Question

The shaft EFG, which is running at constant speed of n (rev/min) and transmitting a power of P (kW), contains gear F. Gear F transmits torque to shaft ABCD through gear C, which drives the chain sprocket at B, transmitting a force P as shown. Sprocket B, gear C, and gear Fhave pitch diameters of a, b, and c, respectively. The contact force between the gears is transmitted through the pressure angle o. Assume no frictional losses and consider the bearings at A, D, E, and to be simple supports. In the project do the followings in the given order. (Reports with wrong data sets will be graded as zero). A. Draw the free body diagrams of the following parts (neglect inertias) 1. The shaft EFG 2. The shaft ABCD B. Stresses in shaft ABCD 1. Draw the shear force and the bending moment diagrams of the shaft ABCD for xy- and xz-planes. 2. Plot torque diagram. 3. Plot combined bending moment diagram. 4. Locate the point on shaft ABCD that contains the maximum tensile bending, maximum torsional shear and maximum transverse shear stresses. 5. From this, determine the maximum tensile and shear stresses in the shaft. Hint: Although this is a three-dimensional problem and vectors might seem appropriate, we will look at the components of the moment vector by performing a two-plane analysis, i.e. the loading in the xy plane, as viewed down the z axis, where bending moments are actually vectors in the z direction. Thus we label the moment diagram as Mz versus x. For the xz plane, we look down the y axis, and the moment diagram is My versus x. The net (combined) moment on a section is the vector sum of the components. That is, M=M; +M

PLEASE CAN YOU SOLVE THE HOMEWORK ACCORDING TO THE ITEMS AND BY STATING WHICH ITEM THE ANSWERS BELONG TO?
THE ANSWER IS EXPECTED AS SOON AS POSSIBLE
EMEGENCY HOMEWORK

The shaft EFG, which is running at constant speed of n (rev/min) and transmitting a power of P (kW), contains gear F. Gear F transmits torque to shaft ABCD through gear C, which drives the chain sprocket at B, transmitting a force P as shown. Sprocket B, gear C, and gear Fhave pitch diameters of a, b, and c, respectively. The contact force between the gears is transmitted through the pressure angle o. Assume no frictional losses and consider the bearings at A, D, E, and to be simple supports. In the project do the followings in the given order. (Reports with wrong data sets will be graded as zero). A. Draw the free body diagrams of the following parts (neglect inertias) 1. The shaft EFG 2. The shaft ABCD B. Stresses in shaft ABCD 1. Draw the shear force and the bending moment diagrams of the shaft ABCD for xy- and xz-planes. 2. Plot torque diagram. 3. Plot combined bending moment diagram. 4. Locate the point on shaft ABCD that contains the maximum tensile bending, maximum torsional shear and maximum transverse shear stresses. 5. From this, determine the maximum tensile and shear stresses in the shaft. Hint: Although this is a three-dimensional problem and vectors might seem appropriate, we will look at the components of the moment vector by performing a two-plane analysis, i.e. the loading in the xy plane, as viewed down the z axis, where bending moments are actually vectors in the z direction. Thus we label the moment diagram as Mz versus x. For the xz plane, we look down the y axis, and the moment diagram is My versus x. The net (combined) moment on a section is the vector sum of the components. That is, M=M; +M

Answer 1

gomm k 75mm # To f be a= Plnion = 125mm dia of on AkcD Shift - 250m JUAABCD B. с ti ca dia of A Sprocket 와 B 75mm 250mm 125mm Isom 12kW Giren dara Power transmitted shaft shered a ( E F G shahr e 1200 opm Es- 2 0 - 20° -450MPa solution : - pomer closque 12kW power 3 12x10x 60 angular speed min) (sec 22 x 1200 (2pm)

2. 95.54 am We know that, Torque = (tangential forie je Carm/rading) T= fx de Id = 1526. Gyal 95.54 0.0625 Mais implier F = Foss - r = 1626. 074n X= 20 фка им о. radial foorce F = F sina 1 F = 556.380]

Ry (165) - (5 2 8:64) (75) = 252.90/ & BD for Efer shaft. xz plane xy-plane 1526. Gyn 556.38N 75m gom gomn 75mm REX Rut Rey day soking for the reaction forcer using equillibrium equatriem. REF 303.48N &= 695N RE= 83401

Xy Plane 4. X2 plane 152B Glyn 556.36N t 75 mm goon (areas 9 gong 1252.9 N 303.48N $34.90 6950112 SED 69501 252.gal BA F 102 63401 303.48a BMD 62550 Nmm 22761 nman بلا plote maximum Reading moment occurs SKD changes it sign. Torque Diong song5i54um

B-20 from the horizontal. This will be Ankenum bending noment at f of given by 2 Manor 1 (62.550) ²+ (2776) 66.515 Nm shahr cross-sechin at r. 62.51 22.76 tamp=2276 62.55 o ß= 20 . the center Maximum sheat force occurs at the the shaft and is uniform throughout hout the shaft. Torque (mexinum) occurs at outer most fibre off the shaft. (F) moment will reneur at point F and outer most gibre and at maximum bending the isitical point to dentn

6. meximum Normal shen theo 3. Calculation for diameter of the shaft using Heory ago 677.52 32X (66.515 32 M Zd² d3 7d3 16X 119.36 486.6 d' axy 2d3 Principal stress can by given by 1699 extry f city substitutiny, z 1 oooh be exy. 338.76 + d3 ( 593.) کے 931.76 450 2 => d3 safety falbon. 'k in our wohing, d= 16 mm

for shaft 7. TI a 75 125 mm of P- 25 48 nl for shaft ABCD 250 mm 125m lanone that dorque transmitted by year torque hauranted by by pruien 3) Tu 250 - Tu = 1910 pln 95.54. pxe Sadocker Ta Da Tp PP 19 we can write 191 dalm Tangential force by gou e per 35 191.1 nm

8. - plane 1526.&N A = 227 7.88N1 tu 1526. 8 al I= 1629.9 n & t 2 = 556.44N fy=fsince 0A 20° xz glave 25480 556.40 75 Roy Borro)-($56.48528) ao (450)-(1828.8832870 -5467as)- RO = 1104.13 RA=424.641 Ro 1 M te o 125 250 1 325 125 Roz Pay RAZ RA TRO = 1528.8N o - 6 26.51N 3

2548 1528.8N 556.4N & 325 1251 42.46 104.13 $16.50 SFD 1104.13N 141.964 w A B B 424.64 2277.BN BMO : 170837.(um 138016.25 Nm 1033125 31898. 75 mm D B C D AI Torque Dragnum Isi. Non B D

fet us check for the markenum Pending moment print @i (10.337² + (138) 138.38 Nm max (1.7o. 83 ge+(3 184² 'hex @ B 173.77 Nm momeny Clearly merin merinum Bi accent af B. 1700183 Bending 31.84 tauff 170.83 31.84 Bt Tayo

at B 11. Maximum principal stren given by can be A og 94 32x178.77 32M H69.3 12 7d3 a o d3 167 Куч ha 16X1911 nd Ć 973-96" d² remy + Txt ay 2 2 7. 1769.3 1 1769.31 -- +(973.26 2d T: 2199.88 d3

Maximum Neramal shers theory using hartu 6,2 199.88 450 da 19 mm