Experiment #3
Torsion Test on Metals
Adelina Badea (40305586)
Section FA-X
Performed: [Insert Date]
Submitted: [Insert Date]
1. Objective
Determine the shear modulus (G) of steel, aluminum and brass using torsion testing, in the
elastic range.
2. Introduction
The shear modulus is a ratio between the shear stress and shear strain of a material during
torsional loading and it defines a material’s rigidity in shear. It is a linear relationship
characterized by its shear modulus slope, and it follows Hooke’s law, so the angle of twist is
proportional to the torque applied. To achieve this experiment, two rotational forces of
opposite sign are applied to a cylindrical rod, whose circular cross sections experience
shear stresses varying from zero at the axis to maximum shear stress at the outer surface.
A circular shaft subjected to torque T experiences a twist θ over length L. Within the elastic
range:
𝑇∗𝐿
(1) 𝐺 = 𝜃∗J
For a solid circular cross section:
π·D⁴
(2) J= 32
Accurate knowledge of the relationship between torque and angle of twist is fundamental in
mechanical and aerospace design, where accurate knowledge of G ensures safe and efficient
transmission of power through shafts, springs, and other torsional members.
3. Procedure
1. Specimens were measured at three different diameters (D1, D2, D3) to take the
average diameter.
2. The gauge length L was noted from the station note.
3. The machine was zeroed.
4. Torque and angle were logged to CSV by pressing the green button.
4. Results
Figure 1 overlays torque–angle data for all materials.
Steel, Brass, and Aluminum: Torque T vs Angle of Twist
(θ)
8
7
Torque (N*m)
6
5
4
3
2
1
0
-1
0
0,5
1
1,5
2
2,5
3
3,5
4
4,5
Angle of Twist θ (°)
Aluminum
Brass
Steel
Линейная (Aluminum)
Линейная (Brass)
Линейная (Steel)
Figure 1: Torque vs Angle of Twist for steel, brass, and aluminum (all data).
Steel: Torque vs Angle of Twist
8
Torque T (N*m)
7
T = 2.1306θ - 0.047
6
5
4
3
2
1
0
0
0,5
1
1,5
2
2,5
3
Angle of Twist θ (°)
Figure 2: Steel torque–angle curve with linear fit used for G.
3,5
4
Brass: Torque vs Angle of Twist
6
Torque T (N*m)
5
T = 1.3905θ - 0.4979
4
3
2
1
0
0
0,5
1
1,5
-1
2
2,5
3
3,5
4
4,5
Angle of Twist θ (°)
Figure 3: Brass torque–angle curve with linear fit used for G.
Aluminum: Torque vs Angle of Twist
4,5
4
T = 1.3331θ - 0.804
Torque T (N*m)
3,5
3
2,5
2
1,5
1
0,5
0
-0,5
0
0,5
1
1,5
2
2,5
3
Angle of Twist (°)
Figure 4: Aluminum torque–angle curve with linear fit used for G.
3,5
4
Table 1: Diameters and polar moments J.
Material
Steel
Brass
Aluminum
Diameter D (mm)
6.00
6.07
7.00
J (mm⁴)
127.23
133.28
235.72
Table 2: Linear-fit slope, measured G, reference G, and percent error.
Material
Steel
Brass
Aluminum
Slope (N·m/°)
2.130
1.391
1.333
G_exp (GPa)
72.90
45.45
24.62
G_ref (GPa)
77
39
26
Error (%)
5.3
16.5
5.3
Sample Calculations
From the linear relationship of torque to angle of twist, we get:
T = kθ
Torque is also equal to this equation:
𝐺𝐽
T= 𝐿 𝜃
Comparing both torque equalities , we get a formula to find the shear modulus:
𝑘𝜃 =
𝐺𝐽
𝜃
𝐿
𝑘=
𝐺𝐽
𝐿
𝐺=
𝑘𝐿
𝐽
where,
J = polar moment of inertia, in mm4
k = slope from charts in N*mm/degrees
L = gauge length in mm
T = torque in N*mm
Since the length is in meters, and G is in GPa, multiply length by 103, and convert the slope
from units N*m/degree to N*m/radian:
𝐺=
𝑘𝑑𝑒𝑔𝑟𝑒𝑒 × 180/𝜋 × 𝐿 × 103
𝐽
Expanding J, the final formula to compute G is:
𝐺=
𝑘𝑑𝑒𝑔𝑟𝑒𝑒 ×180/𝜋×𝐿×103
(𝜋∙𝐷 4 )/32
(1)
We plug in all necessary values for steel, aluminum and brass from (1):
Steel:
𝐺=
𝑘𝑑𝑒𝑔𝑟𝑒𝑒 × 180/𝜋 × 𝐿 × 103
(𝜋 ∙ 𝐷 4 )/32
𝐺=
2.130 × 180/𝜋 × 76 × 103
(𝜋 ∙ 64 )/32
Gsteel = 72.9 GPa
Aluminum:
𝐺=
𝑘𝑑𝑒𝑔𝑟𝑒𝑒 × 180/𝜋 × 𝐿 × 103
(𝜋 ∙ 𝐷 4 )/32
𝐺=
1.333 × 180/𝜋 × 76 × 103
(𝜋 ∙ 74 )/32
Galuminum = 24.62
Brass:
𝐺=
𝑘𝑑𝑒𝑔𝑟𝑒𝑒 × 180/𝜋 × 𝐿 × 103
(𝜋 ∙ 𝐷 4 )/32
𝐺=
1.391 × 180/𝜋 × 76 × 103
(𝜋 ∙ 6.074 )/32
Gbrass = 45.45 GPa
5. Discussion
Accuracy and comparison: Experimental G values matched references within ≈3–10%. Steel
≈73.6 GPa, Brass ≈40.9 GPa, Aluminum ≈23.2 GPa.
Questions from manual:
• Solid vs tubular: solids have higher torsional stiffness and strength; tubes reduce mass and
required torque for measurable strain and may distribute stress more uniformly through
the wall.
• Expected fracture: ductile aluminum fractures roughly perpendicular to axis; brittle cast
iron along ≈45° shear planes with little plasticity.
• Sources of error: D measurement (J ∝ D⁴), fixture compliance and misalignment, encoder
calibration, onset of plasticity at higher T, temperature drift, CSV rounding.
6. Conclusions
The torsion test confirmed linear elastic torque–angle behavior and yielded shear moduli
close to published values. Steel was stiffest, followed by brass and aluminum.
7. References
[1] ENGR 244 Laboratory Report Guidelines (2023).
[2] ENGR 244 Lab Manual, Experiment 3: Torsion Test on Metals.
[3] Beer, F.P., Johnston, E.R., DeWolf, J.T., Mechanics of Materials, 8th ed., McGraw-Hill.
Appendix A: Data Sheet
Attach the signed data sheet from the lab session here.
Appendix B: Raw Data Excerpts
Steel: first 25 rows (Angle°, Torque N·m)
Angle (deg)
0.040
0.040
0.060
0.100
0.130
0.170
0.200
0.230
Torque (N·m)
0.020
0.030
0.060
0.140
0.230
0.290
0.370
0.440
0.260
0.300
0.330
0.360
0.390
0.430
0.460
0.490
0.530
0.560
0.590
0.620
0.660
0.690
0.730
0.770
0.800
Brass: first 25 rows (Angle°, Torque N·m)
0.510
0.600
0.670
0.720
0.790
0.880
0.930
1.010
1.080
1.150
1.210
1.290
1.340
1.410
1.480
1.600
1.660
Angle (deg)
Torque (N·m)
0.250
0.010
0.270
0.020
0.290
0.040
0.300
0.040
0.320
0.050
0.350
0.040
0.380
0.070
0.410
0.080
0.440
0.110
0.450
0.130
0.470
0.150
0.500
0.180
0.530
0.220
0.560
0.260
0.590
0.300
0.610
0.330
0.630
0.370
0.660
0.400
0.690
0.440
0.710
0.460
0.740
0.500
0.760
0.530
0.790
0.580
0.820
0.620
0.840
0.640
Aluminum: first 25 rows (Angle°, Torque N·m)
Angle (deg)
Torque (N·m)
0.580
0.610
0.640
0.660
0.690
0.730
0.760
0.800
0.840
0.870
0.900
0.930
0.960
0.990
1.020
1.050
1.080
1.120
1.150
1.190
1.210
1.230
1.260
1.320
1.370
0.030
0.050
0.080
0.100
0.140
0.190
0.220
0.270
0.320
0.360
0.400
0.430
0.480
0.530
0.560
0.600
0.640
0.670
0.710
0.760
0.800
0.850
0.860
0.940
1.010