Shafts. Power transmission shafting Continuous mechanical power is usually transmitted along and etween rotating shafts. The transfer etween shafts is accomplished y gears, elts, chains or other similar means for matching the torque/speed characteristics of the interconnected shafts, e.g. a car needs gears etween the engine crankshaft and drive wheel half-shafts. Shafts rotating only at constant speed N (rev/s) are considered here. Power = force ( N) linear velocity ( m/s) in translational applications and Power = torque ( Nm) angular velocity ( = π N rad/s) in rotational applications, then it follows that torque is a major load component in power transmitting rotating shafts.. Torque transmission Torque may e transferred to or from the end of one shaft y a second coaxial shaft - this is a pure torque, a twist aout the shaft axis. The transfer is carried out y a shaft coupling, see fig.(.). Torque may e transferred also at any point along a shaft y a gear, elt pulley, or chain sprocket for example, mounted on the shaft. These common elements apply forces offset from the shaft axis, and therefore the torque (T) is accompanied y a radial load which results in ending A spur gear and a elt pulley are Fig.(.) sketched in fig.(.), each sujected to loading tangential to its effective or pitch diameter D. The load on the spur gear arises from inter-tooth contact with its mating gear and comprises two components, the useful tangential component F t and the unwanted ut unavoidale radial component F r (commonly 0.6 F t ). Gear forms other than spur give rise also to a load component parallel to the shaft axis - ut for all gears, shifting the offset force as aove, T = F t D/. See fig. (.) A elt, eing flexile, cannot withstand compression - the pulley is therefore sujected to two strand tensions F max and F min oth of which must exceed zero. The net torque T = ( F max - F min ) D/ is clockwise here. A chain sprocket is similar though the minimum tension may drop to zero due to the positive drive not relying on friction. Dr. Salah Gasim Ahmed YIC
Fig. (.). Design of shafts under various types of loading The following equations can e used to otain the size of the shaft under various types of loadings Shafts under torsion only: 5.K T t D B (.) Ss 000K P S N s Shafts under pure ending 0.K M S Shafts under ending and torsion: D When using power When using power t D B (.) m D B (.) 5. B ( KmM ) ( KtT ) (.a) pt 5. 6000K P D ( p N t Short shafts under transverse shear only t B ( K M ) (.) m.7v D (.) S s When using metric system then equation (.) ecomes.7k P t D B (.5) S N s And equation (.) ecomes ) Dr. Salah Gasim Ahmed YIC
5. 9.55K P D B ( ) (.6) N t ( KmM ) pt Where, D = external diameter of shaft in inch D = internal diameter of shaft in inch K = D (for hollow shafts) D B ( K ) K m = comined factor of shock and fatigue under ending K t = comined factor of shock and fatigue under torsion M = maximum ending moment (in l) T = maximum torque (in l) N = rotational speed (rpm) P = Power (hp) p t = maximum allowale shear stress under comined load of ending and torsion (psi) S =maximum allowale ending stress (psi) S s =maximum allowale shear stress (psi) V =maximum allowale transverse shear stress (psi) Tale (.) Comined Shock and Fatigue Factors for Various Types of Load Type of load Constant loads without shocks Stationary shafts K m K t.0.0 Rotating shafts K m K t.5.0 Sudden loads with light shocks.5 --.0.5 --.0.5 --.0 --.5 Sudden loads with heavy shocks.0 --.0.5 --.0 Tale (.) Recommended Maximum Allowale Stresses for Shafts Under Various Types of Loads Material Type of load Bending only Torsion only Torsion + ending Commercial steel Without key way S=6000 psi (0 N/mm ) S s = 000 psi (55 N/mm ) p t = 000 psi (55 N/mm ) Commercial steel With key way S = 000 psi ( N/mm ) S s = 6000 psi ( N/mm ) p t = 6000 psi ( N/mm ) Steel with specific properties Note(a) Note() Note() (a) S= 60% of elastic limit in tension and not more than 6% of ultimate tensile strength Dr. Salah Gasim Ahmed YIC
() S s and p t = 0% of elastic limit in tension and not more than % of ultimate tensile strength Tale (.) Values of the Factor B Corresponding to Various Values of K for Hollow Shafts K D D K B ( ) 0.50.0 0.55.0 0.60.05 0.65.07 0.70.0 0.75. 0.0.9 0.5. 0.95.75 0.90. Example A shaft carrying two pulleys is shown in fig.(.).the shafts transmits a torque of 00 in.l. and carries sudden loads at A and B equivalent to 50 l and 50 l. respectively, with minor shocks. The two pulleys are fixed to the shaft through rectangular keys. The hu length of pulley A is inch and that of pulley B is inch. If total length of the shaft is 0 inch and it is made of commercial steel find the proper diameter of the shaft in all sections. Pulley A Fig.(.) Pulley B Dr. Salah Gasim Ahmed YIC
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General guidelines:. Make the shaft as short as possile while locating the earings as close as possile to the loads. This reduces ending moments and deformation and increases critical speed.. Eliminate stress raisers near highly stresses areas if possile otherwise use large filets and improve surface finish. Also cold rolling and shot peening can e used to improve the mechanical properties of the material at these areas.. If deformation is the main design factor using expensive steel does not solve the prolem as all steels have almost the same modulus of elasticity. If the weight of the shaft is an important factor the using hollow shafts may give satisfactory solution, eg shaft propeller in cars Dr. Salah Gasim Ahmed YIC 7
Exercises:. Figure (.6) shows the loads acting on a shaft. The shaft caries two pulleys a and. The loads on shaft are 00 l at pulley A and 00 l at pulley B and a maximum torque of 50 in l. The shaft is made of steel SAE 00 with Elastic limit of 000 psi and shear strength of 0000 psi. If the shaft drives a compressor and the.5 thick pulley are fixed to the shaft through keys, find the suitale diameters at all sections of the shaft. Take factor of safety =.5. A shaft carries a single gear and transmits a power of.5 kw at a speed of 500 rpm. The gear exerts a tangential load of 500 N and a radial load of 0 N on the shaft. The total length of the shaft is 50 mm and it is carried y two earings at its ends while the gear is fixed at its centre with a square key. The shaft is connected to a centrifugal pump at one of its ends. Determine suitale diameters for the shaft at its sections. Find the diameters of a transmission shaft connected to a six-cylinder 00 hp oil engine through a elt drive, see fig,(.7). The shaft is driving a woodworking and metalworking machinery and runs at 5 rpm. The torque transmitted through the shaft is divided equally etween the working machinery. Take a safety factor of.5 6 in A 00 l 00 l 0 in Fig. (.6) B in Metal working machine pulley 000N Wood working machine pulley 00N 00 N Engine pulley 00 mm 00 mm 00 mm 00 mm Fig. (.7) Dr. Salah Gasim Ahmed YIC
5. A geared industrial roll shown in fig.(.) is driven at 00 rpm y a force F acting on a in diameter pitch circle as shown. The roll exerts a normal force of 0 l/in of roll length on the material eing pulled through. The material passes under the roll. The coefficient of friction is 0.. Develop the moment and shear diagrams for the shaft modelling the roll force as: a) a concentrated force at the centre of the roll and, ) a uniformly distriuted force along the roll. Select material and determine dimensions of the shaft if the roll is fitted to the shaft through a square key F in F 0 0 in in in in Gear in PCD Fig. (.) Dr. Salah Gasim Ahmed YIC 9
Keys and pins 5. Types of keys The main function of a key is to transmit torque etween a shaft and a machine part assemled to it. In most cases keys prevent relative motion, oth rotary and axial. In some construction they allow axial motion etween the shaft and the hu; such keys are called feather keys or spline keys. Keys can e classified according to their shape into straight, and tapered, rectangular, square, round, and dovetail. h / Square Rectangular shallow Tapered Two width Dovetailed Woodruff Flat Saddle Fig. (5.) Shaft keys for light and medium duty Keys are also classified according to their intended duty as:. Light duty keys, square: rectangular key, shallow key. See fig. (5.). Medium duty: taper key.. Heavy duty keys: round tapered key, Barth key. See fig. (5.) Taper /6 in per ft = d/ 90 0 Taper / in per ft h= / = d/5 to d/ = d/6 Fig. (5.) Heavy duty keys Dr. Salah Gasim Ahmed YIC 0
5. Design of square keys: When torque is transmitted through keys, they are sujected to shear and compressive crushing stresses. See fig. (5.) Shearing Crushing Crushing strength: Since a hu is always much more rigid than a shaft, the shaft will e twisted y the torque whereas the hu will remain practically undistorted. As a result the pressure along the key will vary and it will e minimum at the free end of the shaft and maximum on the other side. The maximum pressure can e denoted y P while the minimum pressure P and the pressure at any point along the key y P. So at a distance L o the pressure equals to zero (see fig (5.). The pressure can e expressed y the equation: P P L tan (5.) Where ( P P ) P tan (5.) L L 0 Fig.(5.) h D P P L P L L 0 =.5 D Fig. (5.) Dr. Salah Gasim Ahmed YIC
Torque transmitted dt PxdLx D (5.) Sustituting the value of P from equation (5.) into equation (5.) and integrating etween the limits L = 0 to L = L yields: T P DL DL tan (5.) The pressure /unit length equals the crushing stress x the area of unit length of the key side, (S x0.5hx) then, P 0.5Sh Experiments showed that length of key greater than.5d is not effective. So we can consider that the pressure at L=.5D equals zero and hence, P Sh tan (5.5) L0.5D And the torque transmitted can e expressed y, hl T S hdl S (5.6) The length of the key can e determined from equation (5.6). If the outcome is negative value then one key is not enough, ut if L< D then take L = D. Shear strength The strength of the key can e represented y a diagram similar to fig.(5.) and with P = S s where S s is the maximum shear at the end of the key and hence P Ssh tan (5.7) L0.5D And the torque transmitted can e expressed y: T SSDL L 9 SS (5.) From equation (5.) T SS (5.9) L ( 0.5D 0.L ) Based on the diameter of the shaft the standard dimensions of a square can e determined from tale (5.a) or tale (5.) The maximum length of the key should not exceed.5 D as the extra length, practically, will e useless Dr. Salah Gasim Ahmed YIC
Diameter of Shaft inclusive (mm) Tale (5.a) Standard dimensions of straight key (metric) Key Dimensions (mm) Diameter of Key Dimensions (mm) Shaft inclusive Width Thickness h Width Thickness h (mm) 6-6 95 5 9 0 96 0 6 0 7 5 5 50 6 0 6 6 5 70 0 0 7 7 00 5 5 0 0 0 50 9 60 50 5 50 9 6 90 6 5 5 6 0 9 0 70 6 59 65 0 0 0 66 75 0 00 90 5 76-5 0-500 00 50 Diameter of shaft D (in) (Inclusive) 9 6 5 7 5 6 6 6 6 6 7 5 6 Tale (5.) Standard dimensions of straight keys (inch) Width 6 5 Key dimensions (in) Thickness Standard h 6 5 Flat h 6 7 6 Diameter of shaft D (in) (Inclusive) 7 5 6 6 5 6 6 5 6 7 9 0 Width 7 Key dimensions (in) Thickness Standard h 7 Flat h 5 7 Dr. Salah Gasim Ahmed YIC
Material Tale (5.) Mechanical Properties of Metals (Inch system) Ultimate tensile strength (kpsi) Tensile (kpsi) Elastic limit Compressive (kpsi) Shear (kpsi) Young s Modulus (kpsi) Modulus of rigidity (kpsi) Steel casting SAE 00 60 5 5 9000 00 Steel casting SAE 000 7 0 9 7 9000 00 Steel casting SAE 0050 0 0 9000 600 Alloy steel casting SAE 090, ASTM A- Stainless steel: C 0.0, Cr, Ni Stainless steel: C 0.0, Mn 0., Si 0.5, Cr, Ni 0.6 Stainless steel: SAE 0905 90 60 60 6 9000 00 90 0 0 0 9000 00 05 60 60 6 9000 00 96 0 0000 000 Caron steel SAE00 5 0 0000 700 Caron steel SAE00 Caron steel SAE00 Caron steel SAE00 Caron steel SAE050 Caron steel SAE095 Caron steel SAE0 6 5 5 000 600 75 6 000 500 90 50 50 0 0000 00 95 5 5 5 900 00 0 60 60 6 9700 00 6 000 600 Nickel steel SAE0 70 5 5 7 9700 000 Nickel steel SAE0 0 95 00 60 0000 00 Cr-Ni steel SAE 0 55 95 00 57 0500 500 Cr-Ni steel SAE 0 60 0 0 7 0500 500 Cr-V steel SAE 650 00 70 90 00 000 000 Cr-Ni-V steel 60 0 0 0 0500 500 Nitraalloy Steel 5 90 0 55 9000 600 Wrought iron 7 6 6 7000 0000 Dr. Salah Gasim Ahmed YIC
Tale (5.) Mechanical Properties of Metals (Metric system) Material Ultimate tensile strength Elastic limit )MN/m ( Tensile Compressive Shear Young s Modulus )GN/m ( Modulus of rigidity )GN/m ( Steel casting SAE 00 0 75 0 5 00 76 Steel casting SAE 000 500 0 70 0 00 76 Steel casting SAE 0050 550 0 00 0 00 79 Alloy steel casting SAE 090, ASTM A- Stainless steel: C 0.0, Cr, Ni Stainless steel: C 0.0, Mn 0., Si 0.5, Cr, Ni 0.6 Stainless steel: SAE 0905 60 0 0 50 00 77 50 900 900 550 00 76 70 0 0 50 00 76 660 0 0 0 07 Caron steel SAE00 75 5 5 0 07 Caron steel SAE00 0 0 0 50 0 Caron steel SAE00 50 90 90 0 0 79 Caron steel SAE00 60 5 5 0 07 7 Caron steel SAE050 655 60 60 50 05 77 Caron steel SAE095 50 5 5 5 0 77 Caron steel SAE0 0 55 55 50 0 77 Nickel steel SAE0 0 0 0 90 0 Nickel steel SAE0 50 655 700 5 07 Cr-Ni steel SAE 0,00 660 700 95 0 6 Cr-Ni steel SAE 0,00 0 960 500 0 6 Cr-V steel SAE 650,00,00,00 700 90 Cr-Ni-V steel,00 900 900 550 0 6 Nitraalloy Steel 50 60 0 0 00 79 Wrought iron 0 0 65 0 6 70 Dr. Salah Gasim Ahmed YIC 5
Example: 7 Find suitale dimensions of a square key to fit into 6 inch diameter shaft. The shaft transmits 95 hp at a speed of 00 rpm. The key is made of steel SAE 00. Take safety factor of.5 and stress concentration factor k =.6 Solution: Dr. Salah Gasim Ahmed YIC 6
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Exercises:. A shaft transmits a torque of 50 Nm to a pulley through a square key. The key is made of steel SAE 00. Taking a safety factor of determine suitale dimension for the key if the diameter of the shaft is 50 mm.. A power of 0 kw is transmitted from a 0 mm diameter shaft to a spur gear through a square key. The shaft rotates at 000 rpm. Select a suitale material for the key and determine its dimensions. Take factor of safety =.5. A shaft transmits a torque of 00 lin. to a pulley through a square key. The. Taking a safety factor of select a suitale material and determine dimension for the key if the diameter of the shaft is in. Dr. Salah Gasim Ahmed YIC