How is the roller reaction found in this beam with an internal hinge?

Take moments of the part left of the hinge about the hinge and set M = 0

Use only sum Fy = 0 for the whole beam

Take moments about the fixed support B and ignore the hinge

Assume the roller carries half the total load

Why is H_B = 43.30 kN?

It balances the axial (horizontal) component of the 50 kN force at 30°, i.e. 50 cos30°

It balances the vertical component 50 sin30°

It balances the 80 kN point load

Where does the maximum (hogging) bending moment of 592.50 kN·m occur?

At the fixed support B, x = 11 m

2075 Chaitra · 12 marks

Beam AFD, SFD, BMD with hinge (80 kN)

Short answer

Reactions: roller 41.67 kN up; fixed support HB=43.30 kN left, VB=158.33 kN up, MB=592.50 kN·m clockwise. Maximum sagging moment 8.02 kN·m at 3.56 m; maximum hogging -592.50 kN·m at B. Contraflexure at 2.5 m and 5 m (hinge).

Draw AFD, SFD and BMD for following beam. Also indicate the salient features. [Figure: beam A to B, fixed at B on the right end, with a support at 2 m from A; triangular load of peak 20 kN/m (rising from zero at A to 20 kN/m and falling) on the first 5 m; an 80 kN point load at 1 m further; a 50 kN force at $30^\circ$ to the beam; a 15 kN/m UDL over the last 3 m; an internal hinge; dimensions 2 m, 3 m, 1 m, 2 m, 3 m.]

Assumptions (taken from the figure):

Beam of 11 m: A (x = 0) free end; roller at x = 2 m; internal hinge at x = 5 m; fixed support at B (x = 11 m).
Triangular load on 0–5 m, symmetrical, peak 20 kN/m at x = 2.5 m (total 50 kN); 80 kN downward at x = 6 m; 50 kN at 30° to the beam at x = 8 m, acting down and to the right (components 43.30 kN along the beam towards B and 25 kN down); 15 kN/m downward on 8–11 m (45 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

Hinge condition: $M = 0$ at x = 5 m. Moments of the left part (0–5 m) about the hinge: $3R_{roller} = 50\times2.5$ , which gives the roller reaction. The fixed support then follows from $\Sigma F_x = 0$ , $\Sigma F_y = 0$ and $\Sigma M = 0$ for the whole beam.

$R_{roller}$ = 41.67 kN ↑
$H_B$ = 43.30 kN ←
$V_B$ = 158.33 kN ↑
$M_B$ = 592.50 kN·m (clockwise)

Check: the hinge passes $50 - 41.67 = 8.33$ kN to the right part, and $V_B = 8.33 + 80 + 25 + 45 = 158.33$ kN, $M_B = 8.33(6) + 80(5) + 25(3) + 45(1.5) = 592.5$ kN·m ✓

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

Section	N (kN)	V (kN)	M (kN·m)
At A	0.00	0.00	0.00
Just left of roller	0.00	-16.00	-10.67 (top)
Just right of roller	0.00	25.67	-10.67 (top)
At hinge (x = 5 m)	0.00	-8.33	0.00
Just left of 80 kN	0.00	-8.33	-8.33 (top)
Just right of 80 kN	0.00	-88.33	-8.33 (top)
Just left of 50 kN	0.00	-88.33	-185.00 (top)
Just right of 50 kN	-43.30	-113.33	-185.00 (top)
At B	-43.30	-158.33	-592.50 (top)

The segment from x = 8 m to B is in compression (43.30 kN), because the horizontal component of the 50 kN load pushes it towards the fixed end.

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: zero from A to x = 8 m, then a constant compression of 43.30 kN up to B.
The BMD is zero at the free end A and at the hinge.

Salient features

Maximum positive (sagging) M = 8.02 kN·m at x = 3.56 m
Maximum negative (hogging) M = -592.50 kN·m at x = 11.00 m (fixed end B)
V = 0 at 3.56 m from A: M = 8.02 kN·m (bottom), the local maximum
Point of contraflexure (M = 0) at 2.50 m from A
M = 0 again at 5.00 m from A (the internal hinge), where it changes from sagging to hogging

Sketches (sagging +ve above the line)

SFD:




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BMD:





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✓ Answer checked by Claude Opus 5.5 (high effort)

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