When taking moments about A to find the wire force, why does the horizontal component of the wire force not appear in the equation?

Its line of action passes through the beam axis, so its moment arm about A is zero

Horizontal forces never produce moments

It is cancelled by the vertical component

In span AB the maximum sagging bending moment occurs at 1.67 m from A. Why there?

Because the shear force is zero there, so the moment is at a local maximum

Because the bending moment is zero there

Because that is the midpoint of AB

Because the axial force changes sign there

How is the hogging moment of 2.67 kN·m at B obtained from the overhang BC?

The 4 kN triangular load acting at 2/3 m from B (4 x 0.667)

The 4 kN triangular load acting at 1 m from B

2068 Chaitra · 10 marks

Beam ABC with wire support: BMD and SFD

Short answer

Reactions: H_A=5.78 kN →, V_A=3.33 kN ↑, wire tension T=10.42 kN. Maximum sagging moment 2.78 kN·m at 1.67 m from A, maximum hogging moment -2.67 kN·m at B. Contraflexure at 3.33 m from A. Axial force: 5.78 kN compression from A to B, zero beyond. SFD: linear under UDL, parabolic under triangular load; BMD: parabolic and cubic.

Draw bending moment and shear force diagrams for the beam ABC, which has hinged support at 'A' and other support at B, supported by wire (Tension member) as shown in figure. [Figure: beam A-B-C hinged at A on a vertical wall 6 m high; a wire from the top of the wall (6 m above A) to B; 2 kN/m UDL over A to B (4 m); 4 kN/m triangular load over B to C (2 m + 6 m shown as 4 m, 2 m, 6 m); C is the free end.]

Assumptions (taken from the figure):

A(0,0) hinged on the wall; B at x = 4 m, C at x = 6 m (BC = 2 m).
The wire runs from the top of the wall (0, 6) to B, so its direction from B is $(-4, 6)/\sqrt{52}$ (tension only).
2 kN/m downward on AB; triangular load on BC from 4 kN/m at B to 0 at C (total 4 kN, acting 2/3 m from B).

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

Moments about A (the horizontal wire component passes through the beam axis): $T_y\times4 = 8\times2 + 4\times4.667 = 34.667$ , so $T_y = 8.667$ kN and $T = T_y\sqrt{52}/6$ .

$H_A$ = 5.78 kN →
$V_A$ = 3.33 kN ↑
$T$ = 10.42 kN (tension in the wire; components 5.78 kN ← and 8.67 kN ↑ on the beam)

Check: $\Sigma F_y$ : $3.33 + 8.67 = 8 + 4$ ✓; $\Sigma F_x$ : $5.78 - 5.78 = 0$ ✓

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

Section	N (kN)	V (kN)	M (kN·m)
AB at A	-5.78	3.33	0.00
AB at B	-5.78	-4.67	-2.67 (top)
BC at B	0.00	4.00	-2.67 (top)
BC at C	0.00	0.00	0.00

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: a constant compression of 5.78 kN from A to B and zero from B to C.
The BMD is zero at the hinge A and at the free end C.

Salient features

Maximum positive (sagging) M = 2.78 kN·m at x = 1.67 m
Maximum negative (hogging) M = -2.67 kN·m at x = 4.00 m (at B, where the wire is attached)
V = 0 at 1.67 m from A: M = 2.78 kN·m (bottom), the local maximum
Point of contraflexure (M = 0) at 3.33 m from A

Sketches (sagging +ve above the line)

SFD:


**                                 **
  *****                            | ***
       *****                       |    *****
------------*****----------------------------*******
                 *****             |
                      *****        |
                           *****   |
                                ***|

BMD:

          **********
      ****          ***
   ***                 ***
 **                       **
*---------------------------**-------------*********
                              *        ****
                               **    **
                                 * **
                                  *

✓ Answer checked by Claude Opus 5.5 (high effort)

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