Simple Beam — Two Equal Concentrated Loads Symmetrically Placed

1

Geometry and load

Define the span, the equal loads and their distance from the supports

Span ℓ

m

Concentrated load P

kN

Distance from support a

m

Middle length = ℓ − 2a = 4 m pure-bending region between the two loads

2

Show calculation details

R₁, R₂, M_max

R₁left support (symmetric) = P = — = —kN

R₂right support (symmetric) = P = — = —kN

M_maxa ≤ x ≤ ℓ−a (constant between the loads) = P · a = — = —kN·m

3

Deflection calculation

Define E and I for δ_max

Modulus of elasticity E

MPa

Moment of inertia I

cm⁴

δ_maxx = ℓ/2 (midspan) = Pa(3ℓ² − 4a²)24 EI = — = —mm

Under two symmetric loads, δ_max occurs exactly at midspan (x = ℓ/2). Between the loads the moment is constant (M = Pa) and the shear is zero — this is the "pure bending" region.

Deflected Shape Diagram

4

Bending-stress check

Compare σ_max with the allowable bending stress

Section modulus S

cm³

Allow. bending F_b

MPa

σ_max = M_maxS = — = —MPa

— σ / F_b

5

Point analysis

V_x, M_x, δ_x values at a given point x

Distance x

m

0 a ℓ−a ℓ

x is in the left region (x < a)

V_xx < a = +R₁ = +P = — = —kN

M_xx < a = P · x = — = —kN·m

δ_xx < a = Px6 EI·(3aℓ − 3a² − x²) = —mm