- Compute the maximum live-load moment, with impact included
In accordance with the AASHTO, the distribution factor is DF = 6.75/5.5 = 1.23; IF = 50/
(74.5 + 125) = 0.251, and 16(1.23)(1.251) = 24.62 kips (109.510 kN); 4(1.23)(1.251) =
6.15 kips (270.355 kN); P1x+1 = 2(24.62) + 6.15 = 55.39 kips (246.375 kN). Refer to Fig.
38a as a guide. Then, M1x+1 = 12[(55.39 x 39.58 x 39.58/74.5) - 24.62(14)] = 9840
in-kips (1111.7 kN-m).
For convenience, the foregoing results are summarized here:
M, in-kips (kN-m) Sb, in^3 (cm^3 ) Sts, in^3 (cm^3 ) S^ in^3 (cm^3 )Noncomposite 6,490 (733.2) 879(14,406.8) 570 (9,342.3)
Composite,
dead loads 2,080 (235.0) 1,066(17,471.7) 1,290(21,143.1) 932(15,275.5)
Composite,
moving
loads 9,840 (1,111.7) 1,179(19,323.8) 2,936(48,121.0) 1,826(29,928.1)
- Compute the critical stresses in the member
To simplify the calculations, consider the sections of maximum live-load and dead-load
stresses to be coincident. Then/ 6 = 6490/879 + 2080/1066 + 9840/1179 = 17.68 kips/in
2
(121.9 MPa);/, = 6490/570 + 2080/1290 + 9840/2936 = 16.35 kips/in
2
(112.7 MPa); ftc =
20807(30 x 932) + 98407(10 x 1826) = 0.61 kips/in
2
(4.21 MPa). The section is therefore
satisfactory.
- Determine the theoretical length of cover plate
Let K denote the theoretical cutoff point at the left end. Let Lc = length of cover plate ex-
clusive of the development length; b = distance from left support to K\ m = LJL\ d = dis-
tance from heavier exterior load to action line of resultant, as shown in Fig. 37; r = 2dlL.
From these definitions, b (L - Lc)/2 = L(- m)/2; m = l- 6/(0.5L). The maximum mo-
ment at K due to live load and impact is
(PLL+IL)(l-r + w-m^2 )
M1x+1 = 4 (51)The diagram of dead-load moment is a parabola having its summit at midspan.
To locate K, equate the bottom-fiber stress immediately to the left of K, where the cov-
er plate is inoperative, to its allowable value. Or, (PLL+i)/4 = 55.39(74.5)(12)/4 = 12,380
in-kips (1398.7 kN-m); d = 9.33 ft (2.844 m); r = 18.67/74.5 = 0.251; 6490(1 - w
2
)/503 +
2080(1 - w 2 )/612 + 12,380(0.749 + 0.251in - m 2 )/677 - 18 kips/in^2 (124.1 MPa); m =
0.659; Lc = 0.659(74.5) = 49.10 ft (14.97 m).
The plate must be extended toward each support and welded to the W shape to devel-
op its strength.
- Verify the result obtained in step 9
Thus, b = '/2(74.5 - 49.10) = 12.70 ft (3.871 m). At K: M"DL = 12(^1 X 2 x 74.5 x 0.780 x
12.70 -^1 A x 0.780 x 12.70^2 ) = 3672 in-kips (414.86 kN-m); MCDL = 3672(250/780) = 1177
in-kips (132.98 kN-m). The maximum moment at K due to the moving-load system occurs
when the heavier exterior load lies directly at this section. Also M1x+1 = 55.39(74.5 -
12.70 - 9.33)(12.70)(12)/74.5 = 5945 in-kips (671.7 kN-m);/d = 3672/503 + 1177/612 +
5945/677 = 18.0 kips/in^2 (124.11 MPa). This is acceptable.