Analysis

 Angle of Twist

q = T*L/C2*a*b^3*G

T = Torque applied

L = Length of neck

C2 = Constant (0.2536667)

a = width of neck

b = height of neck

G = modulus of Rigidity

Using Castigliano's theorem to calculate the deflection for

s = P/A

t = Mc/I

I(prototype) = 1/4P*(.25in)^4 = 0.003068in^4

I(redesign) = 1/4P*[(.375in)^4 – (.3125in)^4] = 0.00804in^4

E(prototype) = 30*10^6 psi

E(redesign) = 2*10^6 psi

M = PR*(1-cosq) + QR*sin q + Mo

d = (1/EI)*[120Lbs*(R)^3*300Lb*in]

(Analyzed as a cantilever beam)

d = (PL^3)/3EI = (ML^2)/3EI

M = 120Lbs*0.5in = 60 Lb*in

I(prototype) = 1/4P*(.25in)^4 = 0.003068in^4

I(redesign) = 1/8*P*(3/4in)^4 + (0.50*(3/4)^3)/12 = 0.14183in^4

E(prototype) = 30*10^6 psi

E(redesign) = 2*10^6 psi

The bending due to string load at the end of the neck = bending of joint to body * (14.25/2.25) + bending of neck.

This yields a deflection of less than one tenth of an inch at the end of the neck in both instances (prototype and redesign).

Shear strength = Force / Area

It took 500Lb’s force to shear a 1.5 square inch bonded area from a test piece of fiberglass.

Therefore shear strength equals 500Lb/1.5in^2 which is 333.3333 psi.

The bridge loading is 120 Lbs/ 4in^2 = 30psi.

The bond strength provides a factor of safety of greater than 10.