WITH KUDOS AND THANKS TO TOOL BOX IMHO what a load of cods...

WITH KUDOS AND THANKS TO
TOOL BOX


IMHO what a load of cods wabble,
They did not take into account the VELOCITY AND THE FOCAL POINT AND VECTOR OF THE INITIAL IMPACT,
If you speed up a soft fly fast enough you could shoot it though a human body.
Useful equations related to acceleration, average velocity, final velocity and distance traveled

Average Velocity

v_a = (v₁ + v₀) / 2 (1)
where
v_a = average velocity (m/s)
v₀ = initial velocity (m/s)
v₁ = final velocity (m/s)
Final Velocity

v₁ = v₀ + a t (2)
where
a = acceleration (m/s²)
t = time taken (s)
Distance Traveled

s = (v₀ + v₁) t / 2 (3)
where
s = distance traveled (m)
Alternative:
s = v₀ t + 1/2 a t² (3b)
Acceleration

a = (v₁ - v₀) / t (4)
Alternative:
a = (v₁² - v₀²) / (2 s) (4b)
Example - Accelerating Motorcycle

A motorcycle starts with an initial velocity 0 km/h (0 m/s) and accelerates to 120 km/h (33.3 m/s) in 5 s.
The average velocity can be calculated with eq. (1) to
v_a = ((33.3 m/s) - (0 m/s)) / 2
= 16.7 m/s
The distance traveled can be calculated with eq. (3) to
s = ((0 m/s) + (33.3 m/s)) (5 s) / 2
= 83.3 m
The acceleration can be calculated with eq. (4) to
a = ( (33.3 m/s) - (0 m/s)) / (5 s)
= 6.7 m/s²
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MORE LINKS BELOW,,,, SHEESH SOME PEOPLE ARE SO IN NEED OF AL FOIL HATS
Beam Loads - Support Force Calculator
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Area Moment of Inertia - Typical Cross Sections I

Area Moment of Inertia, Moment of Inertia for an Area or Second Moment of Area for typical cross section profiles

Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I, is a property of shape that is used to predict deflection, bending and stress in beams.
Area Moment of Inertia - Imperial units

• inches⁴

Area Moment of Inertia - Metric units

• mm⁴
• cm⁴
• m⁴

Converting between Units

• 1 cm⁴ = 10^-8 m⁴ = 10⁴ mm⁴
• 1 in⁴ = 4.16x10⁵ mm⁴ = 41.6 cm⁴

Example - Convert between Area Moment of Inertia Units

9240 cm⁴ can be converted to mm⁴ by multiplying with 10⁴
(9240 cm⁴) 10⁴ = 9.24 10⁷ mm⁴
Area Moment of Inertia (Moment of Inertia for an Area or Second Moment of Area)

​

for bending around the x axis can be expressed as
I_x = ∫ y² dA (1)
where
I_x = Area Moment of Inertia related to the x axis (m⁴, mm⁴, inches⁴)
y = the perpendicular distance from axis x to the element dA (m, mm, inches)
dA = an elemental area (m², mm², inches²)
The Moment of Inertia for bending around the y axis can be expressed as
I_y = ∫ x² dA (2)
where
I_x = Area Moment of Inertia related to the y axis (m⁴, mm⁴, inches⁴)
x = the perpendicular distance from axis y to the element dA (m, mm, inches)
Area Moment of Inertia for typical Cross Sections I

• Area Moment of Inertia for typical Cross Sections II

Solid Square Cross Section

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The Area Moment of Inertia for a solid square section can be calculated as
I_x = a⁴ / 12 (2)
where
a = side (mm, m, in..)

I_y = a⁴ / 12 (2b)
Solid Rectangular Cross Section

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The Area Moment of Ineria for a rectangular section can be calculated as
I_x = b h³ / 12 (3)
where
b = width
h = height

I_y = b³ h / 12 (3b)
Solid Circular Cross Section

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The Area Moment of Inertia for a solid cylindrical section can be calculated as
I_x = π r⁴ / 4
= π d⁴ / 64 (4)
where
r = radius
d = diameter

I_y = π r⁴ / 4
= π d⁴ / 64 (4b)
Hollow Cylindrical Cross Section

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The Area Moment of Inertia for a hollow cylindrical section can be calculated as
I_x = π (d_o⁴ - d_i⁴) / 64 (5)
where
d_o = cylinder outside diameter
d_i = cylinder inside diameter

I_y = π (d_o⁴ - d_i⁴) / 64 (5b)
Square Section - Diagonal Moments

​

The diagonal Area Moments of Inertia for a square section can be calculated as
I_x = I_y = a⁴ / 12 (6)

Rectangular Section - Area Moments on any line through Center of Gravity

​

Rectangular section and Area of Moment on line through Center of Gravity can be calculated as
I_x = (b h / 12) (h² cos a + b² sin² a) (7)
Symmetrical Shape

​

Area Moment of Inertia for a symmetrical shaped section can be calculated as
I_x = (a h³ / 12) + (b / 12) (H³ - h³) (8)
I_y = (a³ h / 12) + (b³ / 12) (H - h) (8b)
Nonsymmetrical Shape

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Area Moment of Inertia for a non symmetrical shaped section can be calculated as
I_x = (1 / 3) (B y_b³ - B₁ h_b³ + b y_t³ - b1 h_t³)

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Beams - Supported at Both Ends - Continuous and Point Loads

Stress in a bending beam can be expressed as
σ = y M / I (1)
where
σ = stress (Pa (N/m²), N/mm², psi)
y = distance to point from neutral axis (m, mm, in)
M = bending moment (Nm, lb in)
I = moment of Inertia (m⁴, mm⁴, in⁴)

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The calculator below can be used to calculate maximum stress and deflection of beams with one single or uniform distributed loads.
Beam Supported at Both Ends - Uniform Continuous Distributed Load

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Maximum moment in a beam with uniform load supported at both ends:
M_max = q L² / 8 (1a)
where
M_max = maximum moment (Nm, lb in)
q = uniform load per length unit of beam (N/m, N/mm, lb/in)
L = length of beam (m, mm, in)
Moment in position x:
M_x = q x (L - x) / 2 (1b)
where
M_x = moment in position x (Nm, lb in)
x = distance from end (m, mm, in)

Maximum Stress

​

Maximum stress in a beam with uniform load supported at both ends:
σ_max = y_max q L² / (8 I) (1c)
where
σ_max= maximum stress (Pa (N/m²), N/mm², psi)
y_max = distance to extreme point from neutral axis (m, mm, in)

• 1 N/m² = 1x10^-6 N/mm² = 1 Pa = 1.4504x10^-4 psi
• 1 psi (lb/in²) = 144 psf (lb_f/ft²) = 6,894.8 Pa (N/m²) = 6.895x10^-3 N/mm²

Maximum deflection:
δ_max = 5 q L⁴ / (384 E I) (1d)
where
δ_max = maximum deflection (m, mm, in)
E = Modulus of Elasticity (Pa (N/m²), N/mm², psi)
Deflection in position x:
δ_x = q x (L³ - 2 L x² + x³) / (24 E I) (1e)
Note! - deflection is often the limit factor in beam design. For some applications beams must be stronger than required by maximum loads, to avoid unacceptable deflections.
Forces acting on the ends:
R₁ = R₂
= q L / 2 (1f)
where
R = reaction force (N, lb)
Example - Beam with Uniform Load, Metric Units

A UB 305 x 127 x 42 beam with length 5000 mm carries a uniform load of 6 N/mm. The moment of inertia for the beam is 8196 cm⁴ (81960000 mm⁴) and the modulus of elasticity for the steel used in the beam is 200 GPa (200000 N/mm²). The height of the beam is 300 mm (the distance of the extreme point to the neutral axis is 150 mm).
The maximum stress in the beam can be calculated
σ_max = (150 mm) (6 N/mm) (5000 mm)² / (8 (81960000 mm⁴))
= 34.3 N/mm²
= 34.3 10⁶ N/m² (Pa)
= 34.3 MPa
The maximum deflection in the beam can be calculated
δ_max = 5 (6 N/mm) (5000 mm)⁴ / ((200000 N/mm²) (81960000 mm⁴) 384)
= 2.98 mm

AND TO DO THE CALCULATIONS PLEASE GO THE LINK BELOW

YOU CAN INPUT DATA A SEE JUST WHAT CONC VEL AND FORCE WILL DO