Structural Engineering: Over-reinforced beam, Under-reinforced beam, Steel beam, Timber beam, Laminated timber beam

Structural Engineering

Question
Over-reinforced beam (cross section: 110 mm by 110 mm, bar diameter is 24 mm)
• Determine moment to cause cracking (Mcrack) from sectional analysis. Use both the conservative
coefficient of 0.6 as well as the realistic coefficient of 0.85 (where f'c = 40 MPa)
M crack= f cr∗I g
y
f cr =0.6∗0.85∗√ f c
'
f cr =0.6∗0.85∗√ 40
f cr =3.225 MPa
50.403

f c
' =40 MPa
f ys=550 Mpa
Es=200 GPa
Ec=4500∗√f c
'
Ec=4500∗√40
Ec=28.46 GPa
Modular ratio n = Es
Ec
n= 200
28.46 =7.027
Area of steel rod AS= π
4 ∗d2
AS=452.3 mm2
Area transform At = ( n−1 )∗AS
At = ( 7.027−1 )∗452.3
At =2726.14 mm2
Calculate y
y=
110∗110∗110
2 +2726.14∗30
110∗110+2726.14
y=50.403 mm
Calculate M.O.I
I g= 110∗1103
12
I g=12.5∗106 m m4
Calculate M crack

M crack= f cr∗I g
y
M crack= 3.225∗12.5∗106
50.403
M crack= 3.225∗12.5∗106
50.403
M crack=800 Nm
• Determine the ultimate moment (Mult) from sectional analysis (where fsy = 550 MPa, bar is 10 mm
diam.) • Compare these theoretically derived values to the recorded load vs deflection graph. You will
need to convert your Mcrack and Mult values into equivalent load value
Solution
From similar triangles property, stress in concrete
f c
3.225 = 59.59
50.403
f c=3.812 MPa
Stress in steel
f s
n
3.225 =50.403−30
50.403
f s=11.44 MPa
Forces act on steel and concrete
Fc=1
2∗3.812∗59.59∗110
Fc=12.49 KN
Force in steel
Fs=11.44∗452.3
Fs=5.174 KN