THE BEAM HAS A CIRCULAR CROSSSECTION WHICH IS TAPERED FROM Z

THE BEAM HAS A CIRCULAR CROSS-SECTION WHICH IS TAPERED FROM ZR TO R IN RADIOS. WHEN IS THE MAXIMUM NORMAL STRESS IN THE BEAM LOCATED?

Let dia at fixed end be d₀ and dia at free end where x = L be d_l.

At any distance x from fixed end, diameter d = d₀ - (x/L)*(d₀ - d_l)

At distance x, Section modulus S_x = (pi/32) *d³

S_x = (pi/32) [d₀ - (x/L)*(d₀ - d_l)]³

Bending moment at x, M_x = P(L - x)

Bending stress at any cross section = M_x / S_x

= P (L-x) / [(pi/32) [d₀ - (x/L)*(d₀ - d_l)]³]

= (32/pi) P (L-x) / [d₀ - (x/L)*(d₀ - d_l)]³

Putting d₀ = 2*2r = 4r and d_l = 2*r = 2r, we get

Bending stress = (32/pi) P (L-x) / [4r - (x/L)*(4r - 2r)]³

= (32/pi) P (L-x) / (4r - 2rx/L)³

For max. stress we differentiate it with respect to x and equate it to zero. We get

(-3) (-2r/L) (L-x) (4r - 2rx/L)^-4 - (4r - 2rx/L)^-3 = 0

(4r - 2rx/L)^-3 [(6r/L)(L-x) / (4r - 2rx/L) -1] = 0

Either 4r - 2rx/L = 0 which means x = 2L which is invalid as x can go only to L.

Or (6r/L)(L-x) / (4r - 2rx/L) -1 = 0

(6r/L)(L-x) = 4r - 2rx/L

3 - 3x/L = 2 - x/L

2x/L = 1

x = L/2