At E, V= 207 - 4.67(4) = 188 kips (836.2 kN); v = 188/29.75 = 6.32 kips/in
2
(43.576 MPa) > 3.45 kips/in
2
(23.787 MPa); therefore, intermediate stiffeners are re-
quired in EB.
- Provide stiffeners, and investigate the suitability of their
tentative spacing
Provide stiffeners at F, the center of EB. See whether this spacing satisfies the Specifica-
tion. Thus [260AMJ]^2 = (260/155)^2 = 2.81; a/h = 122/68 = 1.79 < 2.81. This is accept-
able.
Entering the table referred to in step 6 with a/h = 1.79 and hltw =155 shows vallow =
7.85 > 6.32. This is acceptable.
Before we conclude that the stiffener spacing is satisfactory, it is necessary to investi-
gate the combined shearing and bending stress and the bearing stress in interval EB.
- Analyze the combination of shearing and bending stress
This analysis should be made throughout EB in the light of the Specification require-
ments. The net effect is to reduce the allowable bending moment whenever V >
0.6Fallow. Thus, K 81101 , = 7.85(29.75) = 234 kips (1040.8 kN); and 0.6(234) = 140 kips
(622.7 kN).
In Fig. 86, locate the boundary section G where V= 140 kips (622.7 kN). The allow-
able moment must be reduced to the left of G. Thus, AG = (207 - 140)/4 = 16.75 ft (5.105
m); M 0 = 2906 ft-kips (3940.5 kN-m); ME = 922 ft-kips (1250.2 kN-m). At G, Mallow =
4053 ft-kips (5495.8 kN-m). At £,/allow = 0.825 - 0.375(188/234) = 18.9 kips/in^2
(130.31 MPa); Mallow = 18.9(77,440)/[35.25(12)] = 3460 ft-kips (4691.8 kN-m).
In Fig. 8c, plot points E' and G' to represent the allowable moments and connect these
points with a straight line. In all instances, M< Mallow.
- Use an alternative procedure, if desired
As an alternative procedure in step 8, establish the interval within which M > 0.75Mallow
and reduce the allowable shear in accordance with the equation given in the Specification.
- Compare the bearing stress under the uniform load
with the allowable stress
The allowable stress given in the Specification fb allow = [5.5 + 4/(a//z)^2 ]10,OOOAMJ^2
kips/in^2 (MPa), or, for this girder, fballow = (5.5 +' 4/1.79^2 )10,000/155^2 = 2.81 kips/in^2
(19.374 MPa). Then fb = 4/[12(0.438)] = 0.76 kips/in^2 (5.240 MPa). This is acceptable.
The stiffener spacing in interval EB is therefore satisfactory in all respects.
- Investigate the need for transverse stiffeners
in the center interval
Considering the interval BC, V= 32 kips (142.3 kN); v = 1.08 kips/in^2 (7.447 MPa); a/h =
192/68 = 2.82 « [260/(MJ]^2 -
The Manual table used in step 6 shows that z;allow > 1.08 kips/in^2 (7.447 MPa);/ 6 allow =
(5.5 + 4/2.82^2 )10,000/155^2 = 2.49 kips/in^2 (17.169 MPa) > 0.76 kips/in^2 (5.240 MPa).
This is acceptable. Since all requirements are satisfied, stiffeners are not needed in inter-
val BC.
- Design the intermediate stiffeners in accordance
with the Specification
For the interval EB, the preceding calculations yield these values: v = 6.32 kips/in
2
(43.576 MPa); z;allow = 7.85 kips/in^2 (54.125 MPa). Enter the table mentioned in step 6
with a/h = 1.79 and h/tw = 155 to obtain the percentage of web area, shown in italics in the
table. Thus, Ast required = 0.0745(29.75)(6.32/7.85) = 1.78 in^2 (11.485 cm^2 ). Try two
4 x 1/4 in (101.6 x 6.35 mm) plates; Ast = 2.0 in^2 (12.90 cm^2 ); width-thickness ratio =
4/0.25 = 16. This is acceptable. Also, (/*/50)^4 = (68/5O)^4 = 3.42 in^4 (142.351 cm^4 );
