Shear Center Lab 2015

The Hong Kong Polytechnic University
Department of Civil & Structural Engineering
BEng(Hons) in Civil Engineering
Structural Mechanics II

Laboratory Instruction Sheet: Shear Center

Objective:
(1) To find the shear center of various cross sections, and
(2) To compare them with the theoretical values.

Apparatus:
Three mild steel frames made of different cross sections (channel, equal angle and ‘Z, section). Two dial gauges, end support brackets, load hangers, weights.

Procedure:
1. Set the two dial gauges for measuring the vertical displacement of each top corner of the central loading plate.
2. Measure the dimensions of the beam section and the distance between the two dial gauge positions.
3. Hook the weight hanger onto the first hole of the loading plate.
4. Record the readings of both dial gauges.
5. Apply a load up to 30N in 10N increments and record the readings of both dial gauges at each loading.
6. Check the zero weight dial gauge readings after removal of weights.
7. Move the weight hanger to the next hole and then repeat procedure 3 to 5.

Training on critical thinking (20 marks out of 100):
1. If two dial gauges are placed on one side of the central loading plate as shown in Figure (a), justify whether the objectives of the experiment are still achieved. (8 marks)

We can still determine the shear center while placing two dial gauges are placed on one side of the central loading plate. However, the result may be less accurate as the displacement obtained of two points may be the same. Moreover, the dial gauge is not as sensitive as that of detecting the rotation angle of equal angle section if the gauges are putting on one side. As a result, the error is inversely proportional to the distance of the two gauges and it is recommended to place them on opposite direction to achieve higher accuracy.

2. If an inclined load instead of a vertical load is applied to the holes subsequently as shown in Figure (b), criticize whether the position of the shear centre can be determined. (12 marks)

No. The incline load can be resolved to a vertical and horizontal component. The vertical component is changing while the horizontal component is not. The horizontal component does not act on the axis that shear center lies on so twisting effect is induced. Vertical component acts on the axis that shear center lies on so no twisting effect is induced. It may cause deflection and the dial gauge gives the same reading for the displacement.

Reports and discussion (80 marks out of 100):
(a) Plot the displacement (final – initial) of left dial gauge readings against the (load) for hole number 1, 4, 7, 11.
(b) Derive the angle of rotation of each beam using the data and plot the “rotation” against “hole position” at each loading. Draw the cross section of the beam relative to the hole-position axis on the same graph. Hence determine where the shear center lies. Note that angle of rotation can be negative.
(c) Find the centroids and shear centers of the three cross sections. Draw both the centroids and shear centers onto the three cross sections.
(d) Suppose the load is put right under the shear center. Can you find out the displacement of the shear center at the loading 10, 20 and 30? The displacement measured at hole number 1 in part (a) is definitely different from this result, why?
(e) Compare the theoretical prediction and the experimental determination of the position of the shear center.
(f) Discuss the difference between centroid and shear center.

Laboratory Report – Shear Center

Equal Angle Section

Label all dimensions and draw the centroid and shear center

(a) Plot the displacement (final – initial) of left dial gauge readings against the (load) for hole number 1, 4, 7, 11.

Loading (N)
Reading of A gauge ( mm )
Hole number
1
2
3
4
5
6
7
8
9
10
11
0 (A0)
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
10 (A1)
1.35
1.15
0.95
0.74
0.54
0.35
0.17
-0.02
-0.22
-0.40
-0.59
20 (A2)
2.67
2.29
1.86
1.49
1.11
0.69
0.33
-0.04
-0.44
-0.82
-1.22
30 (A3)
3.97
3.41
2.8
2.23
1.65
1.06
0.49
-0.08
-0.67
-1.23
-1.82

Loading (N)
Reading of B gauge ( mm )
Hole number
1
2
3
4
5
6
7
8
9
10
11
0 (B0)
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
10 (B1)
-0.58
-0.48
-0.4
-0.31
-0.22
-0.1
-0.06
0.00
0.09
0.16
0.26
20 (B2)
-1.15
-0.97
-0.78
-0.63
-0.46
-0.29
-0.13
0.01
0.19
0.33
0.52
30 (B3)
-1.7
-1.46
-1.19
-0.94
-0.69
-0.46
-0.19
0.03
0.29
0.51
0.78

Loading (N)
Displacement of A gauge ( Ai – A0 )
Hole number
1
2
3
4
5
6
7
8
9
10
11
10 (A1 – A0)
1.35
1.15
0.95
0.74
0.54
0.35
0.17
-0.02
-0.22
-0.4
-0.59
20 (A2 – A0)
2.67
2.29
1.86
1.49
1.11
0.69
0.33
-0.04
-0.44
-0.82
-1.22
30 (A3 – A0)
3.97
3.41
2.8
2.23
1.65
1.06
0.49
-0.08
-0.67
-1.23
-1.82

Loading (N)
Displacement of B gauge ( Bi – B0 ))
Hole number
1
2
3
4
5
6
7
8
9
10
11
10 (B1 – B0)
-0.58
-0.48
-0.4
-0.31
-0.22
-0.1
-0.06
0
0.09
0.16
0.26
20 (B2 – B0)
-1.15
-0.97
-0.78
-0.63
-0.46
-0.29
-0.13
0.01
0.19
0.33
0.52
30 (B3 – B0)
-1.7
-1.46
-1.19
-0.94
-0.69
-0.46
-0.19
0.03
0.29
0.51
0.78

I. A graph of dial gauge reading (A) against the loading

II. A graph of displacement of dial gauge reading (A) against the loading

(b) Derive the angle of rotation of each beam using the data and plot the “rotation” against “hole position” at each loading. Draw the cross section of the beam relative to the hole-position axis on the same graph. Hence determine where the shear center lies. Note that angle of rotation can be negative.

Loading (N)
Rotation ( radian )
Hole number
1
2
3
4
5
6
7
8
9
10
11
10
0.50
0.42
0.35
0.27
0.20
0.12
0.06
-0.005
-0.08
-0.15
-0.22
20
0.99
0.85
0.69
0.55
0.41
0.26
0.12
-0.01
-0.16
-0.30
-0.45
30
1.48
1.27
1.04
0.83
0.61
0.40
0.18
-0.03
-0.25
-0.45
-0.68
* Take anti-clockwise positive

(c) Find the centroids and shear centers of the three cross sections. Draw both the centroids and shear centers onto the three cross sections.

The shear center lies on the line where the rotation is equal to 0, which is the dotted line. From the graph, the shear center lies on a line which passes through hole number 7.8, which is 20 +20 +20 +10 + 4 = 74 mm from the right edge of the frame.
Theoretical determination of centroid and Iu and Iv

 (from the bottom)

 (from the left)

Computation of shear center from theory and compare with experimental results

By the method of determining the shear flow, both shear flow lines are intersecting at the edge of the section. That point is the shear center (as shown in the graph) as zero moment will be induced.

(d) Suppose the load is put right under the shear center. Can you find out the displacement of the shear center at the loading 10, 20 and 30? The displacement measured at hole number 1 in part (a) is definitely different from this result, why?

Yes. Reading of the dial gauge shows the twisting and the deflection when loading is added. There is no twisting effect while loading is added under shear center. As a result, the reading is much smaller when compare to other points.

(e) Compare the theoretical prediction and the experimental determination of the position of the shear center.

In theory, the shear center should be at the intersection point of the two legs. As a result, it should be located at 72+1.6/2=72.8mm away from the right edge of the frame, where our result is 74mm. The error is about 1.6%
Z Section

Label all dimensions and draw the centroid and shear center

(a) Plot the displacement (final – initial) of left dial gauge readings against the (load) for hole number 1, 4, 7, 11.

Loading (N)
Reading of A gauge ( mm )
Hole number
1
2
3
4
5
6
7
8
9
10
11
0 (A0)
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
10 (A1)
-0.07
-0.06
-0.04
-0.02
0.00
0.00
0.01
0.02
0.04
0.06
0.07
20 (A2)
-0.15
-0.11
-0.09
-0.05
-0.01
0.00
0.03
0.05
0.08
0.11
0.14
30 (A3)
-0.21
-0.17
-0.13
-0.08
-0.03
0.00
0.04
0.08
0.12
0.17
0.21

Loading (N)
Reading of B gauge ( mm )
Hole number
1
2
3
4
5
6
7
8
9
10
11
0 (B0)
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
10 (B1)
0.07
0.04
0.03
0.01
-0.02
0.01
-0.01
-0.03
-0.04
-0.06
-0.08
20 (B2)
0.14
0.1
0.07
0.04
0.03
0.01
-0.02
-0.05
-0.08
-0.11
-0.14
30 (B3)
0.21
0.15
0.11
0.08
0.05
0.02
-0.04
-0.08
-0.12
-0.16
-0.20

Loading (N)
Displacement of A gauge ( Ai – A0 )
Hole number
1
2
3
4
5
6
7
8
9
10
11
10 (A1 – A0)
-0.07
-0.06
-0.04
-0.02
0
0
0.01
0.02
0.04
0.06
0.07
20 (A2 – A0)
-0.15
-0.11
-0.09
-0.05
-0.01
0
0.03
0.05
0.08
0.11
0.14
30 (A3 – A0)
-0.21
-0.17
-0.13
-0.08
-0.03
0
0.04
0.08
0.12
0.17
0.21

Loading (N)
Displacement of B gauge ( Bi – B0 ))
Hole number
1
2
3
4
5
6
7
8
9
10
11
10 (B1 – B0)
0.07
0.04
0.03
0.01
-0.02
0.01
-0.01
-0.03
-0.04
-0.06
-0.08
20 (B2 – B0)
0.14
0.1
0.07
0.04
0.03
0.01
-0.02
-0.05
-0.08
-0.11
-0.14
30 (B3 – B0)
0.21
0.15
0.11
0.08
0.05
0.02
-0.04
-0.08
-0.12
-0.16
-0.20

I. A graph of dial gauge reading (A) against the loading

II. A graph of displacement of dial gauge reading (A) against the loading

(b) Derive the angle of rotation of each beam using the data and plot the “rotation” against “hole position” at each loading. Draw the cross section of the beam relative to the hole-position axis on the same graph. Hence determine where the shear center lies. Note that angle of rotation can be negative.

Loading (N)
Rotation ( radian )
Hole number
1
2
3
4
5
6
7
8
9
10
11
10
-0.04
-0.028
-0.02
-0.0084
0.0056
-0.003
0.006
0.014
0.022
0.03
0.042
20
-0.08
-0.06
-0.04
-0.025
-0.011
-0.003
0.014
0.028
0.045
0.061
0.078
30
-0.12
-0.09
-0.067
-0.045
-0.022
-0.006
0.022
0.045
0.067
0.092
0.115
* Take anti-clockwise positive

The shear center lies on the line where the rotation is equal to 0, which is the dotted line. From the graph, the shear center lies on a line which passes through hole number 6, which is 20 +20 +20 +20 +20 +10 = 110 mm from the left edge of the frame.

Theoretical determination of centroid and Iu and Iv

 (from the left)
 (from the bottom)

Computation of shear center from theory and compare with experimental results
:

As two legs of the Z section are the same, the shear forces of two legs should be the same when a vertical force is added. Zero moment point, i.e. the shear center, should be located at the centroid.

(d) Suppose the load is put right under the shear center. Can you find out the displacement of the shear center at the loading 10, 20 and 30? The displacement measured at hole number 1 in part (a) is definitely different from this result, why?

Yes. Reading of the dial gauge shows the twisting and the deflection when loading is added. There is no twisting effect while loading is added under shear center. As a result, the reading is much smaller when compare to other points.

(e) Compare the theoretical prediction and the experimental determination of the position of the shear center.

In theory, the shear center should be at the centroid. As a result, it should be located at 110mm away from the right edge of the frame, where our result is also 110mm.

(f) Discuss the difference between centroid and the shear center.

Centroid refers to the geometric center of the area while shear center refers to a point does not have any torsion effect when a shear force is added to it. When I force is added to centroid, it may have torsion effect as centroid is only the center of mass of structures will uniform density. The shear center always lies on the axis of symmetry. For those having two axes of symmetry, the shear center lies on intersection of the axis of symmetry, which is the same point as centroid. For section consist of narrow rectangles which intersect at one point, the junction will be the shear center as the shear flow always pass through that point and cause zero moment. .