Singly-Reinforced Concrete Beams Singly-Reinforced ...

Singly-Reinforced Concrete Beams Uncracked State b

εc

fc C

d (n-1)Ast εs transformed area

cross section

fs

strain

Tc Ts force

stress block

C=Ts+Tc

Singly-Reinforced Concrete Beams In the elastic analysis of RC beams, the concrete in the tensile zone does not resist any tensile force, ie., that it is cracked completely to the neutral axis (NA) b

εc

fc C

kd d

jd nAst εs cross section

transformed section

strain

T fs/n linear stress block

force

Stress in the imaginary concrete equivalent of the steel area. Note fs=nfc

3

Singly-Reinforced Concrete Beams 1. Determine the neutral axis position, concrete stresses at the top and bottom fibers and steel stress produced by a moment of 10kNm for the uncracked state

250

2. Find the cracking moment 500

3. Determine the neutral axis position, concrete stresses at the top and bottom fibers and steel stress produced by a moment of 60kNm for the cracked state

4N24

50

f´c= 32MPa

Singly-Reinforced Concrete Beams 1. Uncracked State (a) Replace tensile steel by concrete area (n-1)Ast (b) Find dg (position of NA of equivalent concrete section) (c) Find moment of inertia It about NA of equivalent concrete section f´c= 32MPa Ec= ρ 1.5 0.043 f c′ = 28602MPa (cl6.1.2 p.38) Es= 200000 MPa n = Es/Ec~ 7 Ast= 1810mm2 (4N24) (n-1) Ast = 10860 mm2

500

4N24

dg NA

(n-1) Ast

4

Singly-Reinforced Concrete Beams 1. Uncracked State Ag = 250x550 = 137500 mm2 Total area = Ag + (n-1)Ast = 137500+10860 = 148360mm2 dg=291 Find dg 148360dg = (137500x275)+(10860x500) dg = 291mm

550

NA

209

I for concrete

Stress σ = My/I IT =

500

I for steel

250 × 5503 + 137500 × (291 − 275) 2 + 10860 × 209 2 12

=3.98x109 mm4

Singly-Reinforced Concrete Beams 1. Uncracked State Concrete stress at the top

10 ×10 6 × 291 = 0.73MPa 3.98 × 109

σct =

291 209

500

259

Concrete stress at the bottom

10 ×10 6 × 259 = 0.65MPa 3.98 ×109

σcb = Steel stress σs =

n

7 ×10 ×10 6 × 209 = 3.68MPa 3.98 ×109

5

Singly-Reinforced Concrete Beams 2. Cracking Moment (a) Find tensile strength of concrete (b) Cracking will occur when concrete stress reaches its tensile strength f´c= 32MPa f´t= 0 . 6 f c′ = 3.39MPa (cl 6.1.1.2 p.38) Stress σ = My/I If σ= f´t, then M=Mcr Mcr = f´t IT/y 9 Mcr = 3.39 × 3.98 ×10 259 ×10 6

= 52.1 kNm

Singly-Reinforced Concrete Beams 250

3. Cracked State Note: concrete below the NA is cracked and does not contribute to the beam strength

500

4N24

nAst= 7*1810 = 12670 mm2 2

(250d cr + 12670)d cr = 250d cr + 12670 * 500 dcr2 + 101dcr - 50680 = 0

dcr

2

2 dcr= − 101 + 101 + 4 × 50680 = 180 mm 2

NA

500

n Ast

6

Singly-Reinforced Concrete Beams 250

3. Cracked State 3 Icr = 250 × 180 + 250 ×180 × 90 2 + 12670 × 320 2 12

= 1.78 mm4

σct =

σs =

180

NA

320

60 ×10 6 ×180 = 6.07MPa 1.78 × 109

nAst =12670

7 × 60 × 106 × 320 = 75.5MPa 1.78 ×109

Doubly-Reinforced Concrete Beam Similar to singly-reinforced concrete beam b

d

dsc

εc

nAsc kd

εsc

Asc Ast

transformed section

T=Cs+Cc Cs=Ascfsc

fsc/n

Cs Cc

nAst εs

cross section

fc

Cc=b kd fc/2

strain

T fs/n linear stress block

force

T=Astfs

7