Intro-Three Hinged Arches

8/13/2019 Intro-Three Hinged Arches

http://slidepdf.com/reader/full/intro-three-hinged-arches 1/14

Lecture 14 - of 14

Lecture 14 – Arches

An arch is an extremely efficient structural element suitable for spanning almostany distance. It was probably first used by the Egyptians as early as 2500 BCand was used widely in ancient buildings and structures in Greece and Rome.

Arches do not necessarily have to take the stereotypical “barrel” form. Examplesof arches are shown below:



Lecture 14 - of 14



Lecture 14 - of 14

Types of Arches:

All arches are characterized by a horizontal “thrust” at the base which isinevitable. The following are various types of arches used in architecture.

1. Masonry “ Voussoir” Arch

This type of arch makes use of wedge-shaped “voussoirs” that are fittedtogether to form a curved profile as shown below (also in the 2nd photoabove):

ThrustThrust



Lecture 14 - of 14

2. Masonry “ Corbeled” Arch

This arch-building technique consists of successive courses of masonrylapped and extending closer and closer to the center until they meet asshown below:

ThrustThrust



Lecture 14 - of 14

3. Funicular Arch

A funicular arch takes the shape of an inverted draped string. As thedraped string is in pure tension – the funicular arch is in purecompression.

4. Zero-Hinged Arch (Fixed Arch)

This type of arch is fixed at the base, thereby carrying moment. It isstatically indeterminate. It is typical of arches that are built into wallopenings, or free-standing arches such as the Gateway Arch in St. Louis.

Fixed base

ThrustThrust



Lecture 14 - of 14

5. Two-Hinged Arch

A two-hinged arch is pinned at the supports, which means that there is nomoment. It is commonly used for semi-circular framed members (seeMcDonalds arch photo above). It is also statically indeterminate.

6. Three-Hinged Arch

A three-hinged arch is pinned at the supports as well as the apex. It istypical of A-frames, and pre-engineered rigid steel and glulam buildingframes. It is statically determinate.

Pinned atbase

Pin (hinge) Pin (hinge)

Pinned atbase

ThrustThrust



Lecture 14 - of 14

Analysis o f 3-Hinged Arch

Regardless of the shape or loading of a 3-hinged arch, the analysis is similar.For example, if a 3-hinged arch were to take the shape of a rigid-frame steelbuilding:

The forces in the 3 pinned joints are then:

Cx

Joint “C”

Joint “A” Joint “B”

Ay

Ax

By

Bx

Cy Cy

Cx



Lecture 14 - of 14

The analysis o f a 3-hinged arch involves these steps:

1. Make F.B.D. of frame and break all of the external loads into vertical &horizontal point loads

2. Sum moments at one of the supports (either Joint A or Joint B) to find the

vertical reaction at the opposite support3. Sum forces vertically to find vertical reaction at other support4. Split the frame at Joint C into two isolated halves and pick one of the two

halves to sum moments at Joint C to determine the horizontal reaction atthe support

5. Sum forces horizontally to find horizontal reaction at other support

ExampleGIVEN: The symmetric pre-engineered steel building frame below hasunbalanced snow and wind loads as shown. Each frame is spaced at 20’-0”apart.

REQUIRED: Determine the vertical & horizontal reactions at the supports, i.e., Ay, By, Ax, Bx.

30’-0”

2 1 ’ - 0 ”

5 ’ - 0 ”

1 6 ’ - 0 ”

Bx Ax

Ay By

Snow = 25 PSFSnow = 40 PSF

W i n d = 1 8 P

S F



Lecture 14 - of 14

Step 1 – Make F.B.D. of frame and break all of the external loads into vertical &horizontal point loads:

P1 = 20’(40 PSF)(15’)= 12000 lbs.

P2 = 20’(25 PSF)(15’)= 7500 lbs.

P3 = 20’(18 PSF)(21’)= 7560 lbs.

C

P1 = 12000 lbs. P2 = 7500 lbs.

P3 = 7560 lbs.

7’-6”7’-6”

1 0 ’ - 6 ”

30’-0”

1 8 ’ - 6 ”

Bx Ax

Ay By



Lecture 14 - of 14

Step 2 - Sum moments at Joint A to find vertical reaction By

ΣM A = 0

= 7560 lbs(10.5’) + 12000 lbs(7.5’) + 7500 lbs(22.5’) – By(30’) = 0

By = 11271 lbs.

Step 3 - Sum forces vertically to find vertical reaction Ay

ΣFv = 0

Forces down = Forces up

12000 lbs. + 7500 lbs. = 11271 lbs + Ay

Ay = 8229 lbs.

Step 4 - Split the frame at Joint C into two isolated halves and pick one of the twohalves to sum moments at Joint C to determine the horizontal reaction Ax

ΣMC = 0

= 12000 lbs(7.5’) + 7560 lbs(10.5’) – 8229 lbs(15’) – Ax(21’) = 0

Ax = 2188 lbs.

C

P1 = 12000 lbs.

P3 = 7560 lbs.

7’-6”

1 0 ’ - 6 ”

Ax

Ay = 8229 lbs.



Lecture 14 - of 14

Step 5 - Sum forces horizontally to find horizontal reaction Bx

ΣFh = 0

Forces from left = Forces from right

7560 lbs = 2188 lbs. + Bx

Bx = 5372 lbs.

Summary:

Usually, multiple analyses are performed on a 3-hinged rigid steel frame such asthe one in this example for various LOAD COMBINATIONS as obtained from theInternational Building Code. The results of the analysis from each loadcombination are then tabulated. The design of the base structural elements suchas footings, anchor bolts and tension rods (or “hairpins”) is based on the largestloads obtained from the analyses.

Bx = 5372 lbs. Ax = 2188 lbs.

Ay = 8229 lbs. By = 11271 lbs.



Lecture 14 - of 14

A typical metal building baseplate and foundation system is shown below.

Intro-Three Hinged Arches

Documents

Transcript of Intro-Three Hinged Arches