Beam AB with tie rod: AFD, SFD, BMD

Assumptions (taken from the figure):

• Origin at A (hinge on the wall). Beam A–F–G–H–C–B: A to F = 1 m, F–G = 3 m, G–H = 3 m, H–C = 1 m, C–B = 1 m (B at x = 9 m).
• Loads (downward): triangle FG rising from 0 at G to 5 kN/m at F; triangle GH with 0 at G and 5 kN/m at H. Each has a total load of $\tfrac12\times3\times5=7.5$ kN, acting 1 m from F and 1 m from H respectively.
• The 5 kN load at B acts at 60° to the beam, downward and away from the wall: $F_x = 5\cos60^\circ = 2.5$ kN (tension), $F_y = 5\sin60^\circ = 4.33$ kN (down).
• Bracket CD is rigid and vertical (D is 1 m above C, at x = 8 m); the tie rod DE joins D to E (0, 4 m), so its direction is $(-8, 3)/\sqrt{73}$.

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.

Support reactions

The tie force $T$ acts at D, 1 m above the beam: $T_x = \frac{8}{\sqrt{73}}T$ (to the left) and $T_y = \frac{3}{\sqrt{73}}T$ (upward). Moments about A: $8T_y + 1\cdot T_x = 7.5(2) + 7.5(6) + 4.33(9) = 98.97$, so $3.745\,T = 98.97$.

• $H_A$ = 22.24 kN →
• $V_A$ = 10.05 kN ↑
• $T$ = 26.43 kN (tension in the rod; components 24.74 kN ← and 9.28 kN ↑ on the beam)

Check: the three equilibrium equations of the complete frame are satisfied to rounding.

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

The jump in M at C ($20.42 + 4.33 = 24.75$ kN·m) is the couple $T_x\times1$ m that the bracket CD transfers to the beam.

Shape of the diagrams

• Between loads V is constant; under a UDL it varies linearly and under a triangular load parabolically. M is one degree higher than V (linear, parabolic, cubic respectively). V jumps at point loads and reactions; M jumps at couples.
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• The 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.

Salient features

• Maximum positive (sagging) M = 28.18 kN·m at x = 5.75 m
• Maximum negative (hogging) M = -4.33 kN·m just right of C (x = 8.00 m)
• V = 0 at 5.75 m from A along AB: M = 28.18 kN·m (bottom), the local maximum

Sketches (sagging +ve above the line)

SFD:

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BMD:

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