three-hinged archstatics.marcks.cc/mod17/pdf/mod17.pdf · as previously mentioned, the three-hinged...
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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
Example 2 - Three-Hinged Arch
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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Example 2 - Three-Hinged Arch
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
7
Example 2 - Three-Hinged Arch
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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Example 3 - Three-Hinged Arch
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
9
Example 3 - Three-Hinged Arch
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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Example 3 - Three-Hinged Arch
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
11
Example 3 - Three-Hinged Arch
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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Example 3 - Three-Hinged Arch
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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Example 3 - Three-Hinged Arch
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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Example 3 - Three-Hinged Arch
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'
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