Plastic Behaviour of Beams and Frames CIVL454 Structures 2 Laboratory Report
Tom Wilkinson 4264009
Table of Contents 1.
Introduction..................................................................................................... 2
2.
Test 1.1 – Tension Coupon Test........................................................................2 2.1.
3.
4.
Discussion................................................................................................. 3
Test 1.2 – Two-Span Continuous Beam............................................................3 3.1.
Aim............................................................................................................ 3
3.2.
Laboratory Test Set-Up..............................................................................3
3.3.
Test Results............................................................................................... 5
3.4.
Analytical Results...................................................................................... 6
3.5.
Comparisons between observations and theoretical predictions..............8
Test 1.3 – Rigid Portal Frame............................................................................9 4.1.
Aim............................................................................................................ 9
4.2.
Laboratory Test Set-Up..............................................................................9
4.3.
Test Results............................................................................................. 11
4.4.
Analytical Results.................................................................................... 13
4.5.
Discussion............................................................................................... 17
5.
Conclusion..................................................................................................... 17
6.
References..................................................................................................... 17
1. Introduction This purpose of this report is to examine the Plastic Behaviour of continuous steel beams and rigid portal steel frames. Referring to the analysis of a structures behaviour after the point at which a plastic hinge has formed, plastic analysis looks at steel that has reached its initial yield stress and continuous to deform plastically. Experimental analysis will be undertaken by using various load ratios on a continuous beam and portal frame respectively. By comparing the results of each load ratio as well as with respect to theoretical predictions, a deeper understanding of the plastic behaviour of structure will be gained.
2. Test 1.1 – Tension Coupon Test In order to determine the yield stress y and plastic moment capacity Mp of the metal rod used in Test 1.2 and Test 1.3, a simple tension coupon test was carried out. This involves loading a portion of the rod in tension with a low loading rate and was conducted prior to the laboratory by a lab staff member. The rod tested was found to have a yield stress of 245MPa and a diameter of 3.2mm. The following calculations outline how to determine the corresponding yield load P y and plastic moment capacity Mp from this experimental information. Yield Load
σ=
F A
F=σA Where: F
=
Force (N)
A
=
Area (mm2)
=
Stress (N/mm2)
σ
In this case the force on the rod is the tensile force equal to the yield load Py, the area is the cross-sectional area of the rod (circle) and the stress is the yield stress y. 2 ˚ ¿=π r A¿
2
3.2 ˚ ¿=π =8.04 mm2 2 A¿
( )
˚ ¿=245 MPa × 8.04 mm2=1969.8 N=1.97 kN P y =❑ y A¿
Plastic Moment Capacity
M p=σ y × S Where: S
=
Plastic section modulus (mm3)
The plastic section modulus for a circle can be determined using the equation: 3
S=
d 6
S=
3.2 =5.46 mm3 6
3
3
M p=245 MPa ×5 .46 mm =1337.7 Nmm=1.34 Nm
2.1.
Discussion
This value for plastic moment capacity of the steel beam represents the point at which the material reaches a fully plastic state and therefore is the maximum bending moment it can resist. When the number of plastic hinges in the structure is one greater than its redundancy, in general the structure will be observed to collapse. This is demonstrated further in tests 1.2 and 1.3.
3. Test 1.2 – Two-Span Continuous Beam 3.1.
Aim
The aim of this test is to determine the plastic collapse load of the metal road examined in Test 1.1, in a two-span continuous beam setup. Observations of the
plastic collapse mechanism that develops under the rods failure is also to be recorded.
3.2.
Laboratory Test Set-Up
The two-span continuous beam experimental arrangement is seen below in Figure 3-1. Each span measures 300mm in length and supports a point load in the middle of its span. These point loads are developed by hanging a spreader bar from the two mid-point locations and then hanging a bucket from the bar which was incrementally filled with pellets until plastic collapse occurred.
Figure 3-1: Test 1.2 Experimental Set-up
The experiment was carried out by different groups, only making adjustments to the ratio of loading on the continuous beam by means of adjusting the position of the bucket hanging from the spreader bar. Figure 3-2 shows the loading components of the experimental setup, including the spreader bar, bucket, pellets and the hangers. The reaction forces on the Hanger at A and Hanger at C are equal in magnitude to the mid-span loadings on the continuous beam and thus are used in determining the buckets positing for the different loading ratios.
Figure 3-2: Test 1.2 Loading Arrangements
*The mass of the hangers and spreader bar were also weighed experimentally to a have a contribution to the collapse load and thus are added to this force. In reality the resultant weight of the spreader bar would be central however for ease of calculating the ratio of loads is assumed to be at the location of the bucket.
The ratio of loads on the continuous beam for Group 9 was equal to 2.5. The following calculations are used to determine distances a and b as per this ratio:
As the load ratio is
2.5=
L
=
300mm
L
=
a+b
a+b =
300mm
a
(300 – b)mm
=
Rc R A , therefore
RC =2.5 × R A .
∑ M B =0 -RAa + Rcb Substituting
=
0
RC =2.5 × R A . -RAa + 2.5RAb
=
0
b
=
0.4a
b
=
0.4 (300 – b)
b
=
120 – 0.4b
b
=
85.71mm = 86 mm
a
=
300 – 86 = 214 mm
As such, the bucket was positioned so that it was 214 mm from the left hanger and 86 mm from the right hanger on the spreader bar. These distances would vary for each group requiring an alternate load ratio however the procedure would be the exact same.
3.3.
Test Results
The bucket was slowly loaded until a point that the beam section E-F collapsed with the formation of a plastic hinge at the mid-span and support at E. Table 3-1 below shows the experimental test results for groups 7-12 conducting Test 1.2 with various load ratios. Table 3-1: Test 1.2 Results
Group Load Ratio Distance from bucket to hanger (mm) Total Load (g)
7 1.5 120
8 2.0 90
9 2.5 86
10 3.0 75
11 3.5 67
12 4.0 60
4711
4100
3557
3820
3806
3373
Group 9 calculations for collapse loads Utilising the total load information from Table 3-1 the total force on the continuous beam can be calculated and further the beams collapse loads. W
=
Load (kg) x 9.81 m/s2
=
3.557 x 9.81
=
34.9N
¯ ∑ F y =0(spreader FBD)
L1 + L 2 - w
=
L1 + 2.5 L1 - w 3.5 L1 = L1 at point A) L2
0 =
0
34.9
=
9.97N
=
9.97 x 2.5
=
24.9N
(Collapse Load
(Collapse Load
at point B) Extrapolating this method across the remaining groups load information, the calculated results are shown below in Table 3-2. Note that the collapse load of
the beam is taken to be the Point B collapse load whereas the load calculated at point A is merely the load carried here when collapse occurred at point B. Table 3-2: Test 1.2 Collapse Loads
Grou p
Load Ratio
Load (kg)
Force (N)
7 8 9 10 11 12
1.5 2.0 2.5 3.0 3.5 4.0
4.711 4.100 3.557 3.820 3.806 3.373
46.2 40.2 34.9 37.5 73.3 33.1
3.4.
Point A @ Collapse load (N) 18.49 13.41 9.97 9.37 8.30 6.62
Point B Collapse load (N) 27.73 26.81 24.92 28.11 29.04 26.47
Analytical Results
In order to make a theoretical prediction of the collapse load of the continuous beam scenario outlined above, the principle of virtual work will be used. As two plastic hinges formed during the experiment, at supports E and F, this component of the continuous beam will be analysed as shown in Figure 3-1.
Figure 3-3:Test 1.2 Virtual Work Diagram
Virtual Work Method
The virtual work method of analysis is simple method that can be used in the plastic analysis of beams. The fundamental principle is that the work done internally by the displacement of the loads must balance with the internal work absorbed by rotation under fully plastic moments at plastic hinges. This can be expressed by the equation:
∑ (W ×δ )=∑ ( M p ×θ) Where:
δ
= the vertical displacement of the load
Mp = plastic moment capacity
θ
= the rotation of the hinge
Substituting experimental specific variables:
L1 δ =2 M p θ+ M p ∅
Figure 3-4: Angles for virtual work method of Test 1.2
From Figure 3-4, it can be seen that by symmetry = and thus:
Lc δ =2 M p θ+ M p θ=3 M p θ Additionally as the tan of a small angles is considered to equal to the angle itself:
θ=δ/ 0.15
δ=0.15 θ Substituting this into the previous equation gives:
0.15 Lc θ=3 M p θ Lc =20 M p Substituting the value for Mp calculated for Test 1.1:
Lc =20 ×1.34=26.8 N ∴Collapse Load for beam=26.8 N
3.5. Comparisons between observations and theoretical predictions Figure 3-5 shows the collapse load values determined by each group against the theoretical prediction for the continuous beams collapse load. From this graph we can see clearly that the experimental collapse loads calculated by each groups various arrangement are relatively close to the predicted collapse load for the beam (maximum 7% difference). It should be noted however that the experimental results found collapse loads both higher and lower than that predicted for the beam. There are many reasons that could result in such differences between the experimental and theoretical result, the key factors being:
Error in measurement of load positon e.g. moving the bucket closer than required to the right hand side for a given ratio would result in an actual higher load ratio, causing it to fail at a lower total load than expected. Inability to correctly detect when failure is occurring and thus continuing to load the beam once failure has begun resulting an overly large value for the load reading. The moment contribution of the bars weight is not considered in the loading ratio causing an incorrect actual loading ratio effecting collapse load calculations.
Collapse Load - Experimental Vs Theoretical 30.00 28.00 Collapse Load (N) 26.00 24.00
Experimental Collapse Load Theoretical Collapse Load
22.00 7 8 9 101112 Group Number
Figure 3-5: Collapse Load - Experimental Vs Theoretical
4. Test 1.3 – Rigid Portal Frame 4.1.
Aim
The aim of this test is to determine the plastic collapse load of a rigid portal frame made of the steel rod examined in Test 1.1. Observations of the plastic collapse mechanism that develops under the frames failure is also to be recorded.
4.2.
Laboratory Test Set-Up
The rigid portal frame experimental arrangement is seen below in Figure 4-6. Each member of the frame measures 300mm in length and the top horizontal member supports a vertical point load in the middle of its span while a horizontal load is also applied at the top right hand corner of the frame. These point loads are developed by hanging a spreader bar from the mid-point span of the vertical point load and the horizontal point load translated into a vertical force by a pulley system. Then a hanging a bucket was attached to the bar which was incrementally filled with pellets until plastic collapse occurred.
Figure 4-6: Test 1.3 Experimental Set-up
The experiment was carried out by different groups, only making adjustments to the ratio of loading (Vertical/Horizontal) on the rigid portal frame by means of adjusting the position of the bucket hanging from the spreader bar. Figure 4-7 shows the loading components of the experimental setup, including the spreader bar, bucket, pellets and hangers. The reaction forces on the Hanger at A and Hanger at C are equal in magnitude to the mid-span loading and the horizontal
loading respectively and thus are used in determining the buckets positing for the different loading ratios.
Figure 4-7: Test 1.3 Loading Arrangements
*The mass of the hangers and spreader bar were also weighed experimentally to a have a contribution to the collapse load and thus are added to this force. In reality the resultant weight of the spreader bar would be central however for ease of calculating the ratio of loads is assumed to be at the location of the bucket. The ratio of vertical to horizontal load on the rigid portal frame for Group 9 was equal to 2.5. The following calculations are used to determine distances a and b as per this ratio: L
=
490mm
L
=
a+b
a+b =
90mm
a
(490 – b)mm
=
2.5=
As the vertical load to horizontal load ratio is
V RA = H RC
R A =2.5 × RC .
∑ M B =0 -RAa + Rcb Substituting
=
0
RC =2.5 × R A . -2.5Rca + Rcb b
=
= 2.5a
0
, therefore
b
=
2.5 (490 – b)
b
=
1225 – 2.5b
b
=
350mm
a
=
490 – 350 = 140 mm
As such, the bucket was positioned so that it was 140 mm from the left hanger and 350 mm from the right hanger on the spreader bar. These distances would vary for each group requiring an alternate load ratio however the procedure would be the exact same.
4.3.
Test Results
The bucket was slowly loaded until a point that the portal frame was observed to fail with the formation of plastic hinges. Table 3 1 below shows the experimental test results for groups 7-12 conducting Test 1.23 with various vertical to horizontal load ratios.
Table 4.1: Test 1.3 Loading Arrangements
Group V/H Load (g) Distance between hangers (mm) Distance between bucket to hanger A (mm)
7 1.5 3499 460
8 2.0 3400 495
9 2.5 3776 490
10 3.0 3687 563
11 3.5 3818 480
12 4.0 3592 530
184
165
140
125
106
106
Group 9 calculations for collapse loads Utilising the total load information from Table 4.1 the total forces on the portal frame can be calculated
W
=
Load (kg) x 9.81 m/s2
=
3.776 x 9.81
=
37.0N
¯ ∑ F y =0(spreadre FBD)
V+H-w
=
0
2.5H + H - w =
0
3.5 H = H = Collapse Load) V
=
= Collapse Load)
37.0 10.6 N
(Horizontal
10.6 x 2.5 26.5 N
(Vertical
Table 4.2: Test 1.3 Loading Arrangements
Group 7 8 9 10 11 12
4.4.
Load Ratio 1.5 2.0 2.5 3.0 3.5 4.0
Load (kg) 3.499 3.400 3.776 3.687 3.818 3.592
Force (N) 34.33 33.35 37.04 36.17 37.45 35.24
H(N) 13.73 11.12 10.58 9.04 8.32 7.05
V(N) 20.60 22.24 26.46 28.13 29.13 28.19
Analytical Results
In order to make theoretical predictions of the collapse load of the continuous beam scenario outlined above, the principle of virtual work will be used, combining the beam collapse mechanism with the sway collapse mechanism to give a total collapse resultant. Figure 4-8: Test 1.3 - Beam Collapse Mechanism below shows the beam collapse mechanism component of the rigid portal frame.
Figure 4-8: Test 1.3 - Beam Collapse Mechanism
∑ (W ×δ )=∑ ( M p ×θ) V c δ=4 M p θ V c 0.15 θ ¿ 4 M p θ
V c=
4Mp 0.15
Substituting the value for Mp calculated for Test 1.1:
V c=
4 × 1.34 =35.7 N (Equation 1) 0.15
Figure 4-9 below shows the sway collapse mechanism component of the rigid portal frame.
Figure 4-9: Test 1.3 - Sway Collapse Mechanism
∑ (W ×δ )=∑ ( M p ×θ) H c δ=4 M p θ
Figure 4-10:Angles for virtual work method of Test 1.3
The tan of a small angles is considered to equal to the angle itself:
θ=δ/ 0.3
δ =0.3 θ H c 0.3 θ ¿ 4 M p θ
H c=
4Mp 0.3
Substituting the value for Mp calculated for Test 1.1:
H c=
4 × 1.34 =17.9 N (Equation 2) 0.3
Figure 4-11 below shows the combined collapse mechanism of the rigid portal frame.
Figure 4-11: Combined Collapse mechanism Test1.3
The combined collapse virtual work equation is simply the sum of external work of the beam and sway mechanisms equated to the sum of internal work of the beam and sway mechanisms minus the internal work of the hinge at the top left hinge of the frame. As seen in Figure 4-11, the combined collapse mechanism does not have a plastic hinge located at the top left hand corner of the frame so both the internal work from the beam and sway are to be subtracted from calculations.
∑ (W ×δ )=∑ ( M p ×θ) V c 0.15 θ + H c 0.3θ=4 M p θ+ 4 M p θ−2 M p θ V c 0.15 + H c 0.3=6 M p Substituting Mp=1.34Nm.
V c 0.15 + H c 0.3=6 M p
V c + 2 H c =53.6 ( Equation3) Equations 1, 2 and 3 can now be used to draw the interaction diagram for combined collapse mechanism as seen below in Figure 4-12.
Permissible Region
Figure 4-12: Theoretical Load Interaction Diagram
The results for each groups vertical and horizontal loading points (H, V) are then plotted onto this Interaction diagram (Figure 4-13) to determine how close these experimental points were to the theoretical solution and failure mechanism. The V/H ratios used were also plotted to show its intersection with the theoretical collapse mechanisms. Vertical Beam Collapse Load Horizontal Sway Collapse Load Exp. data on Theoretical Load Interaction Diagram Combined Collapse Load Mechanism 40 Group 8
Group 7
35
Group 9
30
Group 10
25
Vertical Load (N) Group 12
20
Group 11
15
Group 7 line
10 Group 8 Line 5
Group 9 line
Group 10 Line0 0 Group 12 line
2
4
6
8
Group 11 Line 10 12
14
16
18
Horizontal Load (N)
Figure 4-13: Experimental data on Theoretical Load Interaction Diagram
From this figure it is clear that all vertical and horizontal combined loadings were within the theoretical permissible region of the load interaction diagram. By calculating where the various V/H ratios should intersect with the theoretical combined collapse load mechanism, a comparison can be made between the
theoretical and experimental horizontal and vertical values determined as seen below in table
Table 4.3: Test 1.3 Loading Arrangements
H experime ntal 13.73 11.12 10.58 0.00 8.04 7.05
4.5.
H theory
% error
15.31 13.40 11.91 10.71 9.75 8.93
10.34 17.03 11.14 100.00 17.56 21.08
H experime ntal 20.60 22.24 26.46 0.00 28.13 28.19
H theory
% error
22.97 26.80 29.78 32.16 34.11 35.73
10.34 17.03 11.15 100.00 17.53 21.10
Discussion
From this Figure 12 and Table 4.3 we can see clearly that the experimental collapse loads calculated by each groups various arrangement are all lower than the predicted collapse loads with percentage errors ranging from approximately 10-20%. This shows that the loads under which the frame collapsed should have been permissible in theory. This is concerning in terms of designing a structure which in reality has a small but reasonably lower load capacity. There are many reasons that could result in such differences between the experimental and theoretical result, the key factors being:
Potential incorrect determination of moment capacity of steel. Error in measurement of load positon e.g. moving the bucket closer than required to the left hand side for a given ratio would result in an actual higher load ratio, causing it to fail at a lower total load than expected. Inability to correctly detect when failure is occurring and thus continuing to load the beam once failure has begun resulting an overly large value for the load reading. The moment contribution of the bars weight is not considered in the loading ratio causing an incorrect actual loading ratio effecting collapse load calculations.
5. Conclusion This report examined the Plastic Behaviour of continuous steel beams and rigid portal steel frames. Experimental analysis was undertaken by using various load ratios on a continuous beam and portal frame respectively with results relatively close to the theoretical values obtained for the continuous beam, less so for the portal frame.
6. References Dr Lip Teh (2015), Laboratory Instructions, University of Wollongong Dr Lip Teh (2015), CIVL454 Structures, Tutorial worked solutions week 4, University of Wollongong Dr Lip Teh (2015), CIVL454 Structures, Lecture Notes week 4, University of Wollongong
RA = Reaction force on Hanger at A
W= Weight of bucket and pellets* B
A
RC = Reaction force on Hanger at C
C
Spreader Bar a mm
b mm L mm
300m m
L1=RA = Loading at mid-span RD
L2=RC =2.5 X RA Loading at midspan
RE D
A
300m m
RF
C
E
F
L2
300m m
RE
RF E
C
F
150m m
150m m Mp
Mp
Mp
300m m
150m m
V= RA
Pulley 300m m
Wire Rigid portal frame H = RC
150m m
150m m 490m m
V= RA H = RC
Beam Collapse
Combined Collapse
Beam load = 3.557kg total X = 86mm
Frame load = 3.776kg total X = 140mm
