Recitation 7Three-hinged

Arches

Kun Zhang | ARCH314 003 | 10.21.2020

Three-hinged Arches DemoProcedure Demo01Lab 6Short Demo, Breakout Rooms02

Recitation 7Three-hinged Arches

University of Michigan, TCAUP Structures I Slide 3of 17

Arch Reactions

Unlike a beam, where reactions are principally vertical, reactions of an arch need to resist horizontal thrust as well.

If a reaction is “pinned” only two forces are needed to describe it. When both reactions are pinned, there are a total of 4 unknowns, and the structure is indeterminate to the 1st degree.

If a reaction is “fixed”, three forces are needed to describe it. When both reactions are fixed, there are a total of six unknown components, and the structure is indeterminate to the 3rd degree.

University of Michigan, TCAUP Structures I Slide 4of 17

3-Hinged Arch

The 3-Hinged Arch has a “hinge” at each pinned support plus one more internally. The internal hinge provides one additional statics equation to be written since the moment at C is known (MC = 0). This makes the system statically determinate.

The solution of the end reactions can usually be obtained in two steps. First by finding the vertical reactions by using the diagram of the whole structure. And second by summing moments at the internal hinge on an FBD of half of the structure to find the horizontal forces.

University of Michigan, TCAUP Structures I Slide 5of 17

Characteristics of a 3-Hinged Arch• Statically determinate – can be calculated with

statics• Movement or settling of foundations will not

alter member stresses• Small fabrication errors in length do not effect

internal stresses• Hinge placement can reduce internal stresses

Gallery of the Machines, 1889 Paris Architect: Ferdinand DutertEngineer: Victor Contamin

University of Michigan, TCAUP Structures I Slide 6of 17

3-Hinged Arch analysis procedure

1. Determine all external loads – find resultants of distributed loads (e.g. wind, snow, dead load)

2. Calculate vertical end reactions –sum moments at each reaction.

3. Draw an FBD of each side of the arch split at the hinge.

4. Find the horizontal reactions – sum moments at hinge.

5. Find internal moments – cut additional FBDs (e.g. at the knees).

Recitation 7Three-hinged Arches

Lab 6Three-hinged Arches Structures I Name 1 _______________________

Arch 314 Name 2 _______________________

Name 3 _______________________

3-Hinged Arches Description

This project compares the moments of a three-hinged arch and a simple beam. Goals

To calculate the knee moment in an arch and a simple beam of the same dimensions. To observe the sign of the moment in the three-hinged arch and the beam. To compare the stability of the three-hinged arch and the beam.

Procedure

1. Write the equation for the knee moment of the arch based on the FBD of ½ of the top span. (sum moments at M)

2. Determine whether the top side of the arch is in tension or compression, 3. Write the equation for the center moment of the beam based on the FBD of ½ of

the span. (sum moments at M) 4. Determine whether the top side of the beam is in tension or compression, 5. Is the arch stable with 3 hinges? 6. If the beam had hinged supports at the base of the column, would it be stable.

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KUN ZHANG kunzh@umich.edu

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