Question

BRASS STEEL ads Problem 1. Torque To is applied at the midpoint of a solid steel bar with diameter d, and length 1.5 L which is bonded to two brass collars, each with wall thickness tp and length 0.25 L. The twist angle at point C is specified to be an allowable angle oca To ** B A L 11/25 늦 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 Gp respectively. (a) Find an expression for the twist angle PB at point B in terms of To, L, GS, Gp and the polar moments of inertia Ips and Ipb 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 th of the brass collars in terms of d, 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.
solution T с pe * 44 y 2 point of shaft AC will experience Torsional moment (To/2) Go 1 L $ = Ts (²4) 08 Gs IPs os para Tos (44) Gg. If perrallel 8 As shaft are To GS IPs GB IIB To/=Ts t TB To/=Tst ts. GB IPB Gs. IPs To/2 1+ GRIPA Gs. Ips 98-(72)(44) (Gs.IPs + GB TPB) 883 TO.L 8 (Gg IP ST Gg FPB) Expression for Twist angle

TOL PCA = PBA + Per PBA 8 (Gs IPS tag FPR) PCB = (%) (4) => & CB = IPs Gs TOL •4 Its GS (CA= TOL 8 (Gs. IP + GB IPB) + TOL 4 IPs GS > PCA = TOLS 2 (G FPS G. TPB + lecare ITSGs u IPB 52 {[de tat) - ast} + ds = (a + 2t) => 32 IPB TT 32 IPB 14 po +ds T 2 t=1 32 IPB V 2 tas a to = QIPE + Celkem de / Expression for thickney (to