1 1−− 22µ µ d
d Only one live load
Only one live load (combination factor is not needed)(combination factor is not needed)
q
qdd == 1,35 1,35 ⋅⋅K K FIFI⋅⋅gg++ 1,5 1,5⋅⋅K K FIFI⋅⋅qq≈≈ 38,7 38,7 kN/m kN/m . . (1)(1)
Max bending moment in the middle Max bending moment in the middle
M Mdd== qq ⋅⋅ L L 2 2 ≈ ≈ 8 8 174,2 174,2 kNm kNm . . (2)(2)
The effective height
The effective height d d is is approximaapproximated to be ted to be asasdd ==hh−− 50mm 50mm ≈≈ 430 430 mm mm ..
Relative bending moment is then: Relative bending moment is then:
µ
µ
=
=
22
≈≈ 0,178 0,178 (dimensionless (dimensionless quantity)quantity)
(3) (3)
Mechanical reinforcement ratio is determined from
Mechanical reinforcement ratio is determined from formula:formula:
ω
ω == 11−− ≈≈ 0,198 0,198 == β β . . (4)(4)
Required area of the steel Required area of the steel
((
5) 5)
Select reinforcing bars: for example 2
Select reinforcing bars: for example 2φ φ 2020 ++ 2 2φ φ 16 , when the area of steel As is 1030 mm16 , when the area of steel As is 1030 mm22..
d d
=
=
((1−
1−
22))=
=
990 mm990 mm2 2Beam Design Example 2
Beam Design Example 2
Moment capacity of a given cross-section (byggkonstruktion)
Moment capacity of a given cross-section (byggkonstruktion)
Determinate the load carrying capacity in ultimate limit state of tDeterminate the load carrying capacity in ultimate limit state of t he concrete beam with cross-he concrete beam with cross-section values are shown in figure. Concrete is
section values are shown in figure. Concrete is C 25, Steelbars: 6C 25, Steelbars: 6 f f 16, B500B. Consequence class16, B500B. Consequence class CC2.
CC2.
Amount of steel: As = 6
Amount of steel: As = 6 ππ 8 822 = 1206 mm= 1206 mm22 is the beam normally reinforced? Calculate is the beam normally reinforced? Calculate ρ ;ρ ;
6,89 10 6,89 10-3-3
Balanced geometric reinforcement is Balanced geometric reinforcement is
ρ
ρ
bal bal=
= 0
0,,8
8
+
+
=
=
18,9 1018,9 10−3−3consequently it is normal reinforced because consequently it is normal reinforced because
ρ ρ < <
ρ
ρ
bal bal OKOK 500 500 350 350
=
=
=
=
Initial values Initial values concrete: fconcrete: f ckck= 25 MPa,= 25 MPa, γγcc= = 1,50 1,50 =>=>
=
= 1
16,
6,7
7
steel f
steel f ykyk= 500 MPa= 500 MPa γγcc= = 1,15 1,15 =>=>
=
= 4
43
35
5
Es = 200 GPa Es = 200 GPa Ultim.strain concrete
Beam Design Example 3.1
Beam Design Example 3.1
Moment capacity of a given cross-section (byggkonstruktion)
Moment capacity of a given cross-section (byggkonstruktion)
The beam with cross-section in the figure isThe beam with cross-section in the figure is constructeconstructed of d of cncreete C 30 and steelbars B500B iscncreete C 30 and steelbars B500B is used. CC2.
used. CC2. a)
a) Determinate the moment carrying capacity at ultimate limitDeterminate the moment carrying capacity at ultimate limit state if t
state if the cross-section is balanced reinforced.he cross-section is balanced reinforced. b)
b) Determinate Determinate reinforcemereinforcement required nt required to resist external desigto resist external designn bending mom
bending moment Ment Med ed = 265 kNm = 265 kNm
Balanced N.A. can be determined from Balanced N.A. can be determined from
= 308 mm = 308 mm ed from either
ed from either moment equilibrium equationsmoment equilibrium equations
Moment capacity of balanced reinforcement can be calculated eiher moment equilibrium equations: Moment capacity of balanced reinforcement can be calculated eiher moment equilibrium equations:
M = F
M = F
ss(d-0,4x) or
(d-0,4x) or
M = F
M = F
cc(d-0,4x) = f
(d-0,4x) = f
cd cd0,8xb (d-0,4x)
0,8xb (d-0,4x)
by using latter m
by using latter moment balanceoment balance 2400 x -1920 x
2400 x -1920 x22 = M = 265 kNm = M = 265 kNm
solving second order equation for x;
solving second order equation for x; the position of N.A. comesthe position of N.A. comes x = 122 mm
x = 122 mm
Checking the strain in reinforcement, if it
Checking the strain in reinforcement, if it has yielded:has yielded:
500 500 300 300
A
A
ss Initial values Initial values concrete: fconcrete: f ckck= 30 MPa,= 30 MPa, γγcc= = 1,50 1,50 =>=>
=
= 2
20
0
steel f
steel f ykyk= 500 MPa= 500 MPa γγcc= = 1,15 1,15 =>=>
=
= 4
43
35
5
Es = 200 GPa Es = 200 GPa Ultim.strain in concrete
Ultim.strain in concrete εεcucu== εεcu3cu3=3,5=3,5 oooo//oo
300 300 As As
+
+
Beam Design Example 3.2
Beam Design Example 3.2
Moment capacity alternative solutions (byggkonstruktion)
Moment capacity alternative solutions (byggkonstruktion)
a)
a) Mechanic reinforcement ratio can be calculated from:Mechanic reinforcement ratio can be calculated from:
Moment capacity of the balanced cross-section ca
Moment capacity of the balanced cross-section can be solved using equation:n be solved using equation:
b)
b) Required reinforcement when MRequired reinforcement when MEd Ed = 265 kNm. = 265 kNm.
µ =
µ =
geometric reinforcement ratio geometric reinforcement ratio
= 0,196 <
= 0,196 <
ω
ω
bal bal check check OK OKRequired steel amount: Required steel amount: