(find Article about training and summarize) Essay

(find Article about training and summarize) - Essay Example Regardless of the sport-required metabolic demands, many coaches have continually stuck to the strategy. Most sporting activities do not require continuous sub-maximal movements. However, these sports require periods of high-intensity effort comprising of maximal bursts of speed and durations of minimal activity. Aerobic interval training concentrates on the provision of more sport-detailed training intensity. It ensures that an athlete enjoys greater training stimuli that can be utilized in the muscles, and applied in specific sporting activities. Aerobic interval training can also apply advanced methods for gaining endurance, particularly for some athletes seeking to achieve continuity in aerobic fitness levels (Ballantyne, 2006). In such cases, intensity of training is higher those of common aerobic training sessions. In fact, during the first attempts, exercise stimulus may be damaging to the muscles, and can lead to muscle soreness as the training sessions continues in later days. It is, therefore, essential that athletes should be advised to develop endurance and muscle strength prior to the inception of full blown training sessions. To be able to achieve that, it is highly suggested that athletes participate in special, short term programs of aerobic exercises before the start of intense exercises. During the first two weeks of intense training, program should be conservative in nature (Ballantyne, 2006). This implies that, the training volume should be between 3 and 5 intervals so as to accustom the athletes’ joints and muscles to the intensity. Additionally, a specific and thorough warm up warm-up should be performed on top of pre and post-training flexibility routines. It is paramount to provide complete and adequate rest intervals between the training sessions. Such periods are necessary so as to enable sufficient recovery, and enhance high power output during the next training

How Apple INC supply issues effect the demand of its products such as Essay

How Apple INC supply issues effect the demand of its products such as Iphone, IPAD - Essay Example There are several types of supply theories abased on the nature of products. Generally we use the supply chain system based on following important components as given below. Factory: The factory is the basic component and the point of origination for any supply chain. If a factory has issues in the production, the whole supply chain will suffer due to lack of the goods or the quality of the supplied products depending upon the problem respectively. Distributor: The distributor has a supporting role for a company in a supply chain. He advises the company about the magnitude of the product to be manufactured. If a distributor is having issues then the supply chain will suffer due to the mismanagement in terms of faulty distribution. Stockist: A stockist acts as a filter between the distributor and the whole sellers as well as show rooms. The basic difference between a distributor and a stockist is that the distributor purchases the items as a bulk directly from the company while the st ockist purchases the items at a relatively smaller scale from the distributor depending upon the local requirements of the market. Whole sale: A whole seller acts as another filter between the retailers and bulk dealers. ... Retailers: They are the source of products for the end users by the help of company support & distributor’s good supply chain. They also play an effective role in satisfying the end user by giving product knowledge and guarantee and warranty as well. End users: End users are the main component of a business. Each and every company designs its marketing and production strategies to attract end users in the form of customers. They do it by facilitating the end users in different ways. CONSEQUENCES OF BREAKING OF SUPPLY CHAIN Companies invest heavily on generating customer demand about their product through advertising. Especially in smart phones, notepad, laptop industry customers have variable choices therefore, once the customer demand have been established the manufacturer has to ensure the continuous supply of the product to maintain the customer loyalty. Companies which fails to do so may loose their customer loyalty. Following are some key issues can be raised if supply li ne break at the time of demand: NEED OF A NEW MARKETING CAMPAIGN: Once the demand developed in the market by the advertising campaigns the supplier is supposed to fulfill the customer demand. Otherwise customer may turned to other vendors. In this competitive era each product has several rivals in the market. Therefore, once the customer would have turned aside towards other product line or manufactures then in such situations the company has to re design the old marketing strategies, because the customer mind may not have any effect from old marketing approaches. This new advertisements and marketing may cause a huge expense over the company sales. Therefore, companies are very careful in maintaining the supply line un-broken. FRANCHISE/DISTRIBUTOR, FIELD FORCE WILL LOOSE TRUST ON

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.

Buddhism Essay -- essays research papers fc

Buddhism   Ã‚  Ã‚  Ã‚  Ã‚  For over 2000 years Buddhism has existed as an organized religion. By religion we mean that it has a concept of the profane, the sacred, and approaches to the sacred. It has been established in India, China, Japan and other eastern cultures for almost 2000 years and has gained a strong foothold in North America and Europe in the past few centuries. However, one might ask; what fate would Buddhism face had Siddartha Guatama been born in modern times; or more specifically in modern day North America? Would his new found enlightenment be accepted now as it was thousands of years ago? Would it be shunned by society as another â€Å"cult† movement? What conflicts or similarities would it find with modern science; physics in particular? The answers to these questions are the aim of this paper, as well as a deeper understanding of modern Buddhism.   Ã‚  Ã‚  Ã‚  Ã‚  Although I will stick with traditional ideas raised by Buddhism, one detail in the story of Siddartha Guatama must be addressed in order for it to be relevant to the main question being asked: What obstacles would Siddartha Guatama face had he been born in modern day North America. Primarily, it must be recognized that rather than being born into the Hindu religion (which in itself is mystical), Siddartha would have most likely been born into a Christian family. This in itself presents the first obstacle, that being that Christianity is a strictly monotheistic and non-mystical faith. Hence from the outset, although in the traditional story Siddartha faced a conflict with his father (Ludwig 137), in the North American scenario the conflict would have been heightened by the fact that his search for enlightenment was not even closely similar to the Christian faith.   Ã‚  Ã‚  Ã‚  Ã‚  As with science, changes in religious thought are often met with strong opposition. It is interesting to note though, that many parallels can be found between modern physics and Eastern Mysticism. As Fritjof Capra writes: The changes, brought about by modern physics . . . all seem to lead towards a view of the world which is very similar to the views held in Eastern Mysticism. The concepts of modern physics often show surprising parallels to the ideas expressed in the religious philosophies of the Far East. (17-18) Thus by examining some of the obstacles imposed by t... ...o overcome the problems of being born into a Christian family/society; a society not used to such abstract ideas of reality, the close- minded nature of western thought, and the problems posed by a media that likes to jump on anything new and unusual and tear it to shreds. However, if it were to overcome these obstacles it is quite probable that it would become a deeply rooted religion in North America due to the likely support it would gain from the scientific community. Bibliography Capra, Fritjof. The Tao Of Physics: An Exploration of the Parallels Between Modern Physics and Eastern Mysticism. Berkley: Shamhala Publications, 1975 Ludwig, Theodore M. The Sacred Paths: Understanding the Religions of the World.   Ã‚  Ã‚  Ã‚  Ã‚  New Jersey: Prentice Hall, 1996 Niwano, Nikky. Buddhism For Today: A Modern Interpretation of the Threefold Lotus   Ã‚  Ã‚  Ã‚  Ã‚  Sutra. New York: WeatherHill, 1980 Richardson, Allen E. East Comes West: Asian Religions and Cultures in North America.   Ã‚  Ã‚  Ã‚  Ã‚  New York: The Pilgrim Press, 1985 Shupe Anson D. Six Perspectives On New Religions: A Case Study Approach.   Ã‚  Ã‚  Ã‚  Ã‚  New York::

Satire: Jean-Baptiste Poquelin (Moliere) and Jonathan Swift

Both Jean-Baptiste Poquelin (Moliere) and Jonathan Swift use satire as a means of conveying their ideas concerning the actions of the characters in their respective works Tartuffe and Gulliver’s Travels. The object of Moliere’s satire is the false religiosity suffused the climate of his time. He parodies the lives of persons who profess Christianity and yet in certain situations behave in a manner non-concurrent with the message they preach. Swift too condemns a sort of hypocrisy in his tale, as the professed rank and honor of the leaders of his time come under attack in his portrayal of them. Swift in particular uses a variety of different metaphors in order to change the scale of humanity and in so doing magnify the problem he seeks to point out. Both novels, therefore, demonstrate the role of satire as â€Å"mediator† between how life actually is and what is ought to be in the eyes of their authors (Bullit, 3). Moliere uses characters to typify the types of persons he wishes to satirize. The title character of his work, Tartuffe himself, represents the type of person in life who professes religion and yet in his action demonstrates himself to be in complete discord with the tenets of that religion. Tartuffe performs actions that amount to fraud and yet acts in the name of the clergy and of Christianity. This man can be seen to stand in the place of the clergy of the Catholic faith (the dominant religion of France at the time) who collected funds (such as indulgences) or other otherwise ingratiated themselves to the masses under false pretences. The person upon whom the fraud is committed represents the masses who willingly give their all to these leaders of the church, whom they believe to be virtuous. However, Moliere indicates that the money being appropriated by the church is being used for personal and non-religious reasons. The situation’s remedy comes in the form of a king who finds out the truth and punishes Tartuffe for his guilt. Moliere’s criticism of the clergy is complete in this description, as he indicates that God (ruler of the earth) is in no way supportive of the actions of these religious persons who claim to be doing His will. Moliere also satirizes the determination of some persons (especially the religious masses) to embrace ignorance and the misfortune that they fall into because of this behavior. The character Orgon is eager to believe not only in the virtue of Tartuffe but also in the particulars of his claims. As a result, he is swindled out of his property and can only be rescued by the royal (divine) intervention of the King. The corrective proposition given by Moliere is that the clergy should seek to truly represent the knowledge and wishes of God by acting in accordance with his teaching. They should also seek to educate the masses, and by promoting education and transparency all round, virtue will increase. Swift in Gulliver’s Travels takes his readers to several different places, and the effect of this is to remove what he consideres the self-imposed grandeur. This grandeur is imposed through the building up of socio-political and religious institutions based upon laws that profess to defend (among other things) a hierarchical view of humanity. In Lilliput and Brobdingnag, for example, the natives give air to Swift’s true ideas concerning these institutions and the form of humanity that obtains within them. The Lilliputians demonstrate the pride and high-mindedness of humans, underscoring how petty this form of behavior is. Such honors as the favor of the Court is demonstrated in the ministers of Lilliput challenge of jumping over a rope and the rewards they are granted. The various heights to which the rope is lifted represent the different titles to which nobles and clergymen might aspire. The Lilliputians who represent such people are small, and their size reflects Swift’s satirical representation of the true size of humans in relation to their opinions of themselves. Likewise, in Brobdingnag, the larger scale of the persons represents the magnification of humans’ foibles and vices in a grotesque manner, as they vainly attempt to decorate themselves with a distinction of rank that does not truly exist. Gulliver’s conversion throughout the tale from a person of naivete to one who is truly skeptical of human behavior represents method in which Swift indicates that humans should correct themselves. In becoming aware of humanity’s own tendency toward pride and pettiness, people will become more likely to recognize and denounce it within themselves and others.

Mother Knows Best

Mother Knows Best There is a quote that says, â€Å"Mothers are angels who teach their children to fly. †Ã‚   Indeed mothers are angels, even though we often think that mothers were meant to torture us especially in our adolescent years when we would rather go out with friends instead of washing the dishes or doing our homework, our mothers nevertheless become our pillar, not just of faith but also of strength. Mothers may often be misunderstood but it does not mean that they don’t know best.My earliest memory of my mother is that of a radiant face. For me she was the most beautiful woman alive and nothing compares with her. Even if she was unlike the celebrity mothers who wear designer clothes or gets to eat in fancy restaurants all the time, my mother is, in other words, real. My mother seemed to be passive and quiet however she was a formidable foe if you cross her. I loved the way she combed her hair getting ready for bed, letting those deep brown locks loose, shiny against the glare of the overhead light.I loved the way she would  caringly wipe my back for perspiration after playing under   the hot sun all afternoon not minding the fact that I smelled like a sweaty sock worn for two weeks. On the incidents that my sibling and I cross my mother, those big green eyes of her would squint in disapproval, mouth pursed. But I came to love that about her. Even if I was born seven years after my brother, this never prompted her to play favorites. But rather, she made it a point to be fair and equal among her children.Whenever I commit a mistake, Mom would not hesitate to point it out to me, punish me if needed. Sometimes, I detested it but over time I came to realize that Mom did that because she only has the best interest at heart. However, she was not one to hold grudges for long. Like most mothers, she is very protective of me that often times I resented her for being so. I did not understand why she has to  constantly ask about my whereabou ts: who I am going with and where I was going. It never occurred to me that this was an essential part of being a mother.A mother cares and worries for her children. Mother taught me that getting reprimanded does not mean that she loves me less. On the contrary, she showed me that when she scolds me, it means that she cares for me and worries about me. Mother has certainly her own flaws but I have come to accept, in fact love every bit of them. When mom starts to call or page  me non-stop, it is either she misses me or she needs something or  I have irked her again. Some may call it nagging, I call it loving. For a person to wear her heart on her sleeves is a vulnerable thing, but mother is not vulnerable at all.In fact she is admirably strong. She never backs off in a fight especially if she knows she is right but she also knows how to compromise too. But compromising for her does not mean that you are wrong. Comprising is just another way of getting things done, later. When I started school, mom was my friend. To me, school was a world where strangers and bad guys lurk. Instead of laughing at my fears, mom showed me instead how great school can be. On my first day of school, she prepared my favorite snacks drove me school.Instead of leaving me at the gates, she got off the golf cart, knelt down and  tenderly cupped my face in both hands and told me, â€Å"Don’t worry honey, there isn’t any bad guys today, Mommy made them go away. †Ã‚   Although these words are pretty naive, but to a child on his first day of school, these words were more than enough to let him conquer his fears. I went to school that day knowing that Mom was there for me and I learned to enjoy school. I made new friends and introduced them to Mom who was more than happy to serve us hot cookies and milk for snacks.Although my mother was not able to achieve her goal of becoming getting a college degree, this never made her bitter but rather she motivated us to pursue our own education and goals in life. She taught us early on that education is very important that is why she never fails to motivate us in school. I know my mother is not perfect. She is only human thus has her own flaws too. But for me, these flaws are tiny imperfections that make her all the more lovable and perfect. These flaws are just evidences of her humanity. Mother is an angel who teaches us to fly and dream.