- Calculate the allowable
stress in each interval
between lateral supports
By applying the provisions of the
Manual, calculate the allowable stress
in each interval between lateral sup-
ports, and compare this with the actual
stress. For A36 steel, the Manual for-
mula (4) reduces to fa = 22,000 -
0.679(Z//r)^2 /Q lb/in^2 (kPa). By Manu-
al formula (S)J 2 = 12,000,000/(L'd/4)
lb/in^2 (kPa). Set the allowable stress
equal to the greater of these values.
For interval AB: L' = 8 ft (2.4 m) <
LC, /./aiiow = 24,000 lb/in^2 (165,480.0
kPa); /max = 148,000(12X109.7 =
FIGURE 2. Dimensions of W21 x 55. 16,200 lb/in^2 (111,699.0 IcPa)-this is
acceptable.
For interval BC: L'Ir = 15(12)/!.99
= 90.5; M 1 XAf 2 = 95/(-148) =
-0.642; Cb = 1.75 - 1.05(-0.642) +
0.3(-0.642)^2 - 2.55; /. set Cb = 2.3; /i = 22,000 - 0.679(90.5)^2 /2.3 = 19,600 lb/in^2
(135,142.0 kPa);/ 2 = 12,000,000/[15(12)(4.85)] = 13,700 lb/in^2 (94,461.5 kPa);/max =
16,200 < 19,600 lb/in
2
(135,142.0 kPa). This is acceptable.
Interval CD: Since the maximum moment occurs within the interval rather than at a
boundary section, Q= 1; ZVr= 16.5(12)71.99 = 99.5;/; = 22,000-0.679(99.5)
2
= 15,300
lb/in
2
(105,493.5 kPa);/ 2 = 12,OOO,000/[16.5(12)(4.85)] = 12,500 lb/in
2
(86,187.5 kPa);
/max = 132,800(12)7109.7 = 14,500 < 15,300 lb/in^2 (105,493.5 kPa). This stress is accept-
able.
Interval DE: The allowable stress is 24,000 lb/in^2 (165,480.0 kPa), and the actual
stress is considerably below this value. The W21 x 55 is therefore satisfactory. Where de-
flection is the criterion, the member should be checked by using the Specification.
DESIGN OFA COVER-PLATED BEAM
Following the fabrication of a Wl 8 x 60 beam, a revision was made in the architectural
plans, and the member must now be designed to support the loads shown in Fig. 3a. Cov-
er plates are to be welded to both flanges to develop the required strength. Design these
plates and their connection to the W shape, using fillet welds of A233 class E60 series
electrodes. The member has continuous lateral support.
Calculation Procedure:
- Construct the shear and bonding-moment diagrams
These are shown in Fig. 3. Also, ME = 340.3 ft-kips (461.44 kN-m).
- Calculate the required section modulus, assuming the built-up
section will be compact
The section modulus S = MIf= 340.3(12)/24 = 170.2 in^3 (2789.58 cm^3 ).
