chapter 7, problem 1P step 4 of 13 -> Zx=64 in^3 is in table 3-2 but why bf=7\'\

Question

chapter 7, problem 1P step 4 of 13 -> Zx=64 in^3 is in table 3-2 but why bf=7'' , tf=0.43??

How can i solve it ..?

Table 3-2 (continued) W-Shapes Selection by Zx Fy = 50 ksi Shape kip-ft kip-ftkip-ft kip-ft kips kips in.ASD LRFD ASD LRFD ASD LRFD ft kips kips ftin.4ASDLRFD W18 3566.5 166 249 101151 8.14 12.3 4.3112.3 510 106 159 W12x4564.2 160 241 1015 3.80 5.80 6.8922.4 34881.1 122 4.0 160 24098.7148 6.249.36 5.37 15.2448 93.8 141 61.5 153 231954 143 5.37 8.205.47 16.2385 874 131 W10x49 60.4151 227 95.4 1432.46 3.71 8.97 31.6 27268.0102 W8x5859.8 149 224 90.8137 .70 2.55 7.42 41.6 228 89.3 134 W12x4057.0 142 214 89.9135 3.66 5.54 6.85 21.1 307 70.2 105 W10x4554.9 137 206 85.8 129 2.59 3.89 7.1026.9 248707 106 W16x36 Step 4 of 13 A With the obtained value of Z, select a suitable W-section from AISC steel construction manual but why..?? table 3-2 Select section W16x36havingZ,64 in.. flange width b,7". ,-0.43 For a compact section, consider the value of pf from table 7.4.1, ''Width/thickness limits for "Noncompact Section Beams to achieve F, at extreme fiber from the text book. Calculate the value of by using the equation 2t Here, b, is width of the flange and , is thickness of the flange Substitute 7 in for b, and 0.43 in for t, 2(0.43) - 8.13

Explanation / Answer

You don't have to solve for the value of width and thickness.

Standard section are manufactured with predefined dimentions,generally we take these values from a standard book or from a standard table.

According to the required section modulus or moment of inertia or area of the section a trial section is selected from these standard tables with the predefined dimensions.

Here also bf = 7" and tf= 0.43" are the standard width and thickness of W16×36 section.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.