Three Hinged Arch - 2 | Structural Analysis - Civil Engineering (CE) PDF Download

Example 32.2 
A three-hinged semicircular arch of uniform cross section is loaded as shown in Fig 32.7. Calculate the location and magnitude of maximum bending moment in the arch.

Solution:

Reactions:
Taking moment of all the forces about hinge B leads to

                                (1)

Bending moment
Now making use of the condition that the moment at hinge C of all the forces left of hinge C is zero gives

M_c = R_ay × 15 - H_a × 15 - 40 × 7 = 0 

Considering the horizontal equilibrium of the arch gives,

The maximum positive bending moment occurs below D and it can be calculated by taking moment of all forces left of D about D.

M_D = R_ay × 8 - H_a × 13.267                            (3)

= 29.33×8 - 10.66×13.267 = 93.213 kN

Example 32.3
A three-hinged parabolic arch is loaded as shown in Fig 32.8a. Calculate the location and magnitude of maximum bending moment in the arch. Draw bending moment diagram.

Solution:

Reactions:
Taking A as the origin, the equation of the three-hinged parabolic arch is given by

                               (1)

Taking moment of all the forces about hinge B leads to,

Now making use of the condition that, the moment at hinge C of all the forces left of hinge C is zero gives,

                               (3)

Considering the horizontal equilibrium of the arch gives,

                                  (4)

Location of maximum bending moment
Consider a section x from end B . Moment at section x in part CB of the arch is given by (please note that B has been taken as the origin for this calculation),

                                  (5)

According to calculus, the necessary condition for extremum (maximum or minimum) is that 

                       (6)

= 40-4x = 0

x = 10 m.

Substituting the value of x in equation (5), the maximum bending moment is obtained. Thus,

M_max = 200 kN.m                                                 (7)

Shear force at D just left of 40 kN load

The slope of the arch at D is evaluated by,

                              (8)

Substituting x =10 m. in the above equation, θ_D = 21.8⁰ 

Shear force S_d at left of D is

S_d = H_a sin θ - R_ay cos θ                                  (9)

S_d = 150sin(21.80) - 80cos(21.80)

= -18.57 kN.

Example 32.4 
A three-hinged parabolic arch of constant cross section is subjected to a uniformly distributed load over a part of its span and a concentrated load of 50 kN, as shown in Fig. 32.9. The dimensions of the arch are shown in the figure. Evaluate the horizontal thrust and the maximum bending moment in the arch.

Solution:

Reactions:

Taking A as the origin, the equation of the parabolic arch may be written as,

y = -0.03x² + 0.6x                                              (1)

Taking moment of all the loads about B leads to,

                         (2)

Taking moment of all the forces right of hinge C about the hinge C and setting M_c = 0 leads to,

                                      (3)

Since there are no horizontal loads acting on the arch,

H_a = H_b = H (say)

Applying ∑Fy = 0 for the whole arch,

85 - 0.15 H + 75 + 0.45 H = 200 

H = 40/0.3 = 133.33 kN                         (4)

From equation (2),

R_ay = 65.0 kN

R_by =135.0 kN                                   (5)

Bending moment
From inspection, the maximum negative bending moment occurs in the region AD and the maximum positive bending moment occurs in the region CB .

Span AD
Bending moment at any cross section in the span AD is

M = R_ay x - H_a (-0.03x² + 0.6 x)          0 < x < 5                               (6)

For, the maximum negative bending moment in this region,

x = 1.8748 m

M = −14.06 kN.m

For the maximum positive bending moment in this region occurs at D ,

M_D = R_ay 5 - H_a (-0.03 X 25 + 0.6 x 5)
= +25.0 kN.m

Span CB
Bending moment at any cross section, in this span is calculated by,

For locating the position of maximum bending moment,

x =17.5 m

M = 56.25 kN.m

Hence, the maximum positive bending moment occurs in span CB.

Summary

In this lesson, the arch definition is given. The advantages of arch construction are given in the introduction. Arches are classified as three-hinged, two-hinged and hingeless arches. The analysis of three-hinged arch is considered here. Numerical examples are solved in detail to show the general procedure of threehinged arch analysis.