three-hinged archstatics.marcks.cc/mod17/pdf/mod17.pdf · as previously mentioned, the three-hinged...

THREE-HINGED ARCH

1

Gateway ArchSt Louis, MS

As previously mentioned, the three-hinged arch is a special class of a simple frame. It consists of two multi force members hinged at ‑their supports and connected at the apex. The frame may be ground mounted or it may be suspended overhead

The three-hinged arch is stable only if both supports are hinges. If one hinge was replaced with a roller, it would collapse. With two hinges, the structure is externally indeterminate. That is, one cannot find all reactions with only a FBD of the whole structure. It must be disassembled to find all four reactions

2

Three-Hinged Arch

Apex

Hinge Hinge Apex

Hinge Hinge

A variation of the three-hinged arch is for the supports to be at different elevations. This can impact how the problem is solved

When the supports are at the same elevation, both y components of ‑reaction are found using the FBD of the frame as a whole. A FBD of either member results in a statically determinate FBD allowing relative ease of complete solution

When the supports are at different elevations, a FBD of the whole structure results in an unsolvable set of three equations. It will be necessary to develop a second FBD and solve all equations simultaneously.

These points are illustrated in the following examples

3

Three-Hinged Arch

Consider the three hinged arch below. It supports a pulley with a radius of 3” and lifting a weight of 60 lb

f. Determine all pin reactions

Take note of the following characteristics and assumptions: Neither member comprising the frame is a two-force member Supports 'A' and 'F' are at the same elevation Line tension is constant around the pulley The system is frictionless

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Example 2 - Three-Hinged Arch

60 lbf

CD

A 12”

3”

B

E

12”

3

5”

5


60 lbf

CD

Ay

12”

3”

Bx

E

12”

3

5”

By

Ax

FBD of whole structure

M B = 0 = A y 24−60 3A y=7.5 lb f

F y = 0 =−7.5−60B yB y=67.5kips

F x = 0 = B x−AxAx=B x

A visual analysis of the structure shows one can draw a FBD of either half of the structure. However, each FBD will have four unknowns

However, even though a FBD of the whole structure also has four unknown reactions, one can solve for the y-components prior to disassembling the structure

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FBD of left member

M D = 0 = 7.5 12−60 3Ax 8

Ax = 11.25 lb f

From previous FBD:Bx = Ax

B x=11.25 kips

Continue with current FBD:F x = 0 = 60−11.25−D x

Dx=48.75 lb f

F y = 0 = D y−7.5D y = 7.5 lb f

60 lbf C

7.5 lb

12”

3” 3

5”

Ax

Dy

Dx

Check solution of internal reactions

M C = 11.25 5−48.75 37.5 12 = 0 CHECKS

The next step is to disassemble the structure and choose a subsystem that allows one to complete the solution

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Check solution of external reactions

M D = 7.5 1267.5 1211.25 8−11.25 8−60 15 = 0 CHECKS

60 lbf

CD

7.5 lbf

12”

3”

E

12”

3

5”

67.5 lbf

11.25 lbf11.25 lbf

Consider the three hinged arch below. It supports a cylinder with a radius of one foot and a weight of 4800 lb

f. Determine all pin

reactions

The weight of the cylinder acts through its center. By inspection, one can determine there are reactions at pins A, C, and E and applied forces at B and D due to the weight of the drum; neither member is a two-force member. In fact, both are three-force members

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4800 lbf

C

B

D

A

E4.5'

6'

7.5'

1'

There are four external unknown reaction components at supports A and E and the support reactions are at different elevations. This is a primary difference between this example and the previous example.

The method of members is required to solve this problem. So what is the approach to solving this problem?

The steps we might take are as follows: Draw a FBD of the whole structure and solve as many unknowns as

possible Disassemble the structure and draw the FBD of each of the two members Determine the forces acting at points 'B' and 'D' due to the weight of the

drum Solve for the remaining unknowns by writing the equations of equilibrium

for each member

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This results in three equations and four unknowns, none of which can be solved. The structure must be disassembled to finish solution. To do so requires finding the force components of the weight vector acting perpendicular to the members as well as the location of points 'B' and 'D'

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4800 lbf

C

B

D

Ax

Ex

4.5'

6'

7.5'

W

Ey

Ay

θ

M A = 00 = E y 6−E x 3−4800 1

F x = 00 = Ax−E x

F y = 00 = A yE y−4800

FBD of whole structure

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4800 lbf

C

B

D

Ax

Ex

4.5'

6'

7.5'

W

Ey

Ay

θ

= arctan 4.56.0 =36.86o

∢BCD = 90− = 53.12o

∢WCD =∢WCB =∢BCD

2= 26.56o

CD=WD

tan 26.56o= 2 ft

BC = CD = 2 ft

Locate points 'B' and 'D'

Resolve 'W' into forces 'B' and 'D'

B

D

W

4800 lbf

θ

FBD of Drum

4800 lbf

D

B

Force Polygon

θ

From above, = 36.86o

cos =4800D

D = 6000 lb f

tan =B

4800B = 3600 lb f

This information allows development of the member FBDs

At this point, we still do not know the values of the y-component of reactions at 'A' and 'C'. Nor do we know the reactions at 'E'. We can find these unknowns in one of two ways: Draw a FBD of member CD and solve Return to the FBD of the whole structure and solve

Since we've already developed a FBD of the whole structure, let's use it to find the remaining unknowns

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3600 lbf

Cx

B

Ax

5.5'

Ay

Cy

2.0'

M A = 00 = C x 7.5−3600 5.5

C x = 2640 lb f

FBD of vertical member AC

F x = 00 = Ax2640−3600Ax = 960 lb f

F y = 00 = A y−C y

A y = C y

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4800 lbf

C

B

D

960

Ex

4.5'

6'

7.5'

W

Ey

Ay

θ

From the FBD on page 20 we know:

F x = 00 = Ax−E x

. . . E x = 960 lb f

M A = 00 = E y 6−960 3−4800 1

E y = 1280 lb f

F y = 00 = A y1280−4800A y = 3520 lb f

Since we also know that A y=C y ,then C y=3520 lb f based on the FBD on page 20.

C x = 2640 lb f

Ax = 960 lb f

A y = C y

Then, from the FBD on page 18:

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4800

C

B

D

960

960

4.5'

6'5.5'

W

1280

3520

θ2.0'

Check the solution of external reactions

M w= −3520 1−960 5.51280 5960 2.5

M w= 0 CHECKS

Check the solution of reactions at 'C'

The value of 'C' was used to determine some externalreactions. Since the check above showed the reactionsto be valid, the reactions at C are likely correct. However,the following is an explicit check for the reactions at 'C'.

M E = 6000 5.5−3520 6−2640 4.5

M E = 0 CHECKS

6000

E

D

960

4.5'

6'

1280

2640

3520

2'

5.5'

three-hinged archstatics.marcks.cc/mod17/pdf/mod17.pdf · as previously mentioned, the three-hinged...

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