + (-8 x 4) = 54 kNmBM 9m from A = (-4 x 4 x 7) + (26 x 8) + (-8 x 5) + (-12 x 1) +(-6 x 1 x 0.5) = 41 kNmBM 9m from A = (-4 x 4 x 8) + (26 x 9) + (-8 x 6) + (-12 x 2) +(-6 x 2 x 1) = 22 kNmBME = (-4 x 4 x 9) + (26 x 10) + (-8 x 7) + (-12 x 3) +(-6 x 3 x 1.5) = -3 kNmBMF = (-4 x 4 x 10) + (26 x 11) + (-8 x 8) + (-12 x 4) +(-6 x 4 x 2) + (34 x 1) = 0 kNmPoint of ContraflexureAt any point where the graph on a bending moment diagram passes through the 0-0 datum line (i.e. where the BM changes sign) the curvature of the beam will change from hogging to sagging or vice versa. Such a point is termed a Point of Contraflexure or Inflexion. These points are identified in the following diagram. It should be noted that the point of contraflexure corresponds to zero bending moment. turn of events PointsThe mathematical relationship between shear force and corresponding bending moment is show on their respective graphs where the cha nge of slope on a BM diagram aligns with zero shear on the complementary shear force diagram. Thus, at any point on a BM diagram where the slope changes direction from upwards to downwards or vice versa, all such Turning Points occur at positions of Zero Shear. Turning points are also identified in the following diagram.Simply Supported Beam with Point and Distributed Loads (2)1 m26 kN12 mEDCFBA8 kN34 kNUDL = 6 kN/mUDL = 4 kN/m12 kN262-422-10Shear Force Diagram (kN)0-28100FFSAGGING (+ve bending)-3224154463418-2Bending Moment Diagram (kNm)00FFHOGGING (-ve bending)Points of ContraflexureThe maximum bending moment is equal to 54 kNm and occurs at point D where the shear force is zero. Turning points occur at -2 kNm and -3 kNm.Cantilever Beam with Point Load6 mFEDCGBARA12 kNFree blockadeFixed EndIn this case there is only one unknown reaction at the frigid end of the cantilever, thereforeUpward Forces = Downward ForcesRA = 12 kNCalculating Shear Forces Starting at point A and looking leftSFA = 0 kNSFA + = 12 kNSFB = 12 kNSFB + = 12 kNSFC = 12 kNSFC + = 12 kNSFD = 12 kNSFD + = 12 kNSFE = 12 kNSFE + = 12 kNSFF = 12 kNSFF + = 12 kNSFG = 12 kNSFG + = 12 12 = 0 kNNote the shear force at either end of a cantilever beam must equate to zero.Calculating Bending Moments NB for simplicity at this stage we shall always look towards the free end of the beam.Starting at fixed end, point A, and looking right towards the free end(the same results may be obtained by starting at point G and looking right)BMA = -12 x 6 = -72 kNmBMB = -12 x 5 = -60 kNmBMC = -12 x 4 = -48 kNmBMD = -12 x 3 = -36 kNmBME = -12 x 2 = -24 kNmBMF = -12 x 1 = -12 kNmBMG = 0 kNmNotesthe maximum bending moment in a cantilever beam occurs at the fixed end. In this case the 12kN force in the beam is essay to bend it downwards, (a clockwise moment). The support at the fixed end must therefore be applying an equal but opposite moment to the beam. This would be 72 kNm in an anti-clockwise direction. See the following diagram.The value of the bending moment at the free end of a cantilever beam will always be zero.-12-24-36-48-60-72Bending Moment Diagram (kNm)0012125Shear Force Diagram (kN)0072 kNm72 kNm6 mFEDCGBA12 kN12 kNThe following shows the line, shear force and bending moment diagrams for this beam.FFHOGGING (-ve bending)Max Tensile StressMax Compressive StressA maximum bending moment of -72 kNm occurs at position A.Cantilever Beam with Distributed LoadUDL = 2 kN/m6 mFEDCGBARATo calculate the unknown reaction at the fixed end of the cantileverUpward Forces = Downward ForcesRA = 2 x 6RA = 12 kNCalculating Shear ForcesStarting at point A and looking leftSFA = 0 kNSFA + = 12 kNSFB = 12 (2 x 1) = 10 kNSFB + = 12 (2 x 1) = 10 kNSFC = 12 (2 x 2) = 8 kNSFC + = 12 (2 x 2) = 8 kNSFD = 12 (2 x 3) = 6 kNSFD + = 12 (2 x 3) = 6 kNSFE = 12 (2 x 4) = 4 kNSFE + = 12 (2 x 4) = 4 kNSFF = 12 (2 x 5) = 2 kNSFF + = 12 (2 x 5) = 2 kNSFG = 12 (2 x 6) = 0 kNSFG + = 12 (2 x 6) = 0 kNNote the shear force at either end of a cantilever beam must equate to zero.Calculating Bending MomentsStarting at fixed end, point A, and looking right towards the free end(the same results may be obtained by starting at point G and looking right)BMA = -2 x 6 x 3 = -36 kNmBMB = -2 x 5 x 2.5 = -25 kNmBMC = -2 x 4 x 2 = -16 kNmBMD = -2 x 3 x 1.5 = -9 kNmBME = -2 x 2 x 1 = -4 kNmBMF = -2 x 1 x 0.5 = -1 kNmBMG = 0 kNmThe following page shows the line, shear force and bending moment diagrams for this beam.Cantilever Beam with Distributed Load864236 kNm36 kNm12105Shear Force Diagram (kN)00-1-4-9-16-25-36Bending Moment Diagram (kNm)006 mFEDCGBA12 kNUDL = 2 kN/mFFHOGGING (-ve bending)Max Tensile StressMax Compressive StressA maximum bending moment of -36 kNm occurs at position A.Cantilever Beam with Point and Distributed LoadsRG2 m10 kNBCDEAFG4 mUDL = 10 kN/mTo calculate the unknown reaction at the fixed end of the cantileverUpward Forces = Downward ForcesRG = (10 x 6) + 10RG = 70 kNCalc ulating Shear ForcesStarting at point A and looking leftSFA = 0 kNSFA + = 0 kNSFB = -10 x 1 = -10 kNSFB + = -10 x 1 = -10 kNSFC = -10 x 2 = -20 kNSFC + = (-10 x 2) + (-10) = -30 kNSFD = (-10 x 3) + (-10) = -40 kNSFD + = (-10 x 3) + (-10) = -40 kNSFE = (-10 x 4) + (-10) = -50 kNSFE + = (-10 x 4) + (-10) = -50 kNSFF = (-10 x 5) + (-10) = -60 kNSFF + = (-10 x 5) + (-10) = -60 kNSFG = (-10 x 6) + (-10) = -70 kNSFG + = (-10 x 6) + (-10) + 70 = 0 kNNote the shear force at either end of a cantilever beam must equate to zero.Calculating Bending MomentsStarting at point A, and looking left from the free end(the same results may be obtained by starting at point G and looking left)BMA = 0 kNmBMB = -10 x 1 x 0.5 = -5 kNmBMC = -10 x 2 x 1 = -20 kNmBMD = (-10 x 3 x 1.5) + (-10 x 1) = -55 kNmBME = (-10 x 4 x 2) + (-10 x 2) = -100 kNmBMF = (-10 x 5 x 2.5) + (-10 x 3) = -155 kNmBMG = (-10 x 6 x 3) + (-10 x 4) = -220 kNmThe following page shows the line, shear force and bending moment diagrams for this beam.70 kN2 m10 kNBCDEAFG4 mUDL = 10 kN/m00Shear Force Diagram (kN)-60-70-10-20-40-50220 kNm220 kNm-30Cantilever Beam with Point and Distributed Loads00Bending Moment Diagram (kNm)-220-5-20-55-100-155FFHOGGING (-ve bending)Max Tensile StressMax Compressive StressA maximum bending moment of -220 kNm occurs at position G.