Question 6 of 6: Spring-applied external double-shoe drum brake

Question 6: Spring-applied external double-shoe drum brake

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) Equilibrium free bodies for each shoe-and-arm assembly, the spring and the drum, with all forces expressed through the spring force F_s; (b) the value of F_s that produces 1200 N·m of braking torque.

Approach. Take moments about the common hinge pin for each arm separately. The spring pulls both arm tops inward with the same force, but the friction force helps one arm and opposes the other, so the two shoes press on the drum with different normal forces. Sum the two friction torques and solve for F_s.

(a) The free bodies

Spring. A two-force member: it is stretched, so it pulls inward on both arm tops with the same magnitude F_s, one force to the left on the right-hand arm and one to the right on the left-hand arm. Its own equilibrium is trivial and it is what makes this a fail-safe brake — the brake is applied whenever hydraulic pressure is lost.

Each shoe-and-arm assembly. Three external actions plus the pin: the spring force F_s acting horizontally inward at the top pin, 400 + 350 = 750 mm above the hinge; the drum’s normal reaction N pushing the shoe outward along the horizontal centreline, applied 150 mm out from the vertical centreline and 350 mm above the hinge; the friction force μN acting vertically on the shoe in the direction the drum surface sweeps; and the pin reaction at O, which is whatever is needed to close force equilibrium and contributes no moment about itself. (The hydraulic release cylinder is inactive when the brake is applied and carries no force in this free body.)

The drum. It carries the two equal-and-opposite reactions: two normal forces N₁ and N₂ pushing inward on opposite sides, which very nearly cancel and so put little net load on the drum bearings, and two friction forces μN₁ and μN₂ both acting against the direction of rotation at radius r. Those two friction forces are what sum to the braking torque, and unlike the normal forces they reinforce each other rather than cancelling.

(b) The required spring force

1. Moment equation for one arm about the hinge pin O. With the pin as origin, the spring acts at height 750 mm, the normal force at (150 mm out, 350 mm up) and the friction force vertically at the same point: $$F_s\,(750) = N\,(350) \mp \mu N\,(150)$$ The friction term subtracts when the drum surface sweeps in the direction that drags the shoe into the drum (self-energising) and adds when it drags the shoe away (de-energising). Solving for the normal force on each shoe, $$N_{\text{se}}=\frac{750\,F_s}{350-150\mu},\qquad N_{\text{de}}=\frac{750\,F_s}{350+150\mu}$$
2. Substitute the assumed friction coefficient. With μ = 0.30 the two denominators are $$350-150(0.30)=305\ \text{mm},\qquad 350+150(0.30)=395\ \text{mm}$$ so N_se = 2.459 F_s and N_de = 1.899 F_s. The self-energising shoe presses about 30 % harder for the same spring force — the two shoes are not equally loaded, which is the point of the question.
3. Sum the braking torque from both shoes. Each friction force acts tangentially at the drum radius, so $$T = \mu r\,(N_{\text{se}} + N_{\text{de}}) = \mu r\,(750F_s)\left(\frac{1}{350-150\mu}+\frac{1}{350+150\mu}\right)$$ Substituting r = 0.150 m, μ = 0.30 and the two denominators in millimetres, $$T = 0.30(0.150)(750F_s)\left(\frac{1}{305}+\frac{1}{395}\right)=0.19610\,F_s\ \ [\text{N}\cdot\text{m per newton of }F_s]$$
4. Solve for the spring force. Setting T = 1200 N·m, $$F_s = \frac{1200}{0.19610} = \boxed{6119\ \text{N}\ (6.12\ \text{kN})}$$
5. Back-substitute for the shoe loads and check the torque. $$N_{\text{se}}=2.459(6119)=15{,}047\ \text{N},\qquad N_{\text{de}}=1.899(6119)=11{,}619\ \text{N}$$ $$T = 0.30(0.150)(15{,}047+11{,}619)=0.045(26{,}666)=1200\ \text{N}\cdot\text{m}\ \checkmark$$ The self-energising shoe contributes 677 N·m and the de-energising shoe 523 N·m — a 56 : 44 split.
6. Note that the answer does not depend on which way the drum turns. Reversing the rotation simply exchanges the roles of the two shoes; the sum 1/(350−150μ) + 1/(350+150μ) is unchanged, so the same 6.12 kN spring gives the same 1200 N·m in either direction. Only the individual shoe loads swap. This symmetry is a genuine design virtue of the two-shoe layout and is worth stating in the answer.
7. Hinge-pin load, for completeness. Closing force equilibrium on the more heavily loaded arm gives a horizontal pin reaction of N_se − F_s = 15,047 − 6119 = 8928 N and a vertical reaction of μN_se = 4514 N, a resultant of 10.0 kN — well above the spring force, so the hinge pin, not the spring anchorage, sizes the pin joint.