Question 6.12: The tee beam of Example Problem 6.2 is made of elastic-plast......
The tee beam of Example Problem 6.2 is made of elastic-plastic material having a stress-strain diagram like Fig. 6.32 with E = 29(10³) ksi and σ_Y = 40 ksi. Determine the following: (a) the yield moment M_Y ; (b) the location of the plastic neutral axis, and the value of the plastic moment M_P ; and (c) the shape factor f.
The section properties calculated in Example Problem 6.2 are shown in Fig. 1.
(a) Determine the yield moment, M_Y. From Eq. 6.47, the yield moment is given by
M_Y = \frac{σ_YI} {c} (1)
The bottom fibers are farthest from the neutral axis, so
M_Y = \frac{(40 ksi)(33.3 in^4)} {(4 in.)} = 333 kip · in
M_Y = 333 kip · in (a) (2)
(b) Locate the plastic neutral axis. The plastic neutral axis divides the cross section into two equal areas (Eq. 6.48).Therefore, for this problem, the plastic neutral axis falls at the flange-web interface, as illustrated in Fig. 2. From Eq. 6.49,
A_C = A_T = \frac{1} {2} A (6.48)
M_P = \frac{σ_YA} {2} (d_C + d_T) (6.49)
M_P = \frac{σ_YA} {2} (d_C + d_T) = \frac{(40 ksi)(10 in^2)} {2} (0.5 in. + 2.5 in.)
or
M_P = 600 kip · in (b) (3)
(c) Calculate the shape factor. The shape factor is given by Eq. 6.50.
f = \frac{M_P} {M_Y} = \frac{Z}{S} (6.50)
f = \frac{M_P} {M_Y} = \frac{600 kip · in} {333 kip · in} = 1.80
f = 1.8 (c) (4)
