Simply Supported And Cantilever Beams

Simply Supported And Cantilever Beams A beam is a structural member which safely carries loads i.e. without failing due to the applied loads. We will be restricted to beams of uniform cross-sectional area. Simply Supported Beam A beam that rests on two supports only along the length of the beam and is allowed to deflect freely when loads are applied. Note see section A of unit. Cantilever Beam A beam that is supported at one end only. The end could be built into a wall, bolted or welded to another structure for means of support. Point or Concentrated Load A load which acts at a particular point along the length of the beam. This load is commonly called a force (F) and is stated in Newtons (N). A mass may be converted into a force by multiplying by gravity whose value 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 is quoted as Newtons per metre (N/m). Beam Failure If excessive loads are used and the beam does not have the necessary material properties of strength then failure will occur. Failure may occur in two ways:- 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 kN SFA + = 6 kN An alternative method of drawing the shear force diagram is to follow the directions of each force on the line diagram.SFB = 6 kN SFB + = 6 kN SFC = 6 kN SFC + = 6 kN SFD = 6 kN SFD + = 6 12 = -6 kN SFE = 6 12 = -6 kN SFE + = 6 12 = -6 kN SFF = 6 12 = -6 kN SFF + = 6 12 = -6 kN SFG = 6 12 = -6 kN SFG + = 6 12 + 6 = 0 kN Note: 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. section F F section F + F + Hogging Beam Sagging Beam Starting at point A and looking left: BMA = 0 kNm BMB = (6 x 1) = 6 kNm BMC = (6 x 2) = 12 kNm BMD = (6 x 3) = 18 kNm BME = (6 x 4) + ( -12 x 1) = 12 kNm BMF = (6 x 5) + ( -12 x 2) = 6 kNm BMG = (6 x 6) + ( -12 x 3) = 0 kNm Note: 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 Point Load 6 m F E D C G B A 6 kN 6 kN F =12 kN Shear Force Diagram (kN) 0 0 -6 6 0 Line Diagram 12 12 18 6 0 6 Bending Moment Diagram (kNm) Max Tensile Stress SAGGING (+ve bending) Max Compressive Stress F F A maximum bending moment of 18 kNm occurs at position D. Note the shear force is zero at this point. Simply Supported Beam with Distributed Load UDL = 2 kN/m F E D C G B A 6 m RA The 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 kNm 1.5m UDL = 2 kN/m D C B A 3 m Taking moments about point D (looking left) We must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RA: Take moments about RG ÃŽ £Clockwise moments (CM) = ÃŽ £Anti-clockwise moments (ACM) RA x 6 = 2 x 6 x 3 RA = 6 kN now, ÃŽ £Upward Forces = ÃŽ £Downward Forces RA + RG = 2 x 6 6 + RG = 12 RG = 6 kN section F + 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 kN SFA + = 6 kN SFB = 6 (21) = 4 kN SFB + = 6 (21) = 4 kN SFC = 6 (22) = 2 kN SFC + = 6 (22) = 2 kN SFD = 6 (23) = 0 kN SFD + = 6 (23) = 0 kN SFE = 6 (24) = -2 kN SFE + = 6 (24) = -2 kN SFF = 6 (25) = -4 kN SFF + = 6 (25) = -4 kN SFG = 6 (26) = -6 kN SFG + = 6 (26) + 6 = 0 kN Note: 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. section F F section F + F + Hogging Beam Sagging Beam Starting at point A and looking left: BMA = 0 kNm BMB = (6 x 1) + (-2 x 1 x 0.5) = 5 kNm BMC = (6 x 2) + (-2 x 2 x 1) = 8 kNm BMD = (6 x 3) + (-2 x 3 x 1.5) = 9 kNm BME = (6 x 4) + (-2 x 4 x 2) = 8 kNm BMF = (6 x 5) + + (-2 x 5 x 2.5 = 5 kNm BMG = (6 x 6) + + (-2 x 6 x 3) = 0 kNm Note: 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 Load 4 2 0 -2 -4 UDL = 2 kN/m 6 m F E D C G B A Shear Force Diagram (kN) 0 0 -6 6 0 Line Diagram 8 8 9 5 0 Bending Moment Diagram (kNm) 5 6 kN 6 kN Max Tensile Stress SAGGING (+ve bending) Max Compressive Stress F F A maximum bending moment of 9 kNm occurs at position D. Note the shear force is zero at this point. Simply Supported Beam with Point Loads 6 m F E D C G B A RA RG F = 15 kN F = 30 kN We must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RA: Take moments about RG ÃŽ £Clockwise moments (CM) = ÃŽ £Anti-clockwise moments (ACM) RA x 6 = (15 x 4) + (30 x 2) RA = 20 kN now, ÃŽ £Upward Forces = ÃŽ £Downward Forces RA + RG = 15 + 30 20 + RG = 45 RG = 25 kN section F + 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 kN SFA + = 20 kN SFB = 20 kN SFB + = 20 kN SFC = 20 kN SFC + = 20 -15 = 5 kN SFD = 20 -15 = 5 kN SFD + = 20 -15 = 5 kN SFE = 20 -15 = 5 kN SFE + = 20 -15 30 = -25 kN SFF = 20 -15 30 = -25 kN SFF + = 20 -15 30 = -25 kN SFG = 20 -15 30 = -25 kN SFG + = 20 -15 30 + 25 = 0 kN Note: 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. section F F section F + F + Hogging Beam Sagging Beam Starting at point A and looking left: BMA = 0 kNm BMB = (20 x 1) = 20 kNm BMC = (20 x 2) = 40 kNm BMD = (20 x 3) + (-15 x 1) = 45 kNm BME = (20 x 4) + (-15 x 2) = 50 kNm BMF = (20 x 5) + (-15 x 3) + (-30 x 1) = 25 kNm BMG = (20 x 6) + (-15 x 4) + (-30 x 2) = 0 kNm Note: 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. 0 20 -25 0 Shear Force Diagram (kN) 5Simply Supported Beam with Point Loads 6 m F E D C G B A 20 kN 25 kN F = 15 kN F = 30 kN Bending Moment Diagram (kNm) 0 0 45 40 20 50 25 Max Tensile Stress SAGGING (+ve bending) Max Compressive Stress F F A 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 m F E D C G B A RA RG 15 kN 30 kN UDL = 10 kN/m We must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RA: Take moments about RG ÃŽ £Clockwise moments (CM) = ÃŽ £Anti-clockwise moments (ACM) RA x 6 = (15 x 4) + (10 x 2 x 3) + (30 x 2) RA = 30 kN now, ÃŽ £Upward Forces = ÃŽ £Downward Forces RA + RG = 15 + (10 x 2) + 30 30 + RG = 65 RG = 35 kN section F + 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 kN SFA + = 30 kN SFB = 30 kN SFB + = 30 kN SFC = 30 kN SFC + = 30 15 = 15 kN SFD = 30 15 (10 x 1) = 5 kN SFD + = 30 15 (10 x 1) = 5 kN SFE = 30 15 (10 x 2) = -5 kN SFE + = 30 15 (10 x 2) 30 = -35 kN SFF = 30 15 (10 x 2) 30 = -35 kN SFF + = 30 15 (10 x 2) 30 = -35 kN SFG = 30 15 (10 x 2) 30 = -35 kN SFG + = 30 15 (10 x 2) 30 + 35 = 35 kN Note: 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. section F F section F + F + Hogging Beam Sagging Beam Starting at point A and looking left: BMA = 0 kNm BMB = (30 x 1) = 30 kNm BMC = (30 x 2) = 60 kNm BMD = (30 x 3) + (-15 x 1) + (-10 x 1 x 0.5) = 70 kNm BME = (30 x 4) + (-15 x 2) + (-10 x 2 x 1) = 70 kNm BMF = (30 x 5) + (-15 x 3) + (-10 x 2 x 2) + (-30 x 1) = 35 kNm BMG = (30 x 6) + (-15 x 4) + (-10 x 2 x 3) + (-30 x 2) = 0 kNm Notes: the 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 unknown (see Shear Force diagram). The distance from point A to this zero SF point must be determined as follows:- x = 2 15 20 x = 1.5 m Total distance from point A = 2 + 1.5 = 3.5 m therefore, BM max = (30 x 3.5) + (-15 x 1.5) + (-10 X 1.5 x 0.75) = 71.25 kNm The following page shows the line, shear force and bending moment diagrams for this beam. 70 71.25 35 30 60 70 0 0 Simply Supported Beam with Point and Distributed Loads (1) 2 m x 30 -5 Shear Force Diagram (kN) 0 -35 15 0 6 m F E D C G B A 30 kN 35 kN 15 kN 30 kN UDL = 10 kN/m 20 kN Bending Moment Diagram (kNm) Max Tensile Stress SAGGING (+ve bending) Max Compressive Stress F F A 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 m RB 12 m E D C F B A 8 kN RE UDL = 6 kN/m UDL = 4 kN/m 12 kN We 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 RE ÃŽ £Clockwise moments (CM) = ÃŽ £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 kN now, ÃŽ £Upward Forces = ÃŽ £Downward Forces RB + RE = (4 x 4) + 8 + 12 + (6 x 4) 26 + RE = 60 RE = 34 kN Calculating Shear Forces Starting at point A and looking left: SFA = 0 kN SFA + = 0 kN SFB = -4 x 1 = -4 kN SFB + = (-4 x 1) + 26 = 22 kN SFC = (-4 x 4) + 26= 10 kN SFC + = (-4 x 4) + 26 8 = 2 kN SFD = (-4 x 4) + 26 8 = 2 kN SFD + = (-4 x 4) + 26 8 12 = -10 kN SFE = (-4 x 4) + 26 8 12 (6 x 3) = -28 kN SFE + = (-4 x 4) + 26 8 12 (6 x 3) + 34 = 6 kN SFF = (-4 x 4) + 26 8 12 (6 x 4) + 34 = 0 kN SFF + = (-4 x 4) + 26 8 12 (6 x 4) + 34 = 0 kN Calculating Bending Moments Starting at point A and looking left: BMA = 0 kNm BMB = (-4 x 1 x 0.5) = -2 kNm BM 2m from A = (-4 x 2 x 1) + (26 x 1) = 18 kNm BM 3m from A = (-4 x 3 x 1.5) + (26 x 2) = 34 kNm BMC = (-4 x 4 x 2) + (26 x 3) = 46 kNm BMD = (-4 x 4 x 6) + (26 x 7) + (-8 x 4) = 54 kNm BM 9m from A = (-4 x 4 x 7) + (26 x 8) + (-8 x 5) + (-12 x 1) + (-6 x 1 x 0.5) = 41 kNm BM 9m from A = (-4 x 4 x 8) + (26 x 9) + (-8 x 6) + (-12 x 2) + (-6 x 2 x 1) = 22 kNm BME = (-4 x 4 x 9) + (26 x 10) + (-8 x 7) + (-12 x 3) + (-6 x 3 x 1.5) = -3 kNm BMF = (-4 x 4 x 10) + (26 x 11) + (-8 x 8) + (-12 x 4) + (-6 x 4 x 2) + (34 x 1) = 0 kNm Point of Contraflexure At 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. Turning Points The mathematical relationship between shear force and corresponding bending moment is evidenced 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 m 26 kN 12 m E D C F B A 8 kN 34 kN UDL = 6 kN/m UDL = 4 kN/m 12 kN 2 6 2 -4 22 -10 Shear Force Diagram (kN) 0 -28 10 0 F F SAGGING (+ve bending) -3 22 41 54 46 34 18 -2 Bending Moment Diagram (kNm) 0 0 F F HOGGING (-ve bending) Points of Contraflexure The 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 Load 6 m F E D C G B A RA 12 kN Free End Fixed End In this case there is only one unknown reaction at the fixed end of the cantilever, therefore: ÃŽ £Upward Forces = ÃŽ £Downward Forces RA = 12 kN Calculating Shear Forces Starting at point A and looking left: SFA = 0 kN SFA + = 12 kN SFB = 12 kN SFB + = 12 kN SFC = 12 kN SFC + = 12 kN SFD = 12 kN SFD + = 12 kN SFE = 12 kN SFE + = 12 kN SFF = 12 kN SFF + = 12 kN SFG = 12 kN SFG + = 12 12 = 0 kN Note: 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 kNm BMB = -12 x 5 = -60 kNm BMC = -12 x 4 = -48 kNm BMD = -12 x 3 = -36 kNm BME = -12 x 2 = -24 kNm BMF = -12 x 1 = -12 kNm BMG = 0 kNm Notes: the maximum bending moment in a cantilever beam occurs at the fixed end. In this case the 12kN force in the beam is trying 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 -72 Bending Moment Diagram (kNm) 0 0 12 125 Shear Force Diagram (kN) 0 0 72 kNm 72 kNm 6 m F E D C G B A 12 kN 12 kN The following shows the line, shear force and bending moment diagrams for this beam. F F HOGGING (-ve bending) Max Tensile Stress Max Compressive Stress A maximum bending moment of -72 kNm occurs at position A. Cantilever Beam with Distributed Load UDL = 2 kN/m 6 m F E D C G B A RA To calculate the unknown reaction at the fixed end of the cantilever: ÃŽ £Upward Forces = ÃŽ £Downward Forces RA = 2 x 6 RA = 12 kN Calculating Shear Forces Starting at point A and looking left: SFA = 0 kN SFA + = 12 kN SFB = 12 (2 x 1) = 10 kN SFB + = 12 (2 x 1) = 10 kN SFC = 12 (2 x 2) = 8 kN SFC + = 12 (2 x 2) = 8 kN SFD = 12 (2 x 3) = 6 kN SFD + = 12 (2 x 3) = 6 kN SFE = 12 (2 x 4) = 4 kN SFE + = 12 (2 x 4) = 4 kN SFF = 12 (2 x 5) = 2 kN SFF + = 12 (2 x 5) = 2 kN SFG = 12 (2 x 6) = 0 kN SFG + = 12 (2 x 6) = 0 kN Note: the shear force at either end of a cantilever beam must equate to zero. Calculating Bending Moments 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 = -2 x 6 x 3 = -36 kNm BMB = -2 x 5 x 2.5 = -25 kNm BMC = -2 x 4 x 2 = -16 kNm BMD = -2 x 3 x 1.5 = -9 kNm BME = -2 x 2 x 1 = -4 kNm BMF = -2 x 1 x 0.5 = -1 kNm BMG = 0 kNm The following page shows the line, shear force and bending moment diagrams for this beam. Cantilever Beam with Distributed Load8 6 4 2 36 kNm 36 kNm 12 105 Shear Force Diagram (kN) 0 0 -1 -4 -9 -16 -25 -36 Bending Moment Diagram (kNm) 0 0 6 m F E D C G B A 12 kN UDL = 2 kN/m F F HOGGING (-ve bending) Max Tensile Stress Max Compressive Stress A maximum bending moment of -36 kNm occurs at position A. Cantilever Beam with Point and Distributed Loads RG 2 m 10 kN B C D E A F G 4 m UDL = 10 kN/m To calculate the unknown reaction at the fixed end of the cantilever: ÃŽ £Upward Forces = ÃŽ £Downward Forces RG = (10 x 6) + 10 RG = 70 kN Calculating Shear Forces Starting at point A and looking left: SFA = 0 kN SFA + = 0 kN SFB = -10 x 1 = -10 kN SFB + = -10 x 1 = -10 kN SFC = -10 x 2 = -20 kN SFC + = (-10 x 2) + (-10) = -30 kN SFD = (-10 x 3) + (-10) = -40 kN SFD + = (-10 x 3) + (-10) = -40 kN SFE = (-10 x 4) + (-10) = -50 kN SFE + = (-10 x 4) + (-10) = -50 kN SFF = (-10 x 5) + (-10) = -60 kN SFF + = (-10 x 5) + (-10) = -60 kN SFG = (-10 x 6) + (-10) = -70 kN SFG + = (-10 x 6) + (-10) + 70 = 0 kN Note: the shear force at either end of a cantilever beam must equate to zero. Calculating Bending Moments Starting 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 kNm BMB = -10 x 1 x 0.5 = -5 kNm BMC = -10 x 2 x 1 = -20 kNm BMD = (-10 x 3 x 1.5) + (-10 x 1) = -55 kNm BME = (-10 x 4 x 2) + (-10 x 2) = -100 kNm BMF = (-10 x 5 x 2.5) + (-10 x 3) = -155 kNm BMG = (-10 x 6 x 3) + (-10 x 4) = -220 kNm The following page shows the line, shear force and bending moment diagrams for this beam. 70 kN 2 m 10 kN B C D E A F G 4 m UDL = 10 kN/m 0 0 Shear Force Diagram (kN) -60 -70 -10 -20 -40 -50 220 kNm 220 kNm -30Cantilever Beam with Point and Distributed Loads 0 0 Bending Moment Diagram (kNm) -220 -5 -20 -55 -100 -155 F F HOGGING (-ve bending) Max Tensile Stress Max Compressive Stress A maximum bending moment of -220 kNm occurs at position G.

How to Change Bad to Good

In a society where children can no longer be spanked because it's considered child abuse, gun violence is at an all time high, and parents fear what ay happen as they are sending their children off to school, it's difficult to watch the evening news anymore. There was a time where people would leave the house to go to the grocery store for a loaf of bread and leave their doors unlocked. Today, if you leave your doors unlocked, there is a high probability you will be burglarized.What is the world coming to and is the Justice system really serving its purpose if there are so many repeated offenses, and in some cases, repeat offenders? For Assignment One for this week, I shared information about Singapore Criminal Justice System. The information I learned through my research was astounding. Singapore has one of the lowest crime rates in the world and while many feel Singapore punishments are inhumane and extreme, some feel Corporal punishment would allow individuals contemplating crimin al activity to think twice before they act out on impulsion.In 1994, American teenager Michael Fay was caned four strokes and sentenced to four months in Jail for vandalizing cars and public property, despite the United States appeals for a different sentence (The Wall Street Journal, 2010). After the caning and turning to the United States, Michael Fay was involved in a butane accident, burning his face and hands, and was subsequently admitted to the Hazarded rehabilitation program, located in Minnesota, for butane abuse (People Magazine, 1994, pig 60). According to People Magazine (1994), Fay blamed his butane abuse on his experience in Singapore.Following the butane incident, Michael Fay continued his troubles with the law. In Florida in 1996, he was cited for several traffic violations (The Atlanta Journal-Constitution, pig 82) and in 1998, Fay was charged with drug rappelling and possession of marijuana (Askew, 1998, pig 1). Did Fay continue to break the law in the United State s because he knew Corporal Punishment was not an option and his punishment would be much lighter than caning, or can his misdoings' really be a direct correlation to his caning experience?Why do people commit crime? Is there a difference between a man stealing a turkey at Thanksgiving because he is homeless and a man stealing clothing so he can then turn around and sell it to support his bad drug habit? It seems as though there should be a difference, et both men could face the same sentence. There are underlying psychological reasons why people commit crimes, however, some may be more reasonable than others. But then again, who defines reasonable? I believe in order to allow Justice to prevail; you must understand the criminal at hand.Why did the person commit the crime? Is the person a threat to society? Can the person be rehabilitated if given the correct direction and opportunities? Sentences should be appropriate to the crime committed. If a man steals a turkey because he is ho meless and is trying to eat, what DOD is it going to do sending him to Jail when he knows he will be giving a warm place to sleep and food to eat? Wouldn't that antagonize him to continue to steal? Instead of sending him to Jail, send him to work.Order him to pay for what he stole and continue to hold a steady Job so he is able to pay for food, contribute to society and show remorse for the crime committed out of disparity. If a man is stealing clothing to sell it for a drug habit, again, the situation should be assessed. Is he a threat to society and would he continue to steal if his substance abuse problem was obliterated? Instead of sending him to Jail, send him to a rehabilitation center. However, do not send the defendant to rehab for ninety days.If someone has a substance abuse problem severe enough to steal, ninety days is not going to rehabilitate the individual. An extensive rehabilitation program is called for. Repeated offenders are a concern. If the individual did not le arn from their prior punishment, it is obvious there must be some sort of attempt to mislead and commit crimes without being caught. For those individuals, they should be sentenced accordingly, and then sent to the military. The enlisting time should depend on the offense committed and their prior criminal history.As a victim of crime, I understand that Justice is not always served in the way some feel it should be served. For example, there are plenty of homicide cases reported in the news where the accused has been convicted before on murder charges. Why is this person out on the streets? If a person murders someone intentionally, there is absolutely no excuse for why they should be given another chance. I am a big proponent for Capital Punishment. If money takes a life, why should they be allowed to continue their life?There is a local municipal Judge in my area that I have much respect for, as do others in the community. His name is Judge Continent. Judge Continent believes in c reative sentencing. He has sentenced a woman who abandoned kittens in a box in the woods during the winter season to a night spent in the woods. He has sentenced a man who called a police officer a â€Å"pig† to stand on a busy city street corner with an actual pig and a sign that read â€Å"Police Officers are NOT pigs. † It is hard not to laugh at some of Hess sentences, but it does seem very close to the â€Å"eye for an eye† mentality, which I thoroughly support.I believe that if this type of mentality was used when handing down all sentences, the element of suspense and criminals thinking they will get a slap on the wrist will soon disappear. If the sentence is known before the crime is committed, and the sentences are uniform, people may think twice before committing a crime, especially if they know they will be subjected to the same treatment they caused someone else to endure. Is there really a quick fix for the United States Criminal Justice System?

Business Strategy Essay

For better understanding, this assignment has been organised in four sub- headings, they are as follows: the general overview of Porter’s Five Forces model; the importance or usefulness of Five Forces model; the criticisms and evaluation of Porter’s model; the recommendation and finally the conclusion. Overview of Porter’s model Johnson et al (2011) described Porter Five Forces as a strategic tool that helps identify the attractiveness of an industry in terms of five competitive forces: the threat of new entry, the threat of substitutes, the power of buyers, the power of suppliers and the extent of rivalry between competitors. Porter (1980) argues that this model is based on the insight that a good business strategy should meet the opportunities and threats in the organizations external environment. Particularly, competitive strategy should be based on an understanding of industry structures and the way they change. From the above explanation we can see that Porter’s Five Forces is a simple tool that supports strategic management in decision making through understanding where strength and weaknesses lie. Importance of Porter’s Five Forces The Porter’s Five Forces is a simple but powerful tool that supports strategic understanding where power lies in a business situation. It also helps to understand both the strength of the firm’s current competitive position, and strength of a position a company is looking to move to. †¢This model also emphasizes extended competition for value rather than just competition among existing rivals, and the simplicity of its application inspired numerous companies as well as business schools to adopt its uses as suggested by Wheelen and Hunger (1998). With clear understanding of where power lies, it will enable a company to take fair advantage of its strengths and improve on its weaknesses and off cause avoid taking wrong steps. Therefore to apply this planning tool efficiently, it is important to understand the situation and look at each of the forces individually. Criticisms and Evaluation of Five Forces Porters Five Forces is no doubt a powerful framework in determining the competitive nature of an industry. However it has been criticised by various commentators regarding its usefulness in today’s dynamic business environment. Although, Wheelen & Hunger (2002) recognised the five force they both thought that Stakeholders influence should have been included as the sixth force. This is because interest groups like the government, local communities, creditors, trade associations, special interest groups, unions and shareholders all have big influence on how the organisation operate. For instance, financial policies such as interest rates are being regulated by the government and may have a negative or positive effect on the organisation. Also as part of the stakeholders if creditors refuse to provide credit facilities for some firms could force them into liquidation. Porter’s model assumed that all businesses are competing against each other while in the actual sense some complement each to provide a better product, which are known as complementors. Complementors in this sense are businesses that provide complementary services to each other. These companies form strategic alliance to enhance the services that they provide either for the purpose of efficiency or cutting of cost. Example of this situation can be seen on iPods and headphones. Apple produced the iPod while Sony produces the headphone which Apple uses and same time Apple is competing with Sony on its brand of MP3 music player, in other words, each of these firm benefits from each other’s presence. Downes (1997) in his article â€Å"Beyond Porter† in the Context Magazine, suggested three new forces which he called; Digitalisation, Globalisation and Deregulation. Digitalization: He claims that organisations are now highly influenced by technological advancement, especially in information technology as well as retail organisations. Most firms no longer depend on high street sales rather online to boost their revenue. Some high street store also trying to play catch up with organisations like eBay and Amazon who have dominated the online. The introduction of credit cards by credit card organisations, loyalty cards by retail stores, all these new digital technologies allows companies to chase competitive advantage on a different level thereby creating new strategy to outweigh its competitors. Globalization: According to Downes (1997) days are gone when firms were only competing with their local rivals. Most organisations are now competing on international level by way of improved distribution channels, businesses can now buy and sell and on a global level. Customers, through the internet have the chance to shop around and compare prices globally. Even most medium size companies find themselves in an international market, even if they do not have branches overseas. Based on these facts it is not enough anymore for any business to position itself as a price-leader. Instead competitive advantages emerge now from the ability to develop long-term relationships to more mobile costumers and to manage extensive networks of customers and associates for mutual benefit. Deregulation: Industries like Communication, Railways, Airlines, now have different ways of competing against each other since firms are now in the hands of entrepreneurs’ unlike in the 1970 when most firms were in the hands of government. Deregulation creates more intense atmosphere for competition because every player in the industry wants to dominate the market. When compared to Porter’s Five Forces model, digitalization, globalization and deregulation have become more powerful forces in the last few decades, which Porter barely takes into consideration in his work. Today’s companies are highly influenced by technological progress, especially in information technology. Therefore, it is not advisable for any business to develop a strategy solely on the basis of Porters model. Shapiro and Varian (1998) claimed that economic laws do not change while technology do then went further to argue that Porter’s models are economic laws and rules that has been around for ages before his study but technology and information has moved and still moving and will continue to be a major force in competitive strategy. Presently, the level of information made available to businesses has intensified competition. For instance, firms can now gather so much information about their customers and how much they are willing to pay for the product. With customer’s shopping available to them they may choose to â€Å"lock them in† which would not have been possible many years ago when five forces was propounded. Despite the huge success of five forces in strategic management, this model did not give exact percentage of these forces to signal when the company is at risk, therefore the model on its own lacks enough information to actually guide decision makers. Although, still applicable in today’s dynamic business environment, it was advocated many years back when the business environment were not that competitive, for instant, when it was proposed it could be that a firm has only one supplier for its material which eliminated the possibility of strong competition among suppliers. This model from all indication assumed identifiable competitors, business partners, and customers who engage in more or less predictable ways. Kippenberger (1998), Haberberg & Rieple (2001) all suggested that it is not advisable to develop a strategy based on Five Forces model alone, it should be used in addition to other frameworks, thus SWOT and PESTEL analysis. This does not mean that Five Forces is invalid, but it is good to adopt a model knowing its merits and demerits. Rather than jump into a framework thinking it has solution to every situation. Recommendations Managers and decision makers in conducting analysis on Porter’s Five Forces should brainstorm on all relevant factors for the firm’s market situations and then check against the factors presented for each force. It will also be relevant to use positive sign to indicate forces that are in favour of the organisation and negative sign for forces strongly against the organisation in question. After identifying the favourable and unfavourable forces of the company’s performance and industry’s attractiveness, the manager should analyse the situation and examine the impact of these forces. Conclusion This assay has given a thorough account of Porter’s Five Forces by drawing attention to some of the criticisms of this model, which includes neglecting the all-powerful forces of Stakeholders, the emergent effects of digitalisation, globalisation and deregulation. Complementors and Collaborators as well as easy access to information were all overlooked by this framework. Through evaluation of this model, we also found that most organisations have managed to avoid the bargaining powers from either the supplier or distributors by way of backward or forward integration. This assignment acknowledged the usefulness of Porter’s model in strategic analyses even in today’s dynamic business environment but must not be used in isolation without considering other forces mentioned above. And finally organisations and decision makers should not assume that all their competitors and business partners are identifiable as suggested by this model.

Essay on Jewish Women in Medieval Ashkenaz - 1547 Words

Medieval Jewish society, like all traditional Jewish culture, was run by patriarchal hierarchy â€Å"Philosophical, medical, and religious views of the time all supported the view that men were superior to women both in nature and in deed† . Women’s position in society was secondary in comparison to that of men. They were characterized as lightheaded, weak, easily seduced, and linked to sorcery. This essay will focus on the Jewish women living in the medieval society of Ashkenaz, a region of northern France and Germany, around the time of 1000-1300 CE. Several questions will be addressed pertaining to the social status, educational opportunities, and their participation in society will be examined. Although not much was written about the†¦show more content†¦Furthermore, they endured violence by their husbands who treated them like children. Their husbands would often go away on business trips to the Muslim world, and abandon their wives while marrying another , and then abandon their second wives after years abroad to return to their first. Jewish laws prevented women from taking action to defend themselves from their unfaithful husbands. Authorities made it difficult for women to obtain a divorce because women were stereotypically characterized as rebellious . Around 1000 CE, Rabbi Gershrom issued two different prohibitions. One was against polygamy, and the other prevented women from being divorced against their will. A woman could initiate divorce and even force it on her husband, if she found the situation fit. The rulings also stated that women wouldn’t lose their economic rights, and were allowed to take back property they had brought to the marriage. As women became more powerful economically, their status transformed . They took advantage of the new power, and the rate of the divorce was estimated by Grossman to have been as high as 20%. They began to see more power with their spouses, family, and society as a whole. The legal sphere transformed the social status of women, as Grossman argues â€Å"for the first time in Jewish history, the women had .the upper-hand† concerning divorce and marriage. It can be assumed that the legal rulings gave women security, peace of mind, and a new place in social status. One ofShow MoreRelatedHalakha, Jewish Religious Law And Religious Commandments1913 Words   |  8 PagesHalakha by definition is Jewish religious law, encompassing both civil and religious commandments and prohibitions. The word Halakha stems from the root meaning to walk. Halakha involves the study of law and customs in the Jewish religion. According to rabbinic law it must be performed to sanctify all life and attain redemption. This idea when introduced was a collection of rabbinic commentaries from the Hebrew bible or the torah. The importance of halakha among Judaism and its trends, ideas, theoriesRead MoreAnalysis Of Elisheva Baumgarten s Mothers And Children Essay1866 Words   |  8 Pagessources from medieval Ashkenazi Jews as well as in its meticulous analysis of the often ambiguous writings. In Mothers and Children, Baumgarten examines a plethora of primary sources to explore the inner dynamics of Jewish families; she then uses this information to draw objective conclus ions about the relationship between the Jewish and Christian communities in the middle ages. In Baumgarten’s own words, â€Å"The premise of this study is that it is impossible to comprehend the history of medieval Jews without