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Q1. Load and Stress Analysis The countershaft with the reverse gears attached to is shown in figu...

Q1. Load and Stress Analysis The countershaft with the reverse gears attached to is shown in figure 2 with the dimension give

LI し2 し3 CB dia. d Driven gear A dia. du Pinion gear iB dia. dB Figure 2. Gear A and B attached to the countershaft.

4 Value Vlne 2 8 12 Unit 2 |!。。。..(L + L.) | 225 1500 × (l+ 은)| 666.4 L9 | mm

machine design question

Q1. Load and Stress Analysis The countershaft with the reverse gears attached to is shown in figure 2 with the dimension given in table 1. Gear A receives power from another gear with the transmitted force Fa applied at the pressure angle au as shown. The power is transmitted through the shaft and delivered through gear B through a transmitted force Fs at the pressure angle shown. You can find the values of the parameters in table 1 [10 points] a) Determine the force Fe, assuming the shaft is running at a constant speed. simple supports. set for the horizontal plane and another set for the vertical plane. b) Find the magnitudes of the bearing reaction forces, assuming the bearings act as c) Draw shear-force and bending-moment diagrams for the shaft. If needed, make one d) At the point of maximum bending moment, determine the bending stress and the torsional shear stress. e) At the point of maximum bending moment, determine the principal stresses and the maximum shear stress.
LI し2 し3 CB dia. d Driven gear A dia. du Pinion gear iB dia. dB Figure 2. Gear A and B attached to the countershaft.
4 Value Vlne 2 8 12 Unit 2 |!。。。..(L' + L.) | 225 1500 × (l+ 은)| 666.4 L9 | mm
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Bending moment, shear force and bending stress all data are submitted in files below:LI し2 し3 CB dia. d Driven gear A Radial comp Tangential component dia. da Tangential component r gar B dia. dB Radial auSolution Part(A) :-FB=2 As per the given table > 0 Nou we havet calculate the tangentia and radial Component :- Tan ential couwhere is the angular speed n radian/sec here - in rpm 6 0 w = 837.76 Yad/s So from equation (2)、 power Ps 88.39 x 837.76 Ps<ince we knolo thot he shaft runs at constan speed means usa=we ço TA-TG because Hence Ta=TB二88.29 N-m Also we know that TaugPart (BI :- Bearina reactions Cxy) plane (FB)z 296 S N LI = 600 mm 17S mm Ro F58.5+N Equilibrium RotRe --992. 09 N Taleiu omeand Roz 1081.84-1557 . ys Answer of part e Part (C):- Bendiuy mowant aud shear force diagram: X 296.S 32.9 о. 175 S9.17 Sheaς.F.D and B.M. D for Xz plane:- 139u.79 N 312 9SN SS7.45N 139u-79 N S.F.D 162 66 N u7S. GN В.М.D A. 285.3 313.83 N- m From boHence max bendij omeut (Bot planz ) Now maximum bendiua stress maX ovevall maximum bendiuastress Part (e):3 al-Pointe 49.16 M

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