Simple Beam — Load Increasing Uniformly to Center

1

Geometry and Load

Define the span and the total triangular load

Span ℓ

m

Total load W

kN

w_peak = 2W/ℓ = 20 kN/m peak load intensity at midspan

2

Show calculation details

R₁, R₂, M_max

R₁left support (symmetric) = W2 = — = —kN

R₂right support (symmetric) = W2 = — = —kN

M_maxx* = ℓ/2 (orta nokta) = Wℓ6 = — = —kN·m

3

Deflection calculation

Define E and I for δ_max

Elastic modulus E

MPa

Moment of inertia I

cm⁴

δ_maxx = ℓ/2 (midspan) = Wℓ³60 EI = — = —mm

For the symmetric triangular load, δ_max occurs exactly at midspan (x = ℓ/2). Since the beam and load are symmetric, the moment and deflection peaks coincide.

Deflected Shape Diagram

4

Stress check

Compare σ_max against the allowable stress

Section modulus S

cm³

Allowable 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 x

Distance x

m

0 ℓ/2 ℓ

V_xsymmetric: for x > ℓ/2 use x'=ℓ−x = W2ℓ²(ℓ² − 4x²) = — = —kN

M_xsymmetric: for x > ℓ/2 use x'=ℓ−x = Wx6ℓ²(3ℓ² − 4x²) = — = —kN·m

δ_xsymmetric: for x > ℓ/2 use x'=ℓ−x = Wx480 EIℓ²·(5ℓ² − 4x²)² = —mm