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• 1. PRESS PROBLEMSPress problems from examination point of view can be classified into two:1. Problems regarding suitability of the press from force and energy point of view2. Problems regarding shout height of the press.Kinematically a mechanical press can be viewed as follows:OA= crank length = r. Therefore Stroke Length is given as S = 2 x OA = 2 r L = length of connecting rod.At some angle , the sliding point on the under face of the ram will be at a distance of H from BDC.Since length of the connecting link AB is large compared to length of crank OA, the inclination of ABwith respect to the vertical central line can be neglected. i.e. = 0.Thus any force reaction R acting at a point B can be taken as acting at point A as shown.Therefore the torque at point O =T = R x OA sin = F x OA sin = F x r sin 1
• 2. [F = force exerted by the press which is equal to the resistance experienced by the ram =R]A press is designed to exert defined amount of force at given crank angle . It means main crankshaft isdesigned keeping in view some value of torque. In other words, torque provided by a press is fixed. T = F * r sin T F r sin 1 F sinWhere ‘r’ is fixed for a particular operation and dependent on the stroke length.Therefore force available from press varies with crank angle. F is minimum when 90 andtheoretically infinite when 0 or 180 i.e. at dead ends. 2
• 3. Though press is capable of exerting infinite force at dead ends, press members such as cranks andconnecting links cannot be designed for infinite load. They have to be designed for finite force. This safefinite force is to be determined by the dimensions, strength and allowable deflections of press componentssuch as press frame, connecting link, crank etc. This safe value is known as press capacity and specifiedas tonnage at crank angle. Let us say press of 63T at 20  .Daylight: The maximum clear distance between the two pressing surfaces of a press when surfaces are inthe usable open position. When a bolster plate is supplied it is considered as the pressing surface.The velocity of ram can be computed using following formula: N S Sin SV 0.105 N h ( 1) 60 hwhere N = Number of strokes per minute (SPM), h = slide location above ( or before) BDC S = stroke, = crank angleShut height: For a press, the distance from the top of the bed to the bottom of the slide with the strokedown and adjustment up. In general, it is the maximum die height that can be accommodated for normaloperation, taking the bolster plate into consideration. Shut height is always measured at BDC. There aretwo shut heights maximum and minimum 3
• 4. Position (1) = position corresponding to maximum stroke where Smax= 2 x r1Position(2) = position corresponding to minimum stroke where Smin = 2 * r2 r1 r2 = (Smax Smin)/2 maximum stroke minimum strokeThus, MaxDLH = MinDLH + Press Adjustment + 2If maximum DLH for a given stroke length, which lies between minimum and maximum stroke length, isto be found following formulation can be used. max stroke min stroke Max DLH stroke = Max DLH 2 DLH STROKE STROKE LENGTH PRESS POSITION ADJUSTMENT MAXIMUM TDC MAXIMUM UP MINIMUM TDC MINIMUM DOWNENERGY OVERLOADING: The energy needed for cutting and forming operations during each strokeis supplied by flywheel which slows down to a permissible percentage, usually 10 20% of its idle speed.For example, if 10% slow down is permissible N1 (N0 N1)/N0 = 0.1, which gives 0 .9 . N0Total energy of the flywheel I 2 I * 2 N2 E FT = (constant) * N2 = K N2 2 2 2 60Where N0 = idle flywheel rpm N1 = flywheel rpm after stroke N = flywheel rpm I = M.I. of the flywheel = angular velocity 4
• 5. The total energy supplied (ES) by flywheel includes energy required to overcome friction and elasticdeflections. The electric motor must restore its ideal speed before stroke. The time available betweentwo strokes depends on type of operation (namely) whether it is continuous or intermittent. In continuoustype of operation, permissible slowdown is 10% and in discontinuous 20%. I * 2 N1 2The energy after stroke = 2 60 2 2 2 ES = EFT(N0) EFT(N1) = K N 0 N1 2 2 2 ES K( N 0 N1 ) N1 1 E FT 2 2 K( N 0 ) N0 N1With a slowdown of 20% = 0.8 N0 ES 1 0.8 2 0.36 E FTThis energy supplied must be more than the required energy.Shut heights for the press Shut Height Stroke Position Stroke Press Adjustment Maximum BDC Maximum UP Minimum BDC Minimum DOWN 5
• 6. Problem 1: A press has a maximum DLH (day light height) of 500 mm and a ram adjustment of 60mm. Calculate maximum and minimum shut height value for a press tooling if the bolster platprovided on the press bed has a thickness of 80 mm. The press has a variable stoke with maximumstroke and minimum stroke of150 mm and 10 mm respectively.SolutionMinimum Shut Height = Maximum DLH Maximum Stroke Press Adjustment(In case of max DLH, press adjustment is UP but in case of minimum shut height, if is down).Minimum Shut Height = 500 150 60 = 290 mmMaximum Shut Height = minimum shut height (SH) + press adjustment + Max stroke Min stroke 2 (150 10 ) = 290 + 60 + = 420 mm 2To calculate shut height for a tooling, thickness of bolster plates must be deducted from press shutheights.Max SH for die = 420 80 = 340 mmMin SH for die = 290 80 = 210 mm 6
• 7. Problem 2: A press has minimum DLH 325 mm and Adjustment of ram is 75 mm. Stroke isvariable and can be varied from 12.5 mm to 100 mm. If the bolster plate provided has a thicknessof 80 mm, calculate maximum and minimum shut height for a die.SolutionMaximum DLH = minimum DLH + (r1 r2) + P.A. (press adjustment) 100 12 .5 = 325 75 = 443.75 mm 2Minimum shut height = Max DLH Max Stroke press adjustment = 443.75 100 75 = 268.75 mm max stroke min strokeMaximum shut height = minimum shut height + PA + 2 = 268.75 + 75 + 43.75 = 387.5 mmMax shut height for tooling = 387.5 bolster plate thickness (= 80 mm) = 387.5 80 = 307.5 mmMin shut height for tooling = 268.75 80 = 188.75 mmProblem 3: A press has maximum DLH 500 mm and a ram adjustment of 60 mm. Stroke can be variedfrom 10 mm to 150 mm. Calculate minimum and maximum shut height for a press tool for stroke of 100mm and bolster plate thickness of 80 mm.Solution: First it is necessary to calculate max DLH for stroke length equal to 100 mm as maximum DLHgiven (500 mm) corresponds to maximum stroke of 150 mm stroke. max stroke(r1 ) given stroke(r ) (150 100 )= MaxDLH = 500 = 475 mm 2 2Min shut height of press = max DLH stroke PA (since PA is up in case of max DLH but down incase of min SH) = 475 100 60 = 315 mmMax SH of press = Min SH + PA = 315 + 60 = 375 mm 7
• 8. ORMax SH of press = Max DLH100 Stroke = 475 100 =375 mmMin SH of press = Max SH PA = 375 60= 315 mmMax Shut height for die = Max SH for press bolster plate thickness = 375 80 = 295 mmMin shut height for die = Min SH for press bolsters plate thickness = 315 80 = 235 mmProblem 4: A press has minimum of DLH 325 mm and press adjustment of slide of 75 mm. Strokecan be varied from 12.5 mm to 100 mm. Bolster plate thickness is 75 mm. Calculate permittedshut height for die if stroke is fixed at 50 mm.Solution Given stroke min strokeMinimum DLH for 50 mm stroke = min DLH (for 12.5 mm stroke) + 2 (50 12 .5) = 325 = 343.75 mm 2 8
• 9. Minimum shut height of the press = min DLH (for 50 mm stroke) stroke = 343.75 50 =293.75 mmMaximum SH of the press = min SH + PA = 293.75 + 75 = 368.75 mmMaximum SH for the die = max SH for press bolster plate thickness = 368.75 75= 293.75 mmMinimum SH for the die = min SH for press bolster plate thickness = 293.75 75 = 218.75 mmProblem 5: A press is designed to offer 90 ton of force at 20  crank angle with a stroke of 15 cm.Stroke is variable from 1 cm to 15 cm. Calculate tonnage available when ram is 3 cm above itsBDC. Take stroke length equal to 10 cm.Solution Given h = 3 cm, s = 10 cm. h 1 cos Where = crank angle from BDC s 2 3 1 Cos 10 2=> 66 .42 As we know that sT = F x r x sin = F* * sin = constant 2 s s=> T F * 1 * sin 1 = F2 * 2 * sin 2 1 2 2F1 = force available 90T at crank angle 1 20 and stroke s1 = 15 cm.F2 = force available at 3 cm above BDC and stroke s2 = 10 cm, 66.42 15 1090 * * sin(20  ) F2 * * sin(66 .42  ) 2 2=>F2 = 50.4T i.e. tonnage available when ram is 3 cm above the BDC is 50.4TProblem 6: Check regarding the suitability of the press with following press data for blanking anddrawing operation with combination die. Capacity of the press tool is 150 T at 20 from BDCStroke length=120mmTotal fly wheel energy = 180T cmPermitted slowdown of speed = 20%Blanking load = 30TDrawing force = 10TBlank holding force = 3TDepth of cup = 35 mm = 3.5 cmStock thickness = 1 mm = 0.1 cmPercent Penetration = 30% 9
• 10. Solution:I. Suitability of press from tonnage point of view:Total load on the press = blanking load + drawing load + blank holding force = 40 + 10 + 3 = 53 TSince depth of drawn cup is 3.5 cm, drawing will commence at height of 3.5 cm from BDC assumingbottom edge of the cup is at BDC. Further, force increases towards the BDC. Now, we will determinetonnage available at a height of 3.5 cm.h = 3.5 cm, assuming stroke = 12 cm.h 1 cosS 23.5 1 cos gives = 65.3712 2Since torque transmitted by crank shaft is constant, S1 S2T F1 sin 1 F2 sin 2 2 2As stroke length is unchanged, S1 S2F2 sin 2 F1 sin 1F2 sin (65.4 ) 150 sin (20 ) F2 56 .44 TConclusion: As available tonnage F2 is greater than the required tonnage (53T), the given press issuitable to carry out the given operations from tonnage point of view.II. Suitability of the press from the energy point of viewTotal energy required = Energy for blanking + Energy for drawing + Energy for blank holdingEnergy required for blanking = Force x distance = F x K x t = 40 x 0.3 x 0.1 = 1.2T.cmEnergy required for drawing = Force x distance (= cup depth) = 10 x 3.5 = 35 T. cmEnergy for blank holding = Blank holding force x distance (= cup depth) = 3 x 3.5 = 10.5 T.cmTotal energy required = 1.2 + 35 + 10.5 = 46.7 T. cm 10
• 11. Energy available from the flywheel:Permitted slowdown of speed = 0.2N0 N1 N1 0.2 gives 0.8 N0 N0 Energy supplied (ES ) 2 N1We know that, 1 1 (0.8) 2 Total energy of the flywheel(E FT ) 2 N0 ES = 0.36 E FT ES E FT * 0.36 = 64.8 T. cmConclusion: Since the energy supplied (64.8 T. cm) is more than energy required (46.7T.cm); the press issuitable from energy point of view. Thus, the press is suitable from both tonnage and energy point ofview.Problem 7: Tonnage rating of a given press is 160 T at 20 crank angle from BDC. Stroke isvariable from 10 mm to 120 mm in steps of 10 mm. The ram can be adjusted by 100 mm.Maximum DLH of the press is 500 mm. Total energy of the flywheel is 180T.cm.Permissibleslowdown in speed is 20%. If it is decided to use stroke of 100 mm, calculate1. The maximum and minimum shut height for a combination die.2. Tonnage available at 25 mm from BDC.3. Check the suitability of press for following operation: Blanking and drawing with the help of combination die. Blanking load is 45T. Drawing force is 10T. Blank holding force is 3T. Cup depth is 20 mm. Stock thickness is 1.5 mm. Percent penetration is 30%.Solution(i) Maximum DLH at 120 mm (max stroke) is equal to 500 mm. Therefore, it is necessary to find the max DLH at 100 mm stroke. (Max stroke given strok) Max DLH at 100 mm stroke = max DLH at 120 mm stroke 2 11
• 12. (120 100) = 500 = 490mm 2Maximum SH of the press = Max DLH at 100 m stroke - stroke = 490 100 = 390 mmMinimum SH of the press = Max SH press adjustment= 390 100 = 290 mmSince there is no mention of bolster plate, we may assume bolster plate thickness of 60mm.Maximum shut height of combination die is =390-60= 330mmMinimum shut height of combination die is =290-60= 230mm(ii) Suitability of the press from the tonnage point of viewDetermining force availability at height of 25 mm from BDC h= 25 mm = 2.5 cm, S = 100 mm = 10 cm h 1 cos 2 S 2 2 .5 1 cos 2 10 2 2 60 S1 S2T F1 sin 1 F2 sin 2 2 2F1 160 T, 1 20 , S1 12 cmF2 = ? 2 = 60 , S 2 = 10 cmSince max tonnage is available at max tonnage 12 10 160 sin (20 ) F2 sin (60 ) 2 2 F2 75 .8 T i.e. Force offered by the press at a height of 25 mm from BDC.Height of the cup is 20 m and total force required for blanking and drawing is 58 T (45 + 10 + 3). Asforce at height of 25 mm from BDC is greater than the required, force at 20 mm from BDC must be fargreater than required (58 T). Hence, the given press is suitable to carry out given operations from tonnagepoint of view. 12
• 13. Suitability of the press from the energy point of viewTotal energy required = blanking + drawing + blank holding = (force x K x t 0 ) + (drawing load x depth of cup)+ (Blank holding force * depthof cup) = (45 * 0.33 * 0.15) + (10 * 2) + (3 * 2) = 28.23 T. cmEnergy supplied by the flywheel when permitted slowdown is 20%…. N N1 1 0.2 that gives 0.8 N0 N0 Energy supplied (E S ) 2 N1 1 1 (0.8) 2 Total energy of the flywheel(E FT ) 2 N0 ES 0.36 E FT ES EFT 0.36 = 180 x 0.36 = 64.8 T.cmEnergy supplied by the flywheel (64.8 T.cm) is substantially more than required for the operation (28.23T.cm). The press is suitable from the energy point of view.In sum, the given press is suitable to carry out given operations from tonnage as well as energypoint of view.Problem 8: The punch force required for a cup drawing operation is 10 T. Depth of the cup is 35mm. An air operated blank holder provides a force of 3 T to prevent wrinkling. A 45 T capacitypress having total flywheel energy = 33 T. cm is available for use. Another press of capacity 150 Tand flywheel energy 180 T. cm is also available. Permissible slowdown of speed is 20%. Calculatetotal energy required and select a press.SolutionTotal tonnage required to draw the cup =10+3=13 TonEnergy required for cup drawing = Force x depth of cup = (10 + 3) x 3.5 = 45.5 T.cmFirst press is unsuitable to carry out the operation from energy point of view. 13
• 14. N1 0.8 From given dataN0 Energy supplied (ES ) 2 N1 1 1 (0.8) 2 =0.36Total energy of the flywheel(E FT ) 2 N0Energy Available from the second Press ES EFT 0.36 = 180 x 0.36 = 64.8 T. cmThe available energy from the second press is more than the energy required for the operation andtherefore this press is suitable for the operation from energy point of view and can be selected.Problem 9: A press is designed for giving 120 t at 30 crank from BDC, when stroke is 20 cm. Prepare amonograph from BDC. From the monograph explain:1. Overloading of torque without overloading capacity.2. Overloading of capacity without overloading of torque.Solution: We know that Sh= r r cos = (1 cos ) 2 h (1 cos ) s 2Torque provided by a press remains constant 14
• 15. S 2T T F sin which gives F ………….(1) 2 S sin In given press T is equal to 20 T 120 sin (30 ) = 600 T. cm 2 2 600 60 F = = …………………………………………(2) 20 sin sinFor equation (1) and (2), common variable is 0 30 45 60 75 90 120 150 180 F 120 85 69 62 60 69 120 h/s 0 0.07 0.15 0.25 0.37 0.5 0.75 0.93 1.0A graph can be drawn to show up the relationship.1. Overloading of torque without overloading capacityPress will get overloaded if load requirement is more than 120 T. Any point above ZZ means overloadingof press and press members will fail. One may select any point on any line like A1 where press will beunder loaded. Any point on curve like A2 will have torque equal to that available on the press i.e. 3600Tcm. Select any point A at = 60 and load equal to F = 80T which is less than rated capacity of120 T meaning that it is under loaded. S 20 T F sin 80 sin (60 ) 2 2 = 692.8 T. cm which is greater than available torque (600 T.cm)Thus, at the point A3 there is overloading of torque without overloading the capacity.Any point between A 1 to A 2 will not cause overloading of torque and obliviously not causingoverloading.Take F = 80T, and = 45 , S 20 T = F sin 80 sin (45 ) = 565.68 T.cm 2 2 15
• 16. 2. Overloading of capacity without overloading torqueTake point B where F = 140 T and = 15Since the load is more than 120 T, the press is overloaded S 20T F sin 140 sin (15 ) = 362.34 T. cm (less than 600T cm.) 2 2This is the case of overloading press capacity (F = 140 T beyond 120 T) without overloading the torque. 16