Structural Engineering: Beam Failure Analysis and Discussion

This report provides a comprehensive analysis of the flexural response of various beams, including over-reinforced and under-reinforced concrete beams, steel beams, timber beams, and laminated timber beams. It details the calculations for determining the moment to cause cracking (Mcrack) and the ultimate moment (Mult) for an over-reinforced beam, comparing theoretical values to recorded load vs deflection graphs. The report also discusses failure locations, failure rates, and warning signs of impending failure for each beam type. Graphs of load versus deflection are presented for each specimen, highlighting key characteristics such as the elastic and plastic stages in steel, and the brittle behavior of timber. The analysis includes calculations for maximum elastic moment and expected deflection, with comparisons to experimental results. The report concludes with discussions on the flexural response of each beam, paying particular attention to yielding, cracking, and crushing during failure.

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)
𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 𝑓𝑓𝑐𝑐𝑐𝑐∗ 𝐼𝐼𝑔𝑔
𝑦𝑦
𝑓𝑓𝑐𝑐𝑐𝑐= 0.6 ∗ 0.85 ∗�𝑓𝑓𝑐𝑐
′
𝑓𝑓𝑐𝑐𝑐𝑐= 0.6 ∗ 0.85 ∗√40
𝑓𝑓𝑐𝑐𝑐𝑐= 3.225 𝑀𝑀𝑀𝑀𝑐𝑐
50.403

𝑓𝑓𝑐𝑐
′ = 40 𝑀𝑀𝑀𝑀𝑐𝑐
𝑓𝑓𝑦𝑦𝑦𝑦= 550 𝑀𝑀𝑀𝑀𝑐𝑐
𝐸𝐸𝑦𝑦= 200 𝐺𝐺𝑀𝑀𝑐𝑐
𝐸𝐸𝑐𝑐= 4500 ∗�𝑓𝑓𝑐𝑐
′
𝐸𝐸𝑐𝑐= 4500 ∗√40
𝐸𝐸𝑐𝑐= 28.46 𝐺𝐺𝑀𝑀𝑐𝑐
Modular ratio n = 𝐸𝐸𝑦𝑦
𝐸𝐸𝑐𝑐
𝑛𝑛 =200
28.46= 7.027
Area of steel rod 𝐴𝐴𝑆𝑆= 𝜋𝜋
4 ∗ 𝑑𝑑2
𝐴𝐴𝑆𝑆= 452.3 𝑚𝑚𝑚𝑚2
Area transform 𝐴𝐴𝑡𝑡= (𝑛𝑛 − 1) ∗ 𝐴𝐴𝑆𝑆
𝐴𝐴𝑡𝑡= (7.027 − 1) ∗ 452.3
𝐴𝐴𝑡𝑡= 2726.14 𝑚𝑚𝑚𝑚2
Calculate y
𝑦𝑦 =
110∗110∗
110
2 +2726.14∗30
110∗110+2726.14
𝑦𝑦 = 50.403 𝑚𝑚𝑚𝑚
Calculate M.O.I
𝐼𝐼𝑔𝑔= 110 ∗ 1103
12
𝐼𝐼𝑔𝑔= 12.5 ∗ 106 𝑚𝑚𝑚𝑚4
Calculate 𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 𝑓𝑓𝑐𝑐𝑐𝑐∗ 𝐼𝐼𝑔𝑔
𝑦𝑦
𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 3.225 ∗ 12.5 ∗ 106
50.403

𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 3.225 ∗ 12.5 ∗ 106
50.403
𝑴𝑴𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄= 𝟖𝟖𝟖𝟖𝟖𝟖 𝑵𝑵𝑵𝑵
• 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
𝑓𝑓𝑐𝑐
3.225= 59.59
50.403
𝑓𝑓𝑐𝑐= 3.812 𝑀𝑀𝑀𝑀𝑐𝑐
Stress in steel
𝑓𝑓𝑦𝑦
𝑛𝑛
3.225= 50.403 − 30
50.403
𝑓𝑓𝑦𝑦= 11.44 𝑀𝑀𝑀𝑀𝑐𝑐
Forces act on steel and concrete
𝐹𝐹𝑐𝑐= 1
2 ∗ 3.812 ∗ 59.59 ∗ 110
𝑭𝑭𝒄𝒄= 𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒 𝑲𝑲𝑵𝑵
Force in steel
𝐹𝐹𝑦𝑦= 11.44 ∗ 452.3
𝑭𝑭𝒔𝒔= 𝟓𝟓. 𝟏𝟏𝟏𝟏𝟒𝟒𝑲𝑲𝑵𝑵
Critical Force
𝐹𝐹𝑐𝑐𝑐𝑐= 1
2 ∗ 3.225 ∗ 50.403 ∗ 110
𝑭𝑭𝒄𝒄𝒄𝒄= 𝟖𝟖. 𝟒𝟒𝟒𝟒 𝑲𝑲𝑵𝑵

Calculate ultimate moment
𝑀𝑀𝑢𝑢𝑢𝑢𝑡𝑡= 𝑀𝑀𝑐𝑐+ 𝑀𝑀𝑦𝑦+𝑀𝑀𝑐𝑐𝑐𝑐
𝑀𝑀𝑢𝑢𝑢𝑢𝑡𝑡= 12.49 ∗
2
3 ∗ 59.59 + 5.174 ∗(80 − 59.59) + 8.94 ∗
2
3 ∗ 50.403
𝑴𝑴𝒖𝒖𝒖𝒖𝒖𝒖= 𝟒𝟒𝟖𝟖𝟏𝟏 𝑵𝑵𝑵𝑵
• 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 .
As the load is tensile in nature , so concrete fail in bending if its value cross the maximum value .
0
5
10
15
20
25
30
35
40
45
50
55
60
0
1.025
1.171
1.305
1.511
1.721
1.931
2.123
2.324
2.545
2.766
2.965
3.168
3.395
3.606
3.826
4.057
4.3
4.53
4.775
5.043
5.322
5.629
5.953
6.335
6.975
8.471
15.978
Load in KN
Load kN Vs Deflection
Load kN

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
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
0
0.056
0.485
0.761
0.978
1.16
1.331
1.708
1.999
2.23
2.469
2.713
2.939
3.166
3.413
3.682
3.924
4.183
4.48
4.793
5.169
5.727
6.666
8.428
11.633
15.599
19.515
23.091
26.512
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
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)
𝑓𝑓𝑦𝑦𝑦𝑦= 400 𝑀𝑀𝑀𝑀𝑐𝑐
𝑦𝑦 =
50
2 = 25 𝑚𝑚𝑚𝑚
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0
1.026
1.927
2.753
3.56
4.322
5.164
6.025
7.012
8.219
9.861
12.347
15.128
17.95
20.589
23.081
25.509
27.796
30.034
32.207
34.405
36.487
38.623
40.868
40.679
37.378
Load KN
Load kN Vs Deflection
Load kN

Moment of inertia 𝐼𝐼 =
50∗503
12 − 46∗463
12
𝐼𝐼 = 14.78 ∗ 104𝑚𝑚𝑚𝑚4
Maximum elastic moment
𝑀𝑀𝑒𝑒= 𝑓𝑓𝑦𝑦𝑦𝑦∗ 𝐼𝐼
𝑦𝑦
𝑀𝑀𝑒𝑒= 400 ∗ 14.78 ∗ 104
25
𝑀𝑀𝑒𝑒= 2.363 𝐾𝐾𝐾𝐾𝑚𝑚
Calculate the ultimate moment
𝑀𝑀𝑢𝑢𝑡𝑡= 𝑊𝑊 ∗
𝐿𝐿
4
𝑀𝑀𝑢𝑢𝑡𝑡= 14.83 ∗ 103 ∗1
4
𝑀𝑀𝑢𝑢𝑡𝑡= 3.707 𝐾𝐾𝐾𝐾𝑚𝑚
• 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
𝛿𝛿 =
𝑊𝑊 ∗ 𝐿𝐿3
48 ∗ 𝐸𝐸 ∗ 𝐼𝐼
𝛿𝛿 = 14.83 ∗ 103 ∗ 10003
48 ∗ 210 ∗ 1000 ∗ 14.78 ∗ 104
𝜹𝜹 = 𝟒𝟒. 𝟒𝟒𝟓𝟓 𝑵𝑵𝑵𝑵
• 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
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 .
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
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
0
0.175
0.767
1.137
1.541
1.922
2.298
2.691
3.062
3.433
3.809
4.24
4.695
5.172
5.663
6.17
6.68
7.217
7.761
8.352
8.958
9.558
10.221
10.919
11.65
12.372
13.203
14.081
15.104
16.685
LOAD IN KN
Load kN Vs Deflection
Load kN

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
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 .
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0
0.311
0.642
1.995
2.529
3.211
3.896
4.58
5.236
5.92
6.627
7.348
8.132
8.912
9.754
10.592
11.435
12.95
15.53
17.965
20.3
22.361
24.428
26.616
29.023
32.099
31.752
31.433
Load kN Vs Deflection
Load kN