Question

BRASS STEEL ti ods Problem 1. Torque To is applied at the midpoint of a solid steel bar with diameter ds and length 1.5 L which is bonded to two brass collars, each with wall thickness to and length 0.25 L. The twist angle at point C is specified to be an allowable angle oca. To I 1 А B с E a L va 2 4 L The steel bar and brass collars are attached to fixed supports at A and B. Equal reaction torques T are applied by the supports to keep the system in equilibrium. The steel and brass have shear moduli Gs and Gb respectively. L (a) Find an expression for the twist angle 0B at point B in terms of To, L, G5, Gp and the polar moments of inertia Ips and 1pb of the steel bar and brass collars, respectively. (b) Find an expression for Iph in terms of Gs, Gb, Ips, To, L, and the allowable twist angle oca: (c) Find an expression for the thickness tp of the brass collars in terms of ds and Ipb

Answer 1

For the given torsional moment at point C, the part of shaft are decomposed and the respective torques are shown on those parts. To calculated φB, the angle of twist on steel shaft due to torque on it and the angle of twist on brass collar due to torque on it are equated. The relation between the torque on steel shaft and brass collar is then substituted into the expression for the total torque so as to obtain torque on steel shaft or brass shaft. This obtained expression is put into the expression of angle of twist and the φB is then obtained.

The φCA will be the sum of φBA and φCB. The angle of twist at point B w.r.t. A has already been calculated. The angle of twist at point C w.r.t. B is then calculated. The final angle of twist can be obtained by adding both the expressions.

The polar moment of inertia can be written in terms of outer diameter and inner diameter. The outer diameter can then be written in terms of inner diameter and the thickness of collar. After simplifying the expression, we can get the expression for thickness t_b.

T To T = Tola B с D A 42 112 lu Tola Tola Tol2 To/ Tolz). Part of Sheft Ac will experience Tordibual onomat T GO ds Ts (414) 4 To (4) T Gs. 7 2 GIBE As Shaft are Parallel 11 ( 14 To To/2 = Tst T3 Gg Ips Gz IBE (Gy. I + G2 573) Io = Ts + To Go Torres TooL 8 (GI + Gg Ipo 1 Supresantion for Twist Angle To/2 1+ Go Free GT = 0 + BA TBA Tol & (GI + Go IPB) (To/2) (1/2) Toh 4776 ToL 4 { 2 (G3 5 + G2 IP) IG Sca ToL 8 (G If + GZ 5.8) 4 Ip Gs e) 14:42) - "} 4 4 32 IP + d = (4 + 2t) " t = 32.77 + do - de 132 IPB dola t = 1 2 de + + u ū 1. Exprecrion for thic Kuera to