Quiz 5 - Solution

Using the method of superposition and beam properties from the attached sheets,

a) Determine the force in the spring at C.

b) Determine the deflection at C.

Let

L = 12 ft

w = 0.2 k/ft

E = 1,000 ksi

I = 150 in⁴

k = 1 kip/in.

Solution

a)

X = force in spring

Compatibility:  Equal deflections at C for beam and spring,

D_beam =  D_spring

Beam deflection at C:

D_beam =  qL/2 - X(L/2)³/3EI =  qL/2 - XL³/24EI

q  =  wL³/24EI -  (XL/2)L/3EI  =  wL³/24EI -  XL²/6EI

D_beam =  wL⁴/48EI -  XL³/12EI - XL³/24EI =  wL⁴/48EI -  XL³/8EI   (upward)

D_beam =  0.2(12)⁴(12)³ /48(1000)(150) -  X(12 x 12)³/8(1000)(150)  inch

D_beam =  0.99533 - 2.48832 X inch

Spring deflection at C:

D_spring = kX = 1X inch (upward)

D_beam =  D_spring :

0.99533 - 2.48832 X =  X

X = 0.2853 kip

b) D_beam =  D_spring = 1X = 0.285 inch (upward)