assignment 1 - rcc

7/30/2019 Assignment 1 - Rcc

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DAR ES SALAAM INSTITUTE OF TECHNOLOGY

DEPARTMENT OF BUILDING AND CIVIL ENGINEERING

ASSIGNMENT No: 1

MODULE: CEU 07317

MODULE NAME: REINFORCED CONCRETE DESIGN

CLASS: B.ENG 11

Name: Adm. No:

Richard Mwanja 110141271071

LECTURER: ENG. B.Y.B. MASANGYA

Structural Engineering Department


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ASSIGNMENT 1 - RCC

Qn. No 1

A series of 400mm deep x 250mm wide of reinforced concrete beam spaced at

5m center and spanning 7.5 m supports a 175mm thick reinforced concrete slab

as shown below. If the imposed floor loading is 3KN/m2 and the load induced bythe weight of concrete is 24KN/m3.

Calculate the total ULS loading condition for the slab and beam

Data given

Consider the diagram bellow

Density of concrete, dc = 24KN/m3

Imposed load (qk) = 3KN/m2

Required: to calculate the total ULS loading condition for the slab and beam

Solution

FOR SLAB

Dead load gk = dc x vol.of slab

=24KN/m3x (5x0.175x7.5) m3

gks=157.5KN

FOR BEAM

Dead load gkb=dcx vol.of beam

=24KN/m3 x (0.4 x 0.25x 7.5) m3


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ASSIGNMENT 1 - RCC

gkb=18KN

Then,

To convert the imposed load into point load

qk = 3KN/m2 x (7.5 x 5) m2

= 112.5KN

Then,

Total gk = gks + gkb

= 157.5KN + 18KN

= 175.5KN

DESIGN LOAD (W) = Total ULS loading condition for the slab and beam

W = 1.4gk + 1.6qk

= 1.4 x 175.5 + 1.6 x 112.5

= 425.7KN

Total ULS loading condition for the slab and beam per meter run is given by

W= 425.7KN /7.5m

= 56.76KN/m

Total ULS loading condition for the slab and beam = 56.76KN/m

Qn. No 2

A singly reinforced concrete beam is required to resist an ultimate moment of550KNm.If the beam is composed of grade 30 and high yield (HY) reinforcement

.Check the size and determine the area of steel required

Data given

Moment M = 550KNm


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ASSIGNMENT 1 - RCC

Concrete grade, fcu = 30 N /mm2

Reinforcement typehigh yield reinforcement

Required: to check the section size

: To determine area of steel, As

Solution

1st- To check for reinforcement type

From, K = M

bd2fcu

Then, K = 550 x 106

350 x 6002 x 30

K = 0.1455

K = 0.1455 < 0.156 (singly reinforcement)

From,

Z = d [0.5 + (0.25k / 0.9)0.5]

= 600[0.5 + (0.250.1455 / 0.9)0.5]

Z = 478.3 mm

2ndTo determine area of steel (As)

From, As = M0.87 fy z

As = 550 x 106

0.87 x 460 x 478.3

As = 2873mm2

From bar table, provided 6T25

3rdTo check the section

From, d = hc/2

h = d + c + /2


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ASSIGNMENT 1 - RCC

h = 600 + 25 + 25/2

h = 638 mm

min allowableAs =0.13%b*h

max allowable As=4.0%b*h

since,actual %As =2873*100 /(350*638) =1.29%

0.13%bh < %As < 4.0%bh, then the section is ok.

Qn. No 3

A reinforced concrete beam is required to transmit an ultimate bending

moment of 140KNm. Inclusive its own weight using the simplified stress block

formula, determine amount of steel needed in a 250mm wide beam for the

following condition.

a) Grade 30 concrete with mild steel reinforcement

b) Grade 35 concrete with high yield reinforced

Solution

Data given

Moment, M = 140KNm

Case I- Concrete grade, fcu = 30

Assumption, Kmax = 0.156


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ASSIGNMENT 1 - RCC

From, K = M

bd2fcu

0.156 = 550 x 106

250 x d2 x 30

d = 346 mm

To determine the area of steel (As)

From,

Z = d [0.50 + (0.25k / 0.9)0.5]

But Kmax = 0.156

Then Z = 0.775d

As = M / 0.87 fy 0.775d

= 140 x 106 / 0.87x250x0.775x346

As = 2464 mm2

From bar table, provided 8R20

To check for section

From, d = hc/2

h = d + c + /2

= 346 + 25 + 20/2

h = 381 mm



actual %As =2464*100 /(250*381) =2.58%

Therefore, since 0.13%bh < %As < 4.0%bh, then the section is ok.


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ASSIGNMENT 1 - RCC

Case II Concrete grade fcu = 35

fy = 460 N/ mm2

From,

K = Mbd2fcu

0.156 = 140 x 106

250 x d2 x 35

d = 346 mm

To determine the area of steel (As)

From, As = M / 0.87 fy 0.775d

= 140 x 106 / 0.87 x 460 x 0.775 x 346

As = 1305 mm2

From bar table, provided 3T25

To check for section

From, d = hc/2

h = d + c + /2

= 346 + 25 + 25 / 2

h = 383.5 mm,(say h=384 mm)



since,actual %As =1305*100 /(250*384) =1.36%

0.13%bh < %As < 4.0%bh, then the section is ok.

assignment 1 - rcc

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