Structural Analysis of Three-hinged Arch

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Arches can be of three types: hingless arch, two-hinged arch and three-hinged arch. Among the three, the three-hinged arch is statically determinate. The distinguishing factor between an arch and a beam is that when a load is applied on a beam, it results into vertical reactions and moments. But when same load is applied to an arch, the arch resists the load by vertical and horizontal components of reaction. The horizontal components of the reaction in effect reduces the moment from that in a simple beam. Thus, the arch supports loads with less moment than a simple beam but considerable stresses are generated in the arch. A three hinged arch which is statically determinate is analysed by the basic equations of statics. The key to this analysis is the fact that there is zero moment of the hinge located out in the span of the arch.

Example

Determine the reactions and construct the moment diagram of a three hinged arch shown below

Solution

Consider the free body diagram of the arch shown below

There are 4 unknowns but we have only three equations of equilibrium. An additional equation must therefore be obtained if the problem is to be statically determinate. This additional equation is obtained from the condition of connection at B. because the connection at B is hinge, the summation of the moment at that point must be zero.

Determination of vertical reactions

Taking moments about point A, we obtain,

(40 x 3) + (60 x 13) + 2 H_C – 17 V_C = 0

>> 2 H_C – 17 V_C = -900                        (1)

Consider the segment of the arch between B and C and take moment about point B for this segment

>> (60 x 6) + 6 H_C – 10 V_C = 0

Or, 6 H_C – 10 V_C = -360                      (2)

Solving equations (1) and (2) simultaneously,

2 H_C – 17 V_C = -900                           (1)

6 H_C – 10 V_C = -360                           (2)

This gives V_C = 57.073 kN; H_C = 35.12 kN

∑F_x = 0 >> H_A – H_C = 0 >> H_A = H_C = 35.12 kN

∑F_y = 0 >> 40 + 60 – V_A – 57.07 = 0 >> V_A = 42.93 kN

Determination of moments

M_A = M_B = M_C = 0

Moment about the 40 kN load = (42.93 x 3) – (35.12 x 3) = 23.43 kNm

Moment about 60 kN load = (-57.03 x 4) + (35.12 x 4) = -87.03 kN

See Figure 3 below for the representation of the reactions

Moment diagrams for arches are commonly constructed on the axis of the arch with the magnitude of moment plotted in a vertical direction as shown in Figure 4 below

The shaded area enclose the arch and the line of moment