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Solution:: 8m 3m A B

1) The beam has a rectangular cross-section and is reinforced with compression and tension bars. It is supported at two points. 2) The ultimate moment capacity of the beam is 402.2791 kN-m. 3) The maximum uniform service live load the beam can carry is 39.5501 kN/m.

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Mark B. Barroga
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562 views4 pages

Solution:: 8m 3m A B

1) The beam has a rectangular cross-section and is reinforced with compression and tension bars. It is supported at two points. 2) The ultimate moment capacity of the beam is 402.2791 kN-m. 3) The maximum uniform service live load the beam can carry is 39.5501 kN/m.

Uploaded by

Mark B. Barroga
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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2.

A reinforced concrete beam having a width of 310 mm and an overall depth of 550mm is reinforced with 2-25 mm
compression bars and 8-25 mm tension bars both having fy = 415 MPa. Concrete protective covering is 50 mm from the
centroid of the bar group. Unit weight of concrete is 23.5 kN/m3. Concrete strength fc’ = 21 MPa. The beams is supported
as shown. L1 = 8m and L2 = 3m. Other than the weight of the beam, the beam will support a uniform live load. 

Calculate the following: 


a)  the balance tension steel area. 
b)  the ultimate moment capacity of the beam. 
c)  maximum uniform service liveload  that the beam can carry. 

SOLUTION :

8m 3m

A B

Fc’ = 21Mpa, fy = 415MPa, γ = 23.5 Kn/m

Continuation:
ρ=
As
=
8 ( π4 ( 25) )
2

bd 310(500)
ρ=0.02534
0.85 ( 21 )( 0.85 ) 120
ρ bal=
415 203 ( )
ρbal=0.021612
ρbal < ρ therefore doubly reinforced concrete beam
C 1+C 2=T

0.85 ( 21 ) a ( 310 )+ 2 ( π4 ( 25 ) ) (415−0.85 ( 21) )=8 ( π4 ( 25 ) )(415)


2 2

a=289.1748
c=281.3821
600
fs= ( 500−281.3821 )=466.16589>fy
281.3821
600
fs= ( 281.3821−50 )=493.3834> fy
281.3821
0.003
ε t= ( 500−281.3821) =0.0023308
281.3821
250 29
∅= ε + =0.6776
3 t 60

239.1748
M 1n=0.85 ( 21 ) ( 239.1748 )( 310 ) 500− ( 2 )( 10001 )
2

M 1n=503.4661 Kn−m

M 2n=2 ( π4 ( 25) )( 415−0.85 (21 )) (218.3821−50)( 10001 )


2
2

M 2n=90.216 Kn−m

M n=503.4461+90.2161 M n=593.682 2

Mu=(0.6776)(593.6822)
Mu=402.2791 Kn−m

b.)

As bal=ρ bal bd
¿ 0.021612(310)(500)

¿ 3349.86 mm2

W W
R 1= ( L 12−L 22) R 2= ( L 12+ L22 )
2L 2L

V1

V3

V2

a2 WL 22
L 1(1− ) M 2=
L12 2

402.2791=¿
W ¿ =39.5501 KN −m

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