Structural Analysis: Flexural Response and Failure of Various Beams

This report presents a comprehensive analysis of the flexural response and failure mechanisms of different beam types, including over-reinforced and under-reinforced concrete beams, steel beams, timber beams, and laminated timber beams. The analysis involves determining the moment to cause cracking (Mcrack) and the ultimate moment (Mult) from sectional analysis for the over-reinforced concrete beam, comparing these theoretical values to experimental load vs. deflection graphs. The report also discusses failure locations, failure rates, and warning signs of impending failure for each beam type. Furthermore, the report includes calculations for maximum elastic moment (Me,max) and expected deflection for the steel beam, comparing these with experimental data. The analysis highlights the distinct behaviors of each material, such as the brittle failure of timber, the ductile behavior of steel, and the combined behavior of laminated timber. The experimental data is presented in the form of load vs. deflection graphs, providing a visual representation of the structural response of each 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=28.46GPa
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 m m2
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

Critical Force
Fcr =1
2∗3.225∗50.403∗110
Fcr =8.94 KN
Calculate ultimate moment
M ult=Mc+ M s + M cr
M ult= 12.49∗2
3 ∗59.59+5.174∗( 80−59.59 ) + 8.94∗2
3 ∗5 0.403
M ult=902 N m
• Discuss failure location, failure rate and any warning of the impending failure
Graph of load and deflection shown below , Initially the deflection increases as load increase then after
achieving highest load value , then load applied is reduced but deflection still increasing .

0.921 1.178 1.527 1.942 2.339 2.779 3.18 3.622 4.068 4.538 5.058 5.648 6.352 8.63
0
5
10
15
20
25
30
35
40
45
50
55
60
Load kN Vs Deflection
Load kN
Load in KN
As the load is tensile in nature , so concrete fail in bending if its value cross the maximum value .
Under-reinforced beam (cross section: 110 mm by 110 mm, bar diameter is 10mm) • Discuss failure
location, failure rate and any warning of the impending failure
From the graph it is clear that load carrying capacity of under reinforced beam is very less as compared
to over reinforced beam .Initially as the load increases the deflection increment is very less but after 2
KN load applied on beam then deflection value keeps on increasing linearly .Maximum load face by this
beam is 23.3 KN , after that by reducing the load , the deflection keeps on increasing .Deflection value in
this beam is high as compared to over reinforced case .

0 0.647 1.095 1.662 2.278 2.835 3.384 3.964 4.625 5.604 8.773 17.26225.674
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Load kN Vs Deflection
Load kN
Steel beam (cross section: 50 mm by 50 mm, wall thickness is 2 mm)
• Present an experimentally recorded load vs deflection graph

0 2.548 3.996 5.863 8.32 13.374 19.638 25.333 30.452 35.341 40.198 37.378
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Load kN Vs Deflection
Load kN
Load KN
From the above graph , the relation between load and deflection is observed in steel .As compare to
reinforced concrete , steel of half dimension section is used but the steel go through eastic stage and
then plastic stage before failure , due to this area under the graph (strain energy) is highest for steel and
same load range .
• From the provided yield stress (400 MPa) use a sectional analysis to determine maximum elastic
moment (Me,max) and ultimate moment (Mult)
f ys=400 MPa
y= 50
2 =25 mm
Moment of inertia I= 50∗503
12 − 46∗463
12

I =14.78∗104 mm4
Maximum elastic moment
M e= f ys∗I
y
M e= 400∗14.78∗104
25
M e=2.363 KNm
Calculate the ultimate moment
M ut= W∗L
4
M ut= 14.83∗103∗1
4
M ut=3.707 KNm
• Determine the expected deflection corresponding to Me,max. You can assume the loading
arrangement represents a simply supported beam with a single point load as the top rollers are quite
close together.
Calculation for deflection
δ= W∗L3
48∗E∗I
δ= 14.83∗103∗10003
48∗210∗1000∗14.78∗104
δ=9.95 mm
• Compare these theoretically derived values to the recorded load vs deflection graph. You will need to
convert your Me,max and Mult values into equivalent load values
From the graph the value of deflection for14.83 KN is 14 mm so there is 40 % deviation from result .
From the graph the value of load at 9.95 mm deflection is 13.7KN load , so there is 16 % deviation from
load result .

• Discuss failure location, failure rate and any warning of the impending failure
From the shear force and bending moment graph , it is clear that bending moment is maximum at
center so beam will try to bens (sagging) and break from centre .

Timber beam (cross section: 90 mm by 90 mm)
• Discuss failure location, failure rate and any warning of the impending failure
0 0.88 1.808 2.691 3.557 4.538 5.663 6.852 8.145 9.558 11.152 12.918 15.104
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
Load kN Vs Deflection
Load kN
LOAD IN KN
The graph between load and deflection is plot for timber beam and all the conclusion mentioned below .
 Initially the applied load cause very less deflection , as after increasing the load from 2 KN then
deflection also increasing linearly .
 Timber beam bear the load of around 40KN then it suddenly fails , this shows the brittle
behavior of timber. No plastic stage occurs , direct fracture occurs .
 As timber can withstand very high load but its deflection value is less than other materials , As
deflection is direction proportional to length of beam . So for timber , long beams shouldn’t be
used as it fail earlier .

Laminated timber beam (cross section: 60 mm in height, 75 mm in width)
• Discuss failure location, failure rate and any warning of the impending failure
0 0.794 2.678 4.09 5.507 6.989 8.6 10.32812.43717.68822.36126.80832.39731.373
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Load kN Vs Deflection
Load kN
The graph between load and deflection is plot for laminated timber beam and all the conclusion
mentioned below .
 Initially the applied load cause very less deflection , as after increasing the load from 2.5 KN then
deflection also increasing linearly .
 Laminated timber beam can withstand the load of around 20KN then it then fails gradually, this
shows the effective combination brittle and ductile behavior of laminated timber. Plastic stage
occurs before failure .
 As laminated timber can withstand heavy load as well as deflection range is also grater than
timber material , As deflection is direction proportional to length of beam . So for laminated
timber , long beams can be used where ever required .