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NATIONAL UNIVERSITY OF IRELAND GALWAY
DEPARTMENT OF CIVIL ENGINEERING
SECOND ENGINEERING
STRENGTH OF MATERIALS LABORATORY
Semester II
STUDENT NAME: Niall Rabbitte
GROUP NO: 17
DATE OF EXPERIMENT: 23 March 2015
SIGNATURE OF SUPERVISOR:
EXPERIMENT NO 2
DEFLECTION OF A SIMPLY SUPPORTED BEAM
BACKGROUND
The midspan deflection v of a simply supported beam of span L subject to a point
load W at midspan can be shown to be
EI
WL
48
1
v
3

The linear relationship between the deflection v and the load W is verified by
measuring the deflections for a series of load values.
PROCEDURE
The beam set-up is shown in the figure
First measure the beam dimensions and calculate the second moment of area (I =
BD3/12). Assuming, E for aluminium is 69 GPa, calculate the theoretical deflections.
Next, place a hanger for loads at midspan. Set up the dial gauge for measuring
deflections at midspan and adjust the reading to zero. Apply a series of loads W to
the hanger and record the deflections.
Plot the theoretical deflection v load curve. On the same graph, mark in the
experimental deflections.
RESULTS
Span L = 1000mm
Width B = 19.5mm
Depth D = 6mm
I = BD3/12
= 19.5mm(6mm)3 / 12
= 351mm4
L
W
Theoretical deflections
EI
WL
48
1
v
3

For 0N
6900(351)
0(1000)
48
1
v
3

v = 0
For 5N
6900(351)
5(1000)
48
1
v
3

v = 4.3mm
For 10N
6900(351)
10(1000)
48
1
v
3

v = 8.6mm
For 15N
6900(351)
15(1000)
48
1
v
3

v = 12.9
For 20N
6900(351)
20(1000)
48
1
v
3

v = 17.2mm
For 25N
6900(351)
25(1000)
48
1
v
3

v = 21.5mm
Table of results
Load
(N)
Deflections
(mm) (loading)
Deflections
(mm)
(unloading)
Deflections (mm)
(average)
Theoretical
deflections
(mm)
0 0 0 0 0
5 3.3 3.3 3.3 4.3
10 6.6 6.5 6.55 8.6
15 9.8 9.9 9.85 12.9
20 13.2 13.2 13.2 17.2
25 16.5 16.5 16.5 21.5
SUMMARY AND CONCLUSIONS
The dimensions are recorded and inertia calculated. These figures and the
given value for E are used to calculate the theoretical deflection values.
The beam is loaded in increments of 5N, starting at 0N. The values were
read off the gauge. This figure was multiplied by 0.1 to give the
deflection in mm.
The results show a linear relationship. This can be seen in the graph
below. Any discrepancy may be due to human error when reading the
gauge, or it may be that the gauge is defective. The beam has been loaded
and unloaded many times which may weaken the beam givingincorrect
results .
Error
Percentage error =
𝑬𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕𝒂𝒍−𝑻𝒉𝒆𝒓𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍
𝑻𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍
X 100
=
16.5−21.5
21.5
X100
= 23%
Experiment No. 2 Load v Deflection Results
0.0
5.0
10.0
15.0
20.0
25.0
0 5 10 15 20 25 30
Load(N)
Deflection (mm)
Load (N) Vs theoretical deflection (avg)
(mm)
Load (N) Vs experimental deflection
(mm)

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1415-CE227_Lab_Assg4_Deflection

  • 1. NATIONAL UNIVERSITY OF IRELAND GALWAY DEPARTMENT OF CIVIL ENGINEERING SECOND ENGINEERING STRENGTH OF MATERIALS LABORATORY Semester II STUDENT NAME: Niall Rabbitte GROUP NO: 17 DATE OF EXPERIMENT: 23 March 2015 SIGNATURE OF SUPERVISOR:
  • 2. EXPERIMENT NO 2 DEFLECTION OF A SIMPLY SUPPORTED BEAM BACKGROUND The midspan deflection v of a simply supported beam of span L subject to a point load W at midspan can be shown to be EI WL 48 1 v 3  The linear relationship between the deflection v and the load W is verified by measuring the deflections for a series of load values. PROCEDURE The beam set-up is shown in the figure First measure the beam dimensions and calculate the second moment of area (I = BD3/12). Assuming, E for aluminium is 69 GPa, calculate the theoretical deflections. Next, place a hanger for loads at midspan. Set up the dial gauge for measuring deflections at midspan and adjust the reading to zero. Apply a series of loads W to the hanger and record the deflections. Plot the theoretical deflection v load curve. On the same graph, mark in the experimental deflections. RESULTS Span L = 1000mm Width B = 19.5mm Depth D = 6mm I = BD3/12 = 19.5mm(6mm)3 / 12 = 351mm4 L W
  • 3. Theoretical deflections EI WL 48 1 v 3  For 0N 6900(351) 0(1000) 48 1 v 3  v = 0 For 5N 6900(351) 5(1000) 48 1 v 3  v = 4.3mm For 10N 6900(351) 10(1000) 48 1 v 3  v = 8.6mm For 15N 6900(351) 15(1000) 48 1 v 3  v = 12.9 For 20N 6900(351) 20(1000) 48 1 v 3  v = 17.2mm For 25N 6900(351) 25(1000) 48 1 v 3  v = 21.5mm Table of results Load (N) Deflections (mm) (loading) Deflections (mm) (unloading) Deflections (mm) (average) Theoretical deflections (mm) 0 0 0 0 0 5 3.3 3.3 3.3 4.3 10 6.6 6.5 6.55 8.6 15 9.8 9.9 9.85 12.9 20 13.2 13.2 13.2 17.2 25 16.5 16.5 16.5 21.5
  • 4. SUMMARY AND CONCLUSIONS The dimensions are recorded and inertia calculated. These figures and the given value for E are used to calculate the theoretical deflection values. The beam is loaded in increments of 5N, starting at 0N. The values were read off the gauge. This figure was multiplied by 0.1 to give the deflection in mm. The results show a linear relationship. This can be seen in the graph below. Any discrepancy may be due to human error when reading the gauge, or it may be that the gauge is defective. The beam has been loaded and unloaded many times which may weaken the beam givingincorrect results . Error Percentage error = 𝑬𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕𝒂𝒍−𝑻𝒉𝒆𝒓𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝑻𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 X 100 = 16.5−21.5 21.5 X100 = 23%
  • 5. Experiment No. 2 Load v Deflection Results 0.0 5.0 10.0 15.0 20.0 25.0 0 5 10 15 20 25 30 Load(N) Deflection (mm) Load (N) Vs theoretical deflection (avg) (mm) Load (N) Vs experimental deflection (mm)