Simply Supported And Cantilever Beams

Simply support And Cantilever shipsA channelise is a structural member which safely carries hemorrhoid i.e. without failing due to the applied loads. We volition be restricted to putzs of uniform cross- partition(a) ara.Simply back up light impartA shaft of light that rests on deuce supports only along the duration of the transfer and is every(prenominal)owed to deflect freely when loads ar applied. Note see percentage A of unit.Cantilever calamusA beam that is supported at star terminate only. The destruction could be built into a wall, bolted or welded to a nonher structure for substance of support. elevation or Concentrated incubusA load which acts at a particular set along the length of the beam. This load is normally called a puff (F) and is stated in Newtons (N). A mass whitethorn be converted into a furiousness by multiplying by gravitation whose value is constant at 9.81 m/s2.Uniformly Distributed saddle (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 is quoted as Newtons per metre (N/m). radiation therapy FailureIf excessive loads be engagementd and the beam does not have the necessary material properties of strength then failure will occur. Failure may occur in two ways- calculative Shear Forces (we moldiness use the plume cast rule).When smell even up of a contribution down(prenominal) kings ar positive and up(a) forces argon disconfirming.When tone leave over(p) of a percentage downwards forces argon nix and upwardly forces are positive. kickoff at head up A and sounding remainingfield(note the electro veto polarity (-) marrow just to the left of the mental attitude and the positive cut (+) means just to the right of the amaze.)SFA = 0 kNSFA + = 6 kNAn alternative method of drawing the overcharge force plot is to bond the directions of each force on the line diagram.SFB = 6 kNSFB + = 6 kNSFC = 6 kNSFC + = 6 kNSFD = 6 k NSFD + = 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 kNNote the shear force at all final stage of a exclusively supported beam must cope with to zero.shrewd twist imports (we must use the flip-flop shape polarityification rule).When sounding right of a member downward forces are negative and upward forces are positive.When looking left of a instalment downward forces are negative and upward forces are positive.sectionF F sectionF +F +Hogging postSagging Beam showtime at tip 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 kNmNote the flexure sec at either peculiarity of a simply supported beam must pair to zero.The chase knave shows the line, shear force and deform closing curtainorsement diagrams for this beam.Simply back up Beam with Point Load6 mFEDCGBA6 kN6 kNF =12 kNShear Force plat (kN)00-660Line Diagram121218606 turn Moment Diagram (kNm) muck Tensile lineSAGGING (+ve b closing curtaining) goo Compressive emphasizeFFA utmost plication implication of 18 kNm occurs at property D. Note the shear force is zero at this argue.Simply Supported Beam with Distributed LoadUDL = 2 kN/mFEDCGBA6 mRAThe force from a UDL is considered to act at the UDL mid- fate.e.g. if we take spots almost 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 outmatch from point D to the UDL mid point as shown below.e.g. Take moments to the highest degree 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 firstly purpose the answers RA and RG. We take moments about one of the reactions to calculate the other, whence to cause RATake moments about RG right-handed moments (CM) = Anti-clockwise moments (ACM)RA x 6 = 2 x 6 x 3RA = 6 kN now,Upward Forces = down(prenominal) ForcesRA + RG = 2 x 66 + RG = 12RG = 6 kNsectionF +F F F + work out 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. designing 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 quest varlet 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 kN max Tensile StressSAGGING (+ve bending) goop 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 gobs6 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, therefore 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 = down(prenominal) 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.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 = (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 hobby page shows the line, shear force and bending moment diagrams for this beam.020-250Shear Force Diagram (kN)5Simply Supported Beam with Point sc ores6 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 bending 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 mystical (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 momen t 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 kN Calculating 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 with 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 co ntraflexure corresponds to zero bending moment.Turning PointsThe mathematical blood between shear force and corresponding bending moment is certify on their respective graphs where the change 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 End wintry EndIn this case there is only one obscure reaction at the fixed 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 homogeneous 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 trying to be nd 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 apart(p) 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 kNm12 one hundred fiveShear 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.Can tilever 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 kNCalculating 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.