(PLZ SOLVE THIS PROBLEM, AND SHOW ALL THE STEPS. PLZ TYPE YOUR ANSWER, OR MAKE Y

Question

(PLZ SOLVE THIS PROBLEM, AND SHOW ALL THE STEPS. PLZ TYPE YOUR ANSWER, OR MAKE YOUR HAND WRITING VERY CLREAR.)

Basic Building Layout 40 I 40 It Problems 2.13/2.14 The roof of the building is flat. It is composed of 2 in. of reinforced concrete on 18-gauge metal decking that weighs 3 psf. A single-ply waterproof sheet of o.7 psf will be used. The ceiling beneath the roof is unfinished, but allowance for mechanical ducts should be provided. Determine a. The roof dead load b. The roof live load in psf to be applied to column A1

Explanation / Answer

Thickness of concrete=2in

Self weight of concrete=150*(2/12)=25 psf

Weight of deck=3 psf

Weight of ply waterproof sheet=0.7 psf

Total dead load = 25+3+0.7=28.7 psf

Triburtary area of column A1=(35/2)*(40/2)=350 ft2

Total dead load on column=350*28.7=10045 lb=10.045kips

b) Roof live load is generally 20 psf per ASCE-7

Therefore,total live load ion column A1=20*350=70000 lb=7 kips

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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