Portal Frame — Pinned Base, Horizontal Point Load at Eave

1

Geometry and load

Span, column height and horizontal point load at B

Span ℓ

m

Column height h

m

Horizontal point load P

kN

Define separate section inertia / elastic modulus for column and beam

Model: supports A and D are pinned (no moment transfer); corners B and C are rigid (the beam-column angle stays fixed). P is applied horizontally exactly at B (the left column-beam joint, eave level). The two columns are assumed IDENTICAL (same h, I and E). The beam is assumed axially rigid (inextensible) — together these two assumptions make the results (H, R_v, M_k) entirely INDEPENDENT of section inertia and elastic modulus (see the Reactions card for the reasoning).

2

Reactions and horizontal thrust

H, R_v

Hat each support, opposing P = P2 = — = —kN

R_vat each support, vertical (force couple) = P·hℓ = — = —kN

k_rfrom your section inputs — does not appear in the H formula at all = E_bI_bhE_cI_cℓ = —

Why it's independent of kr: since the beam is assumed axially rigid (inextensible), B and C undergo the same horizontal displacement — regardless of where P is applied along the beam. Because the two columns are identical (same h, I, E), the horizontal load splits EQUALLY between them: H = P/2, in the same direction at both supports (opposing P). The overturning moment P·h is resisted by an axial force couple spanning ℓ between A and D (R_v = Ph/ℓ) — uplift (tension) on the A side, compression on the D side. Vary k_r above as much as you like — H always stays P/2.

3

Bending moments

Corner moment M_k (B, C) — zero at midspan

M_kcorner (B and C) — equal magnitude, opposite sense = H·h = — = —kN·m

The column moment is zero at the base (pin) and increases linearly to M_k at the corner. In the beam the moment is −M_k at B and +M_k at C — it changes sign at midspan (double curvature). This comes from BOTH beam ends being rotated in OPPOSITE senses by the rigid-corner moment; it differs from a single-sign distribution like wℓ²/8 (which applies to a vertical UDL).

Bending Moment Diagram (M) — on the frame

4

Shear and axial force

V and N in column and beam — on the frame

Vcolumn (both, constant) · beam (constant, no distributed load) = H · R_v

Ncolumn AB (B side): R_v, tension · column CD (C side): R_v, compression · beam: H, compression = R_v · H

The column axial forces form the force couple that resists the overturning moment: the side where P is applied (AB, the B end) is in TENSION (tends to lift), the opposite side (CD) is in COMPRESSION. The beam is axially COMPRESSED by the shear carried in from each column (N = H, compression) — the shear from one column routes through the corner into the beam axis and back out through the other.

Shear Force Diagram (V) — on the frame

Axial Force Diagram (N) — on the frame