5 20 pts A beam has a crosssection shown below Determine a T

5. (20 pts.) A beam has a cross-section shown below. Determine a) The moment that exists on the beam if the beam is elasto-plastic and has the distribution of stress as shown below b) The plastic moment for the beam. The yield stress is 50 ksi. 12 in 50 ksi 1 in l in z-z elasto-plastic 12 in Stress Distributio 4.5 in 3.25 in 1 in 50 ksi , 8 in 5.15 in

Solution

a) Given the stress distribution and cross section of beam moment can be calculated by multiplying the force with the lever arm.

Compressive force in the beam = 50*12*1+25*5.25*1*2 = 862.5 kilopound

Tension force in the beam = 50*8*1+50*2.25*1*2+25*4.5*1*2 = 850 kilopound

Point of application of compressive force is calculated as follows:

= (50*12*1*0.5+25*5.25*2(1+1.75))/862.5 = 1.184in from top

Point of application of tension force is as follows:

= (50*8*1*0.5+50*2.25*1*2*(1+2.25/2)+25*4.5*2(3.25+4.5/3))/850 = 2.055in from bottom

Hence lever arm = 14-(1.184+2.055) = 10.76in

Thus the moment that exists on the beam = 862.5*10.76 = 9281.236 kilopound-inch.

b) In order to calculate the plastic moment for the beam plastic neutral axis needs to be evaluated first

Plastic neutral axis is the section at which top and bottom areas are equal

Let it be at x from the joint between top web and flange

12*1+1*x*2 = 1*(12-x)*2+8*1 Hence x = 5in

Hence location of plastic neutral axis is 6in from top and 8in from bottom. Thus y_p = 8in

Plastic moment = z_p*f_y

Given f_y =50ksi

z_p= I_p/y_p

I_p = 2(1*12³/12+1*12(6-5)²)+12*1³/12+12*1(6-0.5)²+8*1³/12+8*1(8-0.5)² = 1125.67 in⁴

z_p = 1125.67/8 = 140.71in³

M_p = 140.71*50 = 7035.42 kilopound-inch

Thus we obtain that the plastic moment is smaller than the elastoplastic moment which must not be true.

Thus the elastoplastic neutral axis is not correctly marked.