Propped Cantilever Beam — Uniformly Distributed Load

1

Geometry and load

Span and UDL intensity (fixed on the left, pinned on the right)

Span ℓ

m

UDL q

kN/m

Left end fixed (moment + shear), right end pinned (shear only).

2

R_A, R_B, M_A, V_max, M_max+

R_Aat the fixed end = 5qℓ8 = — = —kN

R_Bat the roller support = 3qℓ8 = — = —kN

M_Afixed-end moment = qℓ²8 = — = —kN·m

M_max+at x = 5ℓ/8 = 9qℓ²128 = — = —kN·m

V_x=0= R_A = V_max = —kN

3

Define E and I for δ_max

Elastic modulus E

MPa

Moment of inertia I

cm⁴

δ_maxat x ≈ 0.5785ℓ ≈ qℓ⁴185 EI = —mm

δ(x) = qx²(3ℓ² − 5ℓx + 2x²)48 EI

Deflected Shape Diagram

4

Compare σ_max against the allowable stress

Section modulus S

cm³

Allow. bending F_b

MPa

σ_max = M_maxS = — = —MPa

M_max is taken as the larger absolute value (M_A or M_max+).

— σ / F_b

5

V_x, M_x, δ_x values at a given point x

Distance x

m

0 (fixed) ℓ/4 ℓ (support)

V_x = 5qℓ8 − qx = — = —kN

M_x = −qℓ²8 + 5qℓx8 − qx²2 = —kN·m

δ_x = qx²(3ℓ²−5ℓx+2x²)48EI = —mm