UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
ASSESSMENT 1
Semester 1 – 2026
⋄
Module Code: DYN3701
Module Name: Dynamics of Machines 3B
Assignment No.: Assessment 1
Due Date: 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for DYN3701
at the University of South Africa.
,UNISA | DYN3701 Assessment 1 – 2026
Question 1 [10 marks]
Question: A wooden cantilever beam with a rectangular cross section and a total length L
supports an inclined point load P at its free end. Determine (i) the orientation of the neutral
axis resulting from the inclined loading, and (ii) the maximum tensile stress in the beam.
Given data:
Parameter Value
Width, b 60 mm
Height, h 100 mm
Length, L 1.2 m = 1200 mm
Applied load, P 450 N
Load inclination angle, α 27◦
Step 1: Resolve the Load into Components
The inclined load P is split into two components: one acting along the z-axis (causing bend-
ing about the y-axis) and one acting along the y-axis (causing bending about the z-axis).
Pz = P cos α = 450 cos 27◦ = 450(0.8910) = 400.95 N (1)
Py = P sin α = 450 sin 27◦ = 450(0.4540) = 204.30 N (2)
Step 2: Calculate Bending Moments at the Fixed End
For a cantilever beam, the maximum bending moment occurs at the fixed support, equal to
the load multiplied by the full span.
Moment about the y-axis (caused by Pz ):
My = Pz · L = 400.95 × 1200 = 481 143.52 N mm (3)
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,UNISA | DYN3701 Assessment 1 – 2026
Moment about the z-axis (caused by Py ):
Mz = Py · L = 204.30 × 1200 = 245 154.87 N mm (4)
Step 3: Second Moments of Area
For the rectangular cross-section:
hb3 (100)(60)3 100 × 216 000
Iy = = = = 1.80 × 106 mm4 (5)
12 12 12
bh3 (60)(100)3 60 × 1 000 000
Iz = = = = 5.00 × 106 mm4 (6)
12 12 12
Step 4: Equation of the Neutral Axis
For unsymmetrical bending, the bending stress at any point (y, z) is:
Mz My
σ=− y+ z (7)
Iz Iy
At the neutral axis σ = 0, implying that:
Mz My z Mz /Iz
− y+ z=0 =⇒ = (8)
Iz Iy y My /Iy
Substituting:
z 245 154. 000 000 0.04903
= = = 0.18343 (9)
y 481 143. 800 000 0.26730
Step 5: Orientation of the Neutral Axis
The angle θ that the neutral axis makes with the y-axis is found from:
z
tan θ = = 0.18343 (10)
y
θ = arctan(0.18343) = 10.4◦ (11)
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,UNISA | DYN3701 Assessment 1 – 2026
Therefore the neutral axis lies at 10.4◦ from the y-axis, or equivalently 90◦ − 10.4◦ = 79.6◦
from the z-axis.
Key Distinction
In unsymmetrical bending the neutral axis does not coincide with the bending-moment
vector. Here the load is inclined at 27◦ yet the neutral axis tilts only 10.4◦ from the
y-axis, because Iz > Iy (the section is stiffer about z).
Step 6: Stresses at the Four Corners
The stress equation becomes:
Mz My
σ=− y+ z = −(0.04903) y + (0.26730) z (12)
Iz Iy
The corner coordinates are y = ±h/2 = ±50 mm and z = ±b/2 = ±30 mm.
Corner y (mm) z (mm) σ (MPa)
1 +50 +30 −(0.04903)(50) + (0.26730)(30) = −2.4515 + 8.0190 = +5.57
2 +50 −30 −(0.04903)(50) + (0.26730)(−30) = −2.4515 − 8.0190 = −10.47
3 −50 +30 −(0.04903)(−50) + (0.26730)(30) = +2.4515 + 8.0190 = +10.47
4 −50 −30 −(0.04903)(−50) + (0.26730)(−30) = +2.4515 − 8.0190 = −5.57
Result: Maximum Tensile Stress
The largest tensile (positive) stress is at corner 3, i.e. y = −50 mm, z = +30 mm:
σmax = +10.47 MPa at corner (−50 mm, +30 mm) (13)
Critical Consideration
Corner 2 carries the largest compressive stress of −10.47 MPa. Both extremes are equal
in magnitude here because the section is symmetric; this is not generally true for non-
symmetric sections.
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, UNISA | DYN3701 Assessment 1 – 2026
Question 2 [30 marks]
Question: A simply supported beam of length L = 5.5 m is constructed from a 50 × 200 mm
(actual dimensions) wood beam and two steel angle sections L150×100×10, one on each side
of the wood beam, as shown in Figure Q2. The modulus of elasticity of the steel is 20 times
that of the wood.
The allowable stresses are σs = 110 MPa (steel) and σw = 8.3 MPa (wood).
Question 1.1: Determine the allowable uniformly distributed load q, disregarding the weight
of the beam. [15]
Question 1.2: Repeat Question 1.1 if a 25 × 250 mm (actual dimensions) wood flange is
added on top. [15]
Question 2.1: Composite Beam Without Flange
Step 1: Area of One Steel Angle
The L150×100×10 angle is treated as two rectangles minus the overlapping corner square.
The area of one angle:
Aa = t(a + b − t) (14)
= 10(150 + 100 − 10) (15)
= 10 × 240 (16)
= 2 400 mm2 (17)
Therefore the total steel area for two angles:
As = 2 × 2 400 = 4 800 mm2 (18)
Page 4 of 13
College of Science, Engineering and Technology
⋄
ASSESSMENT 1
Semester 1 – 2026
⋄
Module Code: DYN3701
Module Name: Dynamics of Machines 3B
Assignment No.: Assessment 1
Due Date: 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for DYN3701
at the University of South Africa.
,UNISA | DYN3701 Assessment 1 – 2026
Question 1 [10 marks]
Question: A wooden cantilever beam with a rectangular cross section and a total length L
supports an inclined point load P at its free end. Determine (i) the orientation of the neutral
axis resulting from the inclined loading, and (ii) the maximum tensile stress in the beam.
Given data:
Parameter Value
Width, b 60 mm
Height, h 100 mm
Length, L 1.2 m = 1200 mm
Applied load, P 450 N
Load inclination angle, α 27◦
Step 1: Resolve the Load into Components
The inclined load P is split into two components: one acting along the z-axis (causing bend-
ing about the y-axis) and one acting along the y-axis (causing bending about the z-axis).
Pz = P cos α = 450 cos 27◦ = 450(0.8910) = 400.95 N (1)
Py = P sin α = 450 sin 27◦ = 450(0.4540) = 204.30 N (2)
Step 2: Calculate Bending Moments at the Fixed End
For a cantilever beam, the maximum bending moment occurs at the fixed support, equal to
the load multiplied by the full span.
Moment about the y-axis (caused by Pz ):
My = Pz · L = 400.95 × 1200 = 481 143.52 N mm (3)
Page 1 of 13
,UNISA | DYN3701 Assessment 1 – 2026
Moment about the z-axis (caused by Py ):
Mz = Py · L = 204.30 × 1200 = 245 154.87 N mm (4)
Step 3: Second Moments of Area
For the rectangular cross-section:
hb3 (100)(60)3 100 × 216 000
Iy = = = = 1.80 × 106 mm4 (5)
12 12 12
bh3 (60)(100)3 60 × 1 000 000
Iz = = = = 5.00 × 106 mm4 (6)
12 12 12
Step 4: Equation of the Neutral Axis
For unsymmetrical bending, the bending stress at any point (y, z) is:
Mz My
σ=− y+ z (7)
Iz Iy
At the neutral axis σ = 0, implying that:
Mz My z Mz /Iz
− y+ z=0 =⇒ = (8)
Iz Iy y My /Iy
Substituting:
z 245 154. 000 000 0.04903
= = = 0.18343 (9)
y 481 143. 800 000 0.26730
Step 5: Orientation of the Neutral Axis
The angle θ that the neutral axis makes with the y-axis is found from:
z
tan θ = = 0.18343 (10)
y
θ = arctan(0.18343) = 10.4◦ (11)
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,UNISA | DYN3701 Assessment 1 – 2026
Therefore the neutral axis lies at 10.4◦ from the y-axis, or equivalently 90◦ − 10.4◦ = 79.6◦
from the z-axis.
Key Distinction
In unsymmetrical bending the neutral axis does not coincide with the bending-moment
vector. Here the load is inclined at 27◦ yet the neutral axis tilts only 10.4◦ from the
y-axis, because Iz > Iy (the section is stiffer about z).
Step 6: Stresses at the Four Corners
The stress equation becomes:
Mz My
σ=− y+ z = −(0.04903) y + (0.26730) z (12)
Iz Iy
The corner coordinates are y = ±h/2 = ±50 mm and z = ±b/2 = ±30 mm.
Corner y (mm) z (mm) σ (MPa)
1 +50 +30 −(0.04903)(50) + (0.26730)(30) = −2.4515 + 8.0190 = +5.57
2 +50 −30 −(0.04903)(50) + (0.26730)(−30) = −2.4515 − 8.0190 = −10.47
3 −50 +30 −(0.04903)(−50) + (0.26730)(30) = +2.4515 + 8.0190 = +10.47
4 −50 −30 −(0.04903)(−50) + (0.26730)(−30) = +2.4515 − 8.0190 = −5.57
Result: Maximum Tensile Stress
The largest tensile (positive) stress is at corner 3, i.e. y = −50 mm, z = +30 mm:
σmax = +10.47 MPa at corner (−50 mm, +30 mm) (13)
Critical Consideration
Corner 2 carries the largest compressive stress of −10.47 MPa. Both extremes are equal
in magnitude here because the section is symmetric; this is not generally true for non-
symmetric sections.
Page 3 of 13
, UNISA | DYN3701 Assessment 1 – 2026
Question 2 [30 marks]
Question: A simply supported beam of length L = 5.5 m is constructed from a 50 × 200 mm
(actual dimensions) wood beam and two steel angle sections L150×100×10, one on each side
of the wood beam, as shown in Figure Q2. The modulus of elasticity of the steel is 20 times
that of the wood.
The allowable stresses are σs = 110 MPa (steel) and σw = 8.3 MPa (wood).
Question 1.1: Determine the allowable uniformly distributed load q, disregarding the weight
of the beam. [15]
Question 1.2: Repeat Question 1.1 if a 25 × 250 mm (actual dimensions) wood flange is
added on top. [15]
Question 2.1: Composite Beam Without Flange
Step 1: Area of One Steel Angle
The L150×100×10 angle is treated as two rectangles minus the overlapping corner square.
The area of one angle:
Aa = t(a + b − t) (14)
= 10(150 + 100 − 10) (15)
= 10 × 240 (16)
= 2 400 mm2 (17)
Therefore the total steel area for two angles:
As = 2 × 2 400 = 4 800 mm2 (18)
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