The aim of this report is to demonstrate the ability to apply and calculate the correct formulas for different Scientific and Analytical scenarios like:
- Shear Force & Bending Moment for a Beam.
- The use of software for the calculation of the above.
- Shear Stress for solid shafts.
- Friction & Acceleration.
- Simple Harmonic Motion.
3. David Antuna HNC CAD/CAM
Summary:
3
The aim of this report is to demonstrate the ability to apply and calculate the correct
formulas for different Scientific and Analytical scenarios like:
Shear Force & Bending Moment for a Beam.
The use of software for the calculation of the above.
Shear Stress for solid shafts.
Friction & Acceleration.
Simple Harmonic Motion.
Scientific & Analytical Methods Ass.1 ANT07078917
4. David Antuna HNC CAD/CAM
Shear Force, Bending Moment & Stress for Beams:
4
Determine distribution of shear force, bending moment and stress due to bending
in simply supported beams.
1. A cantilever of rectangular cross-section 74 mm wide and 110 mm deep carries a
concentrated load of 3 kN at its free end. Neglecting the mass of the cantilever: Find
the maximum length if the greatest permissible stress due to bending is 85 MPa.
Draw the shear force and bending moment for the beam.
74mm
BENDING EQUATION
110mm
������ ������
= =
������ ������
When:
σ max=85 Mpa.
������������³ ������.������������������������������.������������³
������ = =
������������ ������������
W= 3Kn
I= 8.21x10¯⁶
������������³
������ =
������������
Y= 0.055m
������· ������������������������ ������.������������������������������¯������ ������ ������������������������������⁶
M= =
������������������������ ������.������������������������¯²
M= 1.27x10⁴
������ ������.������������������������������⁴
M=WL there for L= =
������ ������������������������³
L= 4.23 m
Scientific & Analytical Methods Ass.1 ANT07078917
5. David Antuna HNC CAD/CAM
2. Find the maximum total uniformly distributed load which can be carried by a timber
beam 300 mm deep, 200 mm wide and 3.5 m long which is simply supported at its 5
ends. Neglect the mass of the beam itself and take the maximum stress due to
bending to be 8 MPa. Draw the shear force and bending Moment for the beam.
200mm
300mm
3.5m
BENDING EQUATION
������ ������
= =
������ ������
When:
σ= 8 Mpa = 8x10⁶ N/mm²
Y= 150x10¯³
������·������
M=
������
������������³ ������.������������������.������³
������ = = = 4.50x10¯⁴
������������ ������������
L= 3.5 m
������ ������������ ������ ������������������������������
M= I .·. =I .·. W=
������ ������ ������ ������������������
������.������������������������������¯⁴������������������������������⁶������������
.·. W=
������.������������������������������������������������¯³
W= 5.49x10⁴ N
Scientific & Analytical Methods Ass.1 ANT07078917
6. David Antuna HNC CAD/CAM
Beam Calculation Software:
6
Determine the sectional properties of structural sections.
3. The below screen grabs are an example of how is possible to use different type of
software to calculate Load, Shear Force & Bending Moments for any type of beam.
In this case the software is SuperBeam 4.
Cantilever beam with a 3 KN concentrate load.
Figure 1 Use of Software to Calculate Beams
Scientific & Analytical Methods Ass.1 ANT07078917
7. David Antuna HNC CAD/CAM
UDL Timber Beam 3.5m long with a 8 MPa of bending stress.
7
Figure 2 Use of Software to Calculate Beams
Scientific & Analytical Methods Ass.1 ANT07078917
8. David Antuna HNC CAD/CAM
Shear Stress for Circular Shafts:
8
Determine the distribution of shear stress and the angle and deflection due to torsion
in circular shafts.
4. In a torsion experiment, a specimen is used whose shaft is 30 mm diameter. A torsion
meter, which is clamped to the shaft shows that a 200 mm length of the shaft twists
through an angle of 25° when a twisting moment of 400 Nm is applied.
a. Find the value of the modulus of rigidity of the shaft material.
b. Find the value of the maximum shear stress and hence draw a diagram showing the
shear stress distribution over the cross section of the shaft.
R=0.015
τ=
When: TORSION EQUATION
������ ������ ������������
R= 0.015m = 15x10¯³ = =
������ ������ ������
L= 200mm = 0.2m = 200x10¯³
������������������������
θ= 25° .·. θ= ������������������ = 0.436 rad
T= 400Nm
������������⁴ ������������������.������������⁴
������ = = ������ = =7.95x10¯⁸
������������ ������������
a. Modulus of Rigidity b. Maximum Shear Stress
������ ������������ ������ ������
= = ������
������ ������ ������
������������ ������������
.·. ������ = .·. ������ =
������������ ������
������������������������������.������������������
������ =
������.������������������������������¯⁸
������������������������������.������
������ = τ= 7.54x10⁷
������.������������������������������¯������ ������������.������������������
G= 438742138.4 N/m²
G= 4.38x10⁸ N/m²
Scientific & Analytical Methods Ass.1 ANT07078917
9. David Antuna HNC CAD/CAM
Friction & Acceleration:
9
Determine the behaviour of dynamic mechanical Systems in which uniform
acceleration is present.
5. A cast iron block of mass 30kg is pulled along a surface and accelerates at 0.3m/s2. If the
coefficient of friction is 0.15N, and g=9.81m/s2.
a) Find the necessary tractive force.
If the force in the last example is removed after 20 seconds, and the crate started moving
from rest.
b) Calculate the final velocity, and how far it moved until the force was removed.
Sketch the crate as it accelerates and indicate all the forces acting upon it.
W=30 Kg a=0.3 m/s²
FF=0.15 N F=53.145 N
FRICTION EQUATION
F= FF+FI
F= µMg+Ma
a) Find the necessary tractive force.
F= FF+FI FF=µFN
F= µMg+Ma =µMg
FI =Ma
F= 0.15x30x9.81+30x0.3
F= 53.145 N
a) Calculate the final velocity, and how far it moved until the force was removed.
������
V=u+at ������ = 2 (������ + ������) S= ut+½at²
20
V=at since U=0 ������ = (6 + 0.3) S= ½at²
2
V=0.3x20 S = 63m S= ½x0.3x20²
V=6 m/s = 21.6 Km/Hr S= 60m
Scientific & Analytical Methods Ass.1 ANT07078917
10. David Antuna HNC CAD/CAM
Simple Harmonic Motion:
10
Determine the behaviour of oscillating mechanical systems in which simple harmonic
motion is present.
6. The crank of a scotch-yoke mechanism is 52 mm long and rotates at a uniform 250
rev/mm. Find the slider's: (a) acceleration and velocity when 15 mm from the mid-point
of its travel; (b) maximum acceleration, and state where it occurs; (c) maximum velocity,
and state where it occurs.
ω
SIMPLE HARMONIC MOTION EQUATIONS
X= aCosθ
U= ������ ������² − ������²
F= ω²X
When:
a= 52mm = 52x10¯³ m
250������2������
ω= 250 rev/min = 60 rads/sec
X= 15mm = 15x10¯³ m
Find the Acceleration and Velocity.
250������2������ 2
F= ω² X = X 15x10¯³ .·. F= 10.28 m/s²
60
250������2������ 2
U= ������ ������² − ������² = 52������10¯³ 2 − (15������10¯3 )² .·. U= 1.36 m/s²
60
Find the maximum acceleration, and state where it occurs./F=ω²X and occurs when x=a
250������2������ 2
Fmax=ω²a = X 52x10¯³ .·. Fmax= 35.64 m/s²
60
Find maximum velocity, and state where it occurs. Occurs when X=0 From U= ������ ������² − ������²
250������2������
Umax= ωa = ( 60 ) X 52x10¯³ .·. Umax= 1.36 m/s²
Scientific & Analytical Methods Ass.1 ANT07078917
11. David Antuna HNC CAD/CAM
Appendix:
11
In this section you will find the diagrams for question 1 & 2 of this report.
Scientific & Analytical Methods Ass.1 ANT07078917