For three hinged parabolic arch subjected to u.d.l over entire span, bending moment and radial shear at any section is zero throughout span.

Concept:

The locus of reaction of a two hinged semi-circular arch is a straight line whereas locus of reaction of a two-hinged parabolic arch is a parabolic curve.

For two Hinged Semi-circular Arch:

Reaction locus is straight line parallel to line joining abutments and height at πR/2.

For two Hinged Parabolic Arch:

$y=PE=\frac{1.6h{L}^2}{L^2+Lx-x^2}$

Due to its geometry, bending moment always comes out to be zero in 3 hinged arches.

Concept:

Cable structures:

• Long span structures subjected to tension and use suspension cables for supports.
• Examples of cable structures are suspension bridges, cable-stayed roof.
• Cable structures especially cable of a suspension bridge are in form of catenary. The catenary is shape assumed by a cable freely suspended between two points.
• The central sag or dip of cable varies from(1/10)th to (1/15)th of span.
• Cables can be made of mild steel, high strength steel, stainless steel, or polyester fibers.

Explanation:

The FBD of saddle placed on roller is shown in figure:

 For saddle to be in equilibrium

ΣF_x = 0 ie F_H + T Sinθ = T Sinθ

∴ F_H = 0 → There will be no horizontal reaction developed on saddle.

Furr, Saddle is supported on roller and roller cannot resist moment and horizontal reactions and it can resist only vertical reactions and hence, roller will transfer only vertical reactions to tower.

Therefore, re will not be any horizontal reaction developed on top of tower.

ΣF_y­ = 0 ieT Cosθ + T Cosθ = N

Or

N = 2T Cosθ

This normal or vertical reaction on saddle will be transferred to tower as shown in FBD of tower.

Consider moment equilibrium of tower about its base;

ΣM_base = 0

Or

F_H × H = M

Or

M = 0 (As F_H = 0)

∴ The net horizontal force (F_H) on top of this tower and bending moment (B.M) at base of tower both are zero.

Explanation:

The general equation of a symmetrical parabolic arch as shown in figure can be written as:

$y=\frac{4h}{L^2}(x)(L-x)$

Where

L is horizontal projection of parabolic arch

h is a central rise in arch

$y=\frac{4h}{L^2}\left(x\right)(L-x)$

Given that h = 10m  and we need to find out rise of arch at quarter point i.e. value of y at X = L/4

Substitute $x=\frac{L}{4}$ in above equation

$y=\left(\frac{4}{L^2}\right)\times \left(10\right)\left(\frac{L}{4}\right)\left(\frac{L-L}{4}\right)$

$=\frac{3\times 10}{4}=7.5m$

The rise of arch at Quarter Point is 7.5 m. 

Explanation:

Frames

Frame structures are structures having combination of beam, column, and slab to resist lateral and gravity loads.

Note:

These structures are usually used to overcome large moments developing due to applied loading.

Classification

Frames are classified as:

1. Sway or Non-Sway

A sway frame allows a lateral or sideward movement, while a non-sway frame does not allow movement in horizontal direction. The lateral movement of sway frames is accounted for in ir analysis.

2. Rigid or Flexible

The joints of rigid frames are fixed, whereas those of flexible frames are movable.

Cable when loaded takes shape of bending moment diagram that would be formed if same set of load is applied on to simply supported beam of same length and properties. Thus, shape of cable loaded with uniformly distributed load throughout is parabolic.

Calculation:

ΣV = 0, V_A + V_B = 150 kN

ΣH = 0, H_A = H_B = H

At support B, ΣM_B = 0

V_A × 20 - 150 × 16 = 0

V_A = 120 kN

V_B = 150 - 120 = 30 kN

At central hinge, ΣM_C = 0

V_A × 10 - H × 4 - 150 × 6 = 0

H × 4 = 120 × 10 - 150 × 6

H = 75 kN

So, vertical reaction at A and horizontal thrust are 120 kN and 75kN respectively.