Frame BMD with maximum and minimum moments

Assumptions (taken from the figure):

• A(0,0) hinged at the foot of a 2 m column AC; 2 kN horizontal (→) at 1 m above A. Beam from the column top C: 4 m span to the roller B, then a 1.5 m overhang BD with 5 kN downward at the free end D.
• Load on the 4 m span CB: triangular, symmetrical, peak 6 kN/m at 2 m from the column (total 12 kN).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

C--------------------------------B-----------D
|
|
|
|
|
|
|
A

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

• $H_A$ = 2.00 kN ←
• $\Sigma M_A = 0$: $4R_B = 2(1) + 12(2) + 5(5.5) = 53.5$, so $R_B$ = 13.38 kN ↑
• $V_A = 17 - 13.38$ = 3.62 kN ↑

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AC: V constant (steps at point loads), M linear
• CB: V parabolic, M cubic
• BD: V constant, M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Maximum and minimum bending moments

• Maximum positive (sagging) BM = 5.76 kN·m at 1.55 m from C along CB, where V = 0 (tension on the bottom)
• Maximum negative (hogging) BM = -7.50 kN·m over the roller B (tension on the top); this is the largest BM in magnitude
• Point of contraflexure (M = 0) at 3.05 m from C along CB
• In the column, M rises linearly from 0 at A to 2.00 kN·m at the 2 kN load and stays 2.00 kN·m up to C.