Question: Determine limiting moment of resistance if M20 concrete and Fe 415 steel is used.
0

A tee beam having 1250 mm effective width of flange has a thickness of flange equal to 115 mm. The effective depth of the beam is 550 mm and width of the web is 300 mm. It is reinforced with 8 bars of 20 mm ϕ on the tension side. Determine limiting moment of resistance if M20 concrete and Fe 415 steel is used.

Mumbai University > Civil Engineering > Sem 7 > Limit State Method

Marks: 10M

Year: Dec 2015

ADD COMMENTlink
modified 2.1 years ago  • written 2.1 years ago by gravatar for Juilee Juilee2.1k
0

Data:- bf = 1250mm

bw = 300mm

Ast = $\frac{8Xπ}{4X202}$

=2513.27 $mm^2$

Df = 100mm

Fck = $20N/mm^2$

Fy = $415 N/mm^2$

To find :- Mv

Solution:-

1) Depth of NA

Assume the depth of NA lies within the flange

Xu = 0.87fyAst/0.36Xfckbf

= 0.87X415X2513.27/0.36X20X1250

= 100.82 > 100mm

Xu > Df

.•. Assumption is wrong.

2) NA lies in web.

3) depth of NA xu lies below flange and assume or Df/d <=0.2

.•. 100/550 = 0.18 > 0.2

Yf = 0.15xu + 0.65Df

= 0,15(xud) + 0.65 Df

= 0.15Xx + 0.65X100

= 0.15xu + 65 < Df

.•. OK

Cu1 + Cu2 = Tu

0.36fck.bwxu + 0.45fckYf(bf - bw) = 0.87fy.Ast

0.36X20X300.xu + 0.45X20(0.15xu + 65) X (1250 - 300)

= 0.87 X 415 X 2153.27

Xu = 64.39mm

3/7xu = 27.60mm

.•. safe.

Assumption are correct.

Xu1 max = 0.48d = 264mm

Xu max > xu

Yf = 0.15(64.39) + 65 = 74.65mm

Moment of resistance.

Mu = 0.36xu / d(1 - 0.042xu/d) fckbwd2 + 0.45fck(bf - bw)Yf(d - Yf/2)

= 0.36 X 64.39/550(1 - 0.042.64.39/550) X 20 X 300 X 550 + 0.45 X 20(1250 - 300)74.65(550 - 74.65/2)

= 400KNm

ADD COMMENTlink
written 2.1 years ago by gravatar for Juilee Juilee2.1k
Please log in to add an answer.