Simple Beam — Two Equal Concentrated Loads Unsymmetrically Placed

1

Geometry and load

Define the span, the equal loads and their distance from each support

Span ℓ

m

Concentrated load P

kN

Left distance a

m

Right distance b

m

Middle length = ℓ − a − b = 5 m region between the two loads

2

Show calculation details

R₁, R₂, M_max

R₁left support = Pℓ(ℓ − a + b) = — = —kN

R₂right support = Pℓ(ℓ − b + a) = — = —kN

M_maxunder the larger load = R₁·a (under the left load) = — = —kN·m

3

Deflection calculation

Define E and I for δ_max

Modulus of elasticity E

MPa

Moment of inertia I

cm⁴

δ_maxx = x* = maximum of the δ(x) curve (numerical) ⇒ —mm

Under two unsymmetric loads, δ_max occurs neither at midspan nor under a load — it is found by numerical double integration. AWC gives no closed form for this case.

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−b ℓ

x is in the left region (x < a)

V_xx < a = +R₁ = — = —kN

M_xx < a = R₁ · x = — = —kN·m

δ_x = y(x) at this point (numerical) ⇒ —mm