Group 4, Lab 9
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California State University, Northridge *
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Feb 20, 2024
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Fall 2023
California State University, Northridge
Experiment 9
Two Pinned Arch
Mechanics Lab
AM 317
Written By:
Abstract:
Thrust force refers to the horizontal force exerted by a bridge or an arch on its supports. For engineers, it is critical to understand the stability and equilibrium states of arches and bridges
because it determines the load that they can carry.
One of the main things to consider when analyzing arches is the thrust line, the thrust line
is a line along which the resultant thrust force acts withing the arch. It is not a physical line that is there but on that the forces in the arch act around. The thrust line should be within the middle third of the arch thickness to ensure stability and overturning. Another thing to consider is the stability of an arch. Arch stability depends on the distribution of forces and the construction of the arch. The thrust force must stay within the supports of the arch to prevent it from collapsing. In order to account for arch stability, the foundations of the arch are required to resist the horizontal component of the thrust force. The main consideration that we are going over in the lab is the direction and force of the thrust force. Thrust force on an arch typically pushes horizontally away from the center of the arch and pushes outwards against the supports. In a well built arch, the vertical component and horizontal components are resolved. The vertical components supports the weight of the arch itself, as well as the applied loads. The horizontal component pushes outwards against the supports of the arch.
In the lab, we tested the thrust forces of a fully loaded arch and an arch where readings were taken at key points. What we found was a very similar outcome to what we hoped to see when we calculated the thrust force and when we measured the thrust force.
Introduction:
Analyzing thrust forces in arches is fundamental in the field of engineering and structural engineering. In order to design a same and stable arch, engineers use a combination of analytical and measured methods to ensure that an arch can support its own weight and the loads without experiencing deformations or even failure. One of the main things to consider when analyzing arches is the thrust line, the thrust line
is a line along which the resultant thrust force acts withing the arch. It is not a physical line that is there but on that the forces in the arch act around. The thrust line should be within the middle third of the arch thickness to ensure stability and overturning. Another thing to consider is the stability of an arch. Arch stability depends on the distribution of forces and the construction of the arch. The thrust force must stay within the supports of the arch to prevent it from collapsing. In order to account for arch stability, the foundations of the arch are required to resist the horizontal component of the thrust force. In this lab we hope to find the correlation with the calculated thrust force and the measured thrust force. We will be using the following equipment, the structures test frame, in order to hold the frame of all the equipment, the digital force display and the power supply in order to get a reading of the forces applied to the structures test frame, the aluminum arch itself, in order to actually apply the loads, the hangers and weights, in order to apply the loads onto the aluminum arch, and the scale, which will read out the forces and transfer its readings to the force display.
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Using these items and the formulas in the theory section, we hope to see a correlation between the measured values and the calculated values.
Theory:
There are a handful of equations that we need to know in order to find the calculated thrust forces for this lab. Because this is a controlled experiment, the equation for finding the theoretical thrust force is constant, however, in the real world the equation may become more complex. The equation for finding the horizontal thrust force for a given load is as follows, where H
B
, the horizontal thrust reaction at B (N). is equal to 5 times P, the load (N), times x, the load location, distance from the left-hand side support (m), over 8 times r, the rise of the arch in meters, times L
3
, the span of the arch (m) cubed, all times L
3
again, plus x
3
minus 2 times L times x
2
.
H
B
=
5
Px
8
r L
3
(
L
3
+
x
3
−
2
L x
2
)
The equation above works for the first part of the lab, where the weights were applied one at a time and changed places depending on where the load was to be measured. That was done to simulate a single large load placed on different parts of the arch. The next equation is for when all of the hangers are loaded with weight and simulating an evenly distributed load. The equation for finding the horizontal thrust load when the load is equally distributed is as follows,
H
B
, the horizontal thrust reaction at B (N), is equal to w, the intensity of uniform distributed loads (N/m), times L
2
, the span of the arch (m) squared over 8 times r, the rise of the arch (m). Using those two formulas along with the measurements of the arch, we hope to both measure and calculate the horizontal thrust load on the arch. We also expect to see a curved graph that should be similar between loads.
Procedures:
The procedures in the lab were quite simple but tedious and time consuming. Overall though, we were able to complete the lab without much hassle, and where we did run into issues,
we were able to resolve them quickly.
To start the lab, we first had to plug in the power supply into the force display. We then measured the dimensions of the two-pinned arch. We took the measurements of its length and its height at its apex and recorded them in our table to later calculation. We then went on to load the hangers for the experiment. Initially, all the weights and the hangers were haphazardly placed back into the supply drawer, so it was a bit of an issue sorting that out. Once we had gotten the supplies sorted, we began loading the weight pan. The first part of the experiment had us loading the mid span hanger with a 500-gram preload. We loaded the weight pan and placed it on the mid span hanger. We then zeroed the digital force reader and prepared more weight pans, this time with 100 grams total. We then, with the 500-gram preload still on the midspan hanger, placed the 100-gram weight at each of the hangers, recording what the digital force reader read out. While recording we noticed that as we placed and removed the weight pans, the force of us moving the weight pans would affect the zero reading on the digital force scale. To mitigate this, we ended up zeroing the digital force scale each time a weight pan was removed. Doing this led us to having more consistent reading. Once we had finished this process, we repeated it for a 500-gram weight and recorded the results.
For the next step, we removed the 500-gram preload and loaded all of the hangers at once
with a 70-gram load. Doing this was tedious as we had to count and load all of the weight pans.
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We made sure to zero the digital force reader while there were no weights on an of the hangers. Once all the 70 grams weights were installed, we recorded the digital force reading and repeated the procedure with 140-gram weights. Once all the testing was done, we unplugged all the equipment and put everything back neatly into the supply drawer.
Results:
Table 1: Arch Data
Rise of Arch, r (m)
0.10
Length of arch, L (m)
0.51
Table 2: Weight Data
Added mass (g)
100
500
Hanger mass (g)
15
15
Total mass
115
515
Total load (N)
1.12776475 5.05042475
Table 3: Data for a point load moving across a two-pinned arch.
Load
location
x (m)
Measured thrust
force
for 100 g load (N)
Calculated thrust
force
for 100 g load (N)
Measured thrust
force
for 500 g load (N)
Calculated thrust
force
for 500 g load (N)
0
0.00
0
0.00
0
0.05
0.20
0.345983757
1.10
1.549405521
0.1
0.40
0.6559679448
1.70
2.937595579
0.15
0.60
0.9012591797
2.20
4.036073718
0.2
0.70
1.061134462
2.50
4.75203694
0.25
1.00
1.122841178
2.70
5.02837571
0.3
0.70
1.081597096
2.60
4.843673951
0.35
0.60
0.9405903696
2.00
4.212209046
0.4
0.40
0.7109795375
1.60
3.183951842
0.45
0.20
0.4118935219
1.10
1.844566642
0.5
0.00
0.07043162962
0.00
0.3154112109
Table 4: Data for a uniformly distributed load
Total
mass
(g)
Total mass
of hangers (g)
Uniform distributed
load (N/m)
Measured
thrust force (N)
Calculated
thrust force
(N)
% Error 630
745
14.32540049
4.6
4.657545834
1.235539841
1260
1395
26.82407206
8.3
8.721176428
4.829353374
Graph 1: Measured Thrust Force (100g)
0
0.1
0.2
0.3
0.4
0.5
0.6
0
0.2
0.4
0.6
0.8
1
1.2
Measured Thrust Force (100g)
Location
Thrust Force
Graph 2: Calculated Thrust Force (100g)
0
0.1
0.2
0.3
0.4
0.5
0.6
0
0.2
0.4
0.6
0.8
1
1.2
Calculated Thrust Force (100g)
Location
Thrust Force
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Graph 3: Measured Thrust Force (500g)
0
0.1
0.2
0.3
0.4
0.5
0.6
0
0.5
1
1.5
2
2.5
3
Measured Thrust Force (500g)
Location
Thrust Force
Graph 4: Calculated Thrust Force (500g)
0
0.1
0.2
0.3
0.4
0.5
0.6
0
1
2
3
4
5
6
Calculated Thrust Force (500g)
Location
Thrust Force
Discussion and Conclusions:
For the most part, we saw in our graphs and tables the results that we expected to see from the lab.
In Table 1, we recorded the values of the rise of the arch in meters as well as the length of
the arch in meters. While the length of the arch was easy enough to measure, the rise of the arch was a little difficult to measure as we were measuring the top of the arch to the “bottom” which was empty space below the middle. To remedy this, we placed a ruler at either end of the supports and measured the vertical distance to from the top of the arch to the meter stick. In Table 2, we recorded the values that we got from the first part of the lab, using one hanger and multiple different points. There we also found the weight of the hanger and took that into account when doing calculations.
In Table 3, we recorded our measured values as well as our calculated values. For the most part, our calculated values were higher than our measured values. They did, however, see a similar rise and fall with as the location on the two pinned arch changes. The graphs of the measured and calculated values are later in the report. In Table 4, we recorded the values of force when the arch was fully loaded. Our values for each were very close. The lighter loaded arch had a 1.24% error while the heavier loaded arch
had a 4.83% error.
In the graphs based on Table 3, we saw a general curve on all the locations of the arch. In all the cases save for one in the measured thrust force for the 100-gram load, we saw a consistent
curve.
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Appendix:
Related Documents
Related Questions
Physics 121 Spring 2021 - Document #11: Homework #04 & Reading Assignment page 4 of 8
Problem 1: Gnome Ride - This from a Previous Exam
I.
A Gnome of given mass M goes on the Gnome Ride as follows: He stands on a horizontal
platform that is connected to a large piston so that the platform is driven vertically with a position
as a function of time according to the following equation:
y(t) = C cos(wt)
Here w is a constant given angular frequency, C is a given constant (with appropriate physical
units) and y represents the vertical position, positive upward as indicated.
Part (a) - What is the velocity of the Gnome at time t = 0? Explain your work. Present your
answer in terms of the given parameters
Part (b) – What is the net force on the Gnome at time t = 0? Explain your work. Present your
answer in terms of the given parameters
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9:EV IA 0 O
46
Engineering Mechanics I / Fir.
docs.google.com
Q4: (B)\\ Each propeller of the twin-
screw ship develops a full speed thrust
of 300 kN. In maneuvering the ship, one
propeller is turning full speed ahead and
the other full speed in reverse. What
thrust P must each tug exert on the ship
to counteract the effect of the ship's
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- 50 m-
F
12 m
F
120 m
None
51.4
41.5
15.4
of the
KN
KN
KN
above
P =
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- ID
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R.
Actor
Sandbag
SOLUTION
Conceptualize Look back at the Example "A Grand Entrance," where we analyzed a cable used to support an actor as he swung
onto the stage. We ignored any stretching of the cable there, but we wish to address this phenomenon in this example.
Categorize We perform a simple calculation involving the equation
F
A
Y =
AL
Li
so we categorize this example as
--Select---
v problem.
Solve the Young's modulus equation for the cross-sectional area of the cable:
FL;
A =
YAL
Assuming the cross section is circular, find the diameter of the cable from d = 2r and A = ar (Use the following as necessary: F, Y,
Li, AL, and r.):
FL
A
d = 2r = 2 - = 2
Substitute numerical values. (Enter your answer…
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foot in one minute. Power, in
physics, is defined simply as
the rate of doing work. The
formula below gives the
horsepower at 5,252 radians
per second.
https://philkotse.com/toyota-corona-ior-sale-in-baguio/1991-for-sale-in-aid7017151
625T
1313
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File | C:/Users/ignor/Downloads/Assignment%203%20(2).pdf
T❘ Read aloud
Draw
Ask Copilot
a) Use Newton's second law
b) Use the Torque equation
a
Search
TWO
Particle P is sliding along a circular path of radius r . The circular path is placed on top of a
truck that is moving to the left with a velocity of v and an acceleration of a. The friction is
negligible. Find the EOM of particle P.
O
0
2
r
of 6
| CD
P
eac
СО da | +
o
ENG
60
T
6L
N
{0}
7:39 PM
1/17/2024
x
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I thought the letters represented the connection between the forces and magnitude. Can you help me truly understand how to evaluate this correctly? Thanks
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Instead of 0.q, you have to use my student ID
using the last 3 digits. If the last 3 digits of my
student ID are 495.q use 495 instead.
Question 2: Equivalent Nodal Forces of 2-D Plane Finite Elements (
For the four-node linear plane element shown below with a uniform surface traction along its side 2-3.
You are required to evaluate the force matrix by using the energy equivalent nodal forces. Let the thickness of
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T = 20 + 0.q N/mm2
(0, 4)
(5, 4)
(0, 0)
(8, 0)
1
2
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Note:
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