Workout # 1 (DUE: Monday, October 15, 2012)

WO#1

Answer the following problems

2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 2.10, 2.11, 2.12, 2.13, 2.14 and 2.18

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**All images are from the book Design of Reinforced Concrete by Jack McCormac (Seventh Edition)

**Also take note on how to compute for the area of steel with designation #(number here). You may refer to ConMet notes or the area can be obtain by dividing the number by 8. And, the result gives you the area of the rebar in in^2.

andami naman sir >.<

Sir, sa 2.8-2.11, 1.5in po ba yung concrete cover & wala po ba talagang stirrups (kung meron ano po yung diameter)???

di considered si stirrups sa prob. but since ang definition kay effective depth, d, ay distance from the centroid of the rebars to the extreme comp fiber, di na kailangan si concrete cover at stirrups since shown na si effective depth sa figure.

HIIHIHHIHIHI.... BOOM PRACTICE PRACTICE DIN!!!

TAPOS NA KO! HOOOO.

engr.c wrote:TAPOS NA KO! HOOOO.

Pakopya HAHA! =))))

sir bakit ganun, di ko mapalabas sagot sa 2.9 xD Diba As = 3in^2, since ( 6 / 8 )*4 = 3?

at tsaka ano gagawin pag may kasamang beam weight?

Mag-iiba po yung sagot dahil dun sa pag-compute nila ng As. but we stick sa pagcompute natin na #/8.

Yung beam weight ay given naman si weight ni concrete.

engr.c wrote:Mag-iiba po yung sagot dahil dun sa pag-compute nila ng As. but we stick sa pagcompute natin na #/8.

Yung beam weight ay given naman si weight ni concrete.

ahh bale di na po 1450 ang sagot dun?

IraBalmoris wrote:sir bakit ganun, di ko mapalabas sagot sa 2.9 xD Diba As = 3in^2, since ( 6 / 8 )*4 = 3?

at tsaka ano gagawin pag may kasamang beam weight?

As = [ pi/4 * .75^2 ] * 4

Sir! pwede niyo po i-post yung answers sa seatwork #2 yung ginamitan namen ng probabilty. salamat po