Problem 59 The W10 X 12 beam is made from A36 steel and is c

Problem 59

The W10 X 12 beam is made from A-36 steel and is cantilevered from the wall at B. The spring mounted on the beam has a stiffness of k = 1000 lb/in .

Part A

If a weight of 8 lb is dropped onto the spring from a height of h = 1 ft , determine the maximum bending stress developed in the beam.

Express your answer to three significant figures and include appropriate units.

I got 6.2 ksi and IT WAS WRONGZO.

B 8ft TIll-hill-k A kA

The max bending stress in the beam can be found using the relation,

sigma max = y_max * M / I ---------- eqn 1

The second moment of inertia for the beam W10x12 has Area = 3.54 in2 and d = 9.87 in with t = 0.13 inch

Due to symmetry of the cross section, the centroid of the section is X = 6 inch and Y = 5 inch

I = Ix + Iy

I_xx = H³b/12 + 2[h³B/12 + hB(H+h)²/4] = 9.87^3*0.13 + 2[ 0.13^3*12/12 + 0.13*12 (9.87+0.13)^2/4]

I_xx = 124.99 + 2*39.00 = 202.994 in4

Iyy = b³H/12 + 2(B³h/12)

Iyy = 0.13^3 * 9.87 /12 + 2*10^3*0.13 / 12 = 21.77 in 4

I = 202.994 + 21.776 = 224.770 in4

Now the moment due to loading of the beam is gven by,

M = Fapplied x d

The applied forces on the beam end depends on the net energy resulting from the collision,

Inital potential energy = mgh

Due to the consevation of energy the kinetic energy of the mass equals the inital potential energy,

KE = 1/2 m v^2 = mgh = 8*32.17*1 = 257.36 J

Due to impact on the spring the energy is assumed to be absorbed by the spring without any losses,

KE of spring = KE mass

1/2 k x^2 = 257.36

Solve for displacement x = 0.0428 ft or 0.0035 in

Now the spring force on the beam is F = kx = 1000 * 12 * 0.0428 = 513.6 lb

The moment acting on the beam is M = F x d = 513.6 * 8 = 4108.8 lb-ft

Also the max deflection of the beam is ymax = 0.0428 ft

The max bending stress on the beam is caculated from eqn 1

sigma max = y * M / I

sigma max = 0.0428 * 4108.8 / 224.770 = 0.78 / 144 = 0.00543 lb/in2