Staircase Design Report
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Transcript of Staircase Design Report
Contents Need .............................................................................................................................................................. 3
Specs ............................................................................................................................................................. 3
Concepts ....................................................................................................................................................... 4
Introduction .............................................................................................................................................. 4
How it works ............................................................................................................................................. 6
Features .................................................................................................................................................... 6
Analysis & Optimization ................................................................................................................................ 7
Evaluation ................................................................................................................................................... 22
References .................................................................................................................................................. 22
Need Preliminary Need:
There is a need for easy access between two floors within a household.
Design Considerations:
1. Safety
2. Standards and building codes
3. Load rating
4. Overall size
5. Stress conditions (nominal worst case)
6. Overall Factor of Safety
7. Ease of use
8. Material choice
9. Ease of manufacture
10. Shape and size (geometry)
Specs Performance Specs
1. Capacity (max weight of user): 300 lbf at the center of the step
2. Constraint imposed:
a. 270Β° sweep for garden
3. Staircase Code:
a. Overall height: 10ft
b. Minimum step length: 26in
c. Maximum rise between steps: 9.5 in
d. Each tread having a 7.5 inch minimum tread depth at12 inches from
the narrower edge.
e. Minimum headroom of 6β6ββ
Concepts
Introduction
The design concept is a fairly simple design composing of 12 steps constructed out of
steel with a rise on 9.5 in. Each step in the concept had on overall step length of 26 in and
a tread depth of 8.04 in at a distance of 12 in from the narrower edge. The overall
assembly is broken down into four parts: The steps/center pole, the under step bracket
(including two ΒΎβ-10 fasteners), the supporting pipe and the pole bracket (including two
ΒΎβ-10 fasteners). The center pole diameter is 6 in
step
Support pipe
Pipe to pole
bracket
Center pole
Figure 1: Step assembly
Figure 2: Bracket Assembly
Figure 3: Step assembly exploded
ΒΎβ-16 bolts
ΒΎβ-10 bolts
ΒΎβ-16
nuts
ΒΎβ-10 nuts
Under step
bracket
How it works
Each step is supported at the line of action of the applied nominal force of 300 lbf by a
bracket using two ΒΎββ-10 fasteners. The bracket is welded to a 2ββ steel pipe with a 90Β°
elbow (that supports the weight of the step user) that is then welded to another bracket
(that supports the bending moment created by the step user). This bracket attaches the
entire step assembly to the 6β diameter center pole by use of two ΒΎββ-10. There is 12
steps total.
Features
Due to the pre-welding performed by the manufacturer, using this design installing each
step is as easy as attaching the under step bracket to each step using two provided
fasteners and then attaching the pole bracket to the center pole using the two provided
fasteners. Also, the all steel construction is very durable.
Figure 4: Isometric view of
staircase
Figure 5: Front view of staircase
Analysis & Optimization
The purpose of the following analysis is to determine whether or not our design will
support the given nominal load of 300 lbf. With a factor of safety of n > 1. The line of
action of the applied nominal load will act in the middle of the step (half the distance to
the step edge.
In the following, the staircase is going to be analyzed using free body diagrams. Free
body diagrams will allow us to solve for unknown reaction forces, moments, bolt forces,
weld stresses that will occur at these critical locations. This will allow us to ensure that
we have created a safe and reliable design. We will choose two critical locations to
analyze, the weld that connects the pipe to the pole mounting bracket and the weld right
before the elbow on the pipe.
critical locations
Table 1
Location Failure concern Rational
a buckling (compressive) Weaker material, high contact stress
b tear (tensile) High bending moment , weaker material (weld)
c tear (tensile) High bending moment , weaker material (weld)
d tear (tensile) High bending moment, high shear stress, small
cross sectional area
f buckling (compressive) High contact stress, small cross-sectional area
loads on parts, free body diagrams and load verification
In the above free body diagram, we can find the resultant moment due to the
nominal load of 300 lbf by performing the following.
β πΉπ¦ = 0
300 πππ β 300 πππ = 0
14 in Mp
300 lbf 300 lbf
Figure 7: Free body diagram of the Step
Figure 6: critical locations
a
b c
e d
β ππ = 0
ππ β (14ππ)(300πππ ) = 0
ππ = 4200 πππ β ππ
Therefore, we got a reactionary moment of 4200 lbf-in.
Next, we want to ensure that the chosen fasteners are going to be strong enough
to support our nominal load. This is done below.
β πΉπ¦ = 0
ππ1 + ππ2 β 300 πππ = 0
Assuming Pb1=Pb2
2πππ‘ = 300
ππ1 = ππ2 = 150 πππ
The fasteners being used are ΒΎβ-16 that are constructed out steel with the following properties
π΄π = ππ2 = π(0.375)2 = 0.4418ππ2
π΄π‘ = 0.3345ππ2
πΏπ = 1.75ππ
ππ = 1.75ππ
πΏ = 2.25ππ
Pbt=Pb1+Pb2
300 lbf
Figure 8: Free body diagram of pipe to step bracket
ππ‘ = π β ππ = 0.75ππ
πΈ = 30 πππ π
First, we found the member stiffness constant πΎπ
πΎπ =0.5774ππΈπ
2ln (5(0.5774π + 0.5π)0.5774π + 2.5π
πΎπ =0.5774π(30πππ π)(0.75ππ)
2ln (5(0.5774(1.25ππ) + 0.5(0.75ππ))0.5774(1.25ππ) + 2.5(0.75ππ)
πΎπ = 27.30 ππππ
ππβ
Next, we found the bolt stiffness constant πΎπ
πΎπ = π΄ππ΄π‘πΈ
π΄πππ‘ + π΄π‘ππ
πΎπ = (0.442ππ)(0.373ππ)(30πππ π)
(0.373ππ)(0.75ππ) + (0.373ππ2)(0.5ππ)
πΎπ = 9.55 ππππ
ππβ
Then we found the joint stiffness constant C
πΆ =πΎπ
πΎπ + πΎπ
πΆ =9.55
ππππππβ
9.55 ππππ
ππβ + 27.30 ππππ
ππβ
πΆ = 0.2592
Then we calculated the desired preload πΉπ of the nut
πΉπ =3
4πππ΄π‘
πΉπ =3
4(120 πππ π)(0.373 ππ2)
πΉπ = 33.57 ππππ
Then the fastener torque T
π = 0.2πΉππ
π = (0.2)(33.57 ππππ)(0.75ππ)
π = 5.036 ππππ β ππ
Then since we have 2 bolts the P value is:
π =300 πππ
2 ππππ‘π
π =150 πππ
ππππ‘
From this we can calculate the bolt force πΉπ
πΉπ = πΉπ + π
πΉπ = 33.57 ππππ + 0.150 ππππ
πΉπ = 33.72 ππππ
Then we calculated the yield factor of safety
ππ =πππ΄π‘
(πΆπ + πΉπ)
ππ =(120 πππ π)(0.373ππ2)
(0.2592)(0.150ππππ) β (33.72ππππ)
ππ = 1.33
Since this ππ>1, the fasteners are strong enough to support the 300 lbf load. Next we calculated
the load factor ππΏ
ππΏ =πππ΄π‘ β πΉπ
πΆπ
ππΏ = (120 πππ π)(0.373ππ2) β (33.72 ππππ)
(0.2592)(0.150ππππ)
ππΏ = 284
Finally, we calculated the joint separation factor ππ
ππ = πΉπ
π(1 β πΆ)
ππ = (33.72 ππππ)
(0.150 ππππ)(1 β 0.2592)
ππ = 303
Since ππ > 1 we know that the bracket and the step are not going to separate. Next we analyze
the weld that connects the supporting pipe to the pole bracket to determine whether or not the
weld is going to be strong enough for the given nominal load.
Since the pipe is in pure bending we will use the following formulas for a weld experiencing a
bending stress. πβ² is the horizontal shear stress and π" is the vertical shear stress. The horizontal
shear stress is calculated using the following formula
πβ² =π
π΄
Where V is the shear force and A=1.414πβπ (h=height of weld (0.25in), r= radius of pipe (1in)
π" π"
πβ²
πβ²
πβ²
π" π"
πβ²
V=300 lbf
Figure 9: Free body diagram of the support pipe weld c
πβ² =300 πππ
1.1106ππ2
πβ² = 270 ππ π
To calculate the vertical shear stress, we use the following formula where M=the bending
moment, c=pipe radius and I=1.414βπΌπ’ where h= height of the weld and πΌπ’ = ππ3
π" =ππ
πΌ
π" =(4200 πππ π₯ ππ)(1ππ)
3.14159ππ2
π" = 1204 ππ π
Then we calculated the magnitude of π
|π| = βπ‘β²2 + π"2
|π| = β2702 + 12042
|π| = 1234 ππ π
With the |π| value calculated we can find the factor of safety ππ (ππ¦ was found assuming we
used weld electrode E60xx )
ππ = ππ¦
π=
50,000 ππ π
1234 ππ π= 40.5
Since ππ >1 the weld will not fail. Next we determined whether or not the member force could
carry our nominal load (since we donβt want to load the fasteners in shear).
To start off with, we assumed that both bolt forces are equivalent (ππ3= ππ4
). Because of this we
found an equivalent distance of the two bolts that is equidistant from the two true bolt distances.
This was done using the following formula.
ππ =3.125ππ + 1.125ππ
2= 2.125ππ
After we found the equivalent bolt distance ππ then we summed the moments around point A
β ππ΄ = 0
β(14 ππ)(300 πππ) + (2.125ππ)(πππ‘) = 0
πππ‘= 1976 ππ π
Since we are assuming that ππ3= ππ4
we can divide πππ‘ by 2
ππ2= ππ2
=πππ‘
2 ππππ‘π =
1976 ππ π
2 ππππ‘π = 988 ππ π
We can calculate the reaction force P caused by the bending moment created by the 300 lbf and
the two bolts by using the following formula.
π =ππΏ
ππ=
(300 πππ)(14ππ)
(2.125ππ)= 1976 ππ π
Figure 10: Free body diagram of the entire step assembly
Pb3
3
Pb4
1.125in
3.125in
A
300 lbf 14 in
P
Since we are trying to determine whether the bolts we chose are strong enough for the nominal
load we calculated the preload πΉπ. The value of ππ was a look up value.
πΉπ =3
4πππ΄π‘
πΉπ =3
4(120 πππ π)(0.373 ππ2) = 33.57 ππππ
Then we calculated the joint stiffness constant C
πΆ =ππ
π
πΆ =988 ππ π
1976 ππ π= 0.5
Next, we calculated the force applied to the member ππ
ππ = (1 β πΆ)π = (1 β 0.5)(1976 ππ π) = 988 πππ
Then we calculated the force in the member πΉπ
πΉπ = ππ β πΉπ = 0.988 ππππ β 30.1 ππππ = β29.1 ππππ
Since πΉπ < 0 we know that the member is in compression.
Next we have to see if the |π| (0.3 klbf ) < ππ |πΉπ| because if this criteria isnβt met, then the pipe
to pole bracket will fail
ππ |πΉπ| = (0.25)(29.1) = 7.3 ππππ
0.3 ππππ < 7.28 ππππ
The above inequality shows the step is going to be supported by only the frictional force between
the bracket and the pole and the fasteners are only being used to clamp the two parts together and
thus experience no shear stress.
For the factors of safety we can calculate the bolt force πΉπ
πΉπ = πΉπ + π
πΉπ = 30.1 ππππ + 0.150 ππππ
πΉπ = 30.65 ππππ
Then calculated the yield factor of safety
ππ =πππ΄π‘
(πΆπ + πΉπ)
ππ =(120 πππ π)(0.3345ππ2)
(0.5)(1.976ππππ)(30.1ππππ)
ππ = 1.29
Since this ππ>1, the fasteners are strong enough to support the 300 lbf load. Next we calculated
the load factor ππΏ
ππΏ =πππ΄π‘ β πΉπ
πΆπ
ππΏ = (120 πππ π)(0.3345ππ2 ) β (30.1 ππππ)
(0.5)(1.976ππππ)
ππΏ = 10.15
Finally, we calculated the joint separation factor ππ
ππ = πΉπ
π(1 β πΆ)
ππ = (30.5 ππππ)
(0.150 ππππ)(1 β 0.5)
ππ = 30.5
We can calculate slip factor of safety
ππ = π|πΉπ|
|π|=
(0.25)(29.1 ππππ)
(0.300 ππππ)= 24.3
Since all the factors of safety are >1 the bolts used to fasten the pipe to pole bracket will not fail.
and that the shear force within these bolts is going to equal 0.
Next we are going to analyze the weld that occurs right before the 90Β° elbow on the pipe. We use
the same FBD as the previous weld as only the applied bending moment changes
Figure 11: Free body diagram of the support pipe weld
h right before the 90Β° angle
Note: πβ² = horizontal shear stress and π" = vertical shear stress
As before, this weld is experiencing pure bending. The horizontal shear stress is calculated using
the following formula
πβ² =π
π΄
Where V is the shear force and A=1.414πβπ (h=height of weld (0.25in), r= radius of pipe (1in)
πβ² =300 πππ
1.1106ππ2
πβ² = 270 ππ π
To calculate the vertical shear stress, we use the following formula where M=the bending
moment, c=pipe radius and I=1.414βπΌπ’ where h= height of the weld and πΌπ’ = ππ3
π" =ππ
πΌ
π" π"
πβ²
πβ²
πβ²
π" π"
πβ²
V=300 lbf
π = (2ππ)(300 πππ) = 600 πππ π₯ ππ
π" =(600 πππ π₯ ππ)(1ππ)
3.14159ππ2
π" = 191 ππ π
Then we calculated the magnitude of π
|π| = βπ‘β²2 + π"2
|π| = β2702 + 1912
|π| = 331 ππ π
With the |π| value calculated we can find the factor of safety ππ (ππ¦ was found assuming we
used weld electrode E60xx )
ππ = ππ¦
π=
50,000 ππ π
331 ππ π= 151
Since ππ >1 the weld will not fail.
Solid Works verification of load capacity
A static simulation was performed in Solid Works on the bracket assembly represented by Figure
2. To begin, the shell features of the connecting pipe were set to a thickness of 0.154β. This is
the wall thickness of a 2β schedule 40 steel pipe. Next the default global connection bond was
deleted in order to define the exact weld connections. Edge welds were placed on the pipe at
critical location points c, h, and a in Figure 3. Edge welds at c and a were 0.25β single fillet
welds with an E60 electrode, the weld at h was a 0.25β grove edge weld also with an E60
electrode. The bracket assembly was fixed by using a βfixed geometryβ fixture on the back face
of the curved bracket that connects the pipe to the central staircase pole. Finally a 300lb vertical
forced was placed on the top face of the under-step bracket to simulate the force of someone
stepping on the nominal worst case of the step.
Table 2
Variable Weld c Weld h
Shear (psi) π 988.1 β 400
Factor of Safety n 4.2 4.2
Figure 12 illustrates a section clipping of the cross section between the weld c and the curved bracket. Three total
points were probed where the weld is located giving an average shear of 988.1 psi.
Figure 12
Figure 13
Figure 13 illustrates the section clipping where weld h is located. The geometry of the thin pipe restricted probing
of the weld area. However, from the image it can be assumed that the shear is approximately 400 psi when
comparing the green/yellow pipe edge to the key on the right.
Figure 14 represents the Factor of Safety of the weld areas. From this image it can be determined from the
description in the top left that the minimum FOS in the entire assembly is 4.2 which is illustrated by the darkest red
on the scale to the right. Therefore when comparing the shade of red on the pipe where the welds are located, it is
clear that the entire pipe is the minimum FOS of 4.2.
Verification comparison
The table shown below compares the theoretical values with the SW model
Table 3
Weld c
variable Theoretical Solid Works Relative error
π 1234 psi 988 psi 24.9%
ππ 40.5 4.2 864%
Figure 14
Table 4
Weld h
variable Theoretical Solid Works Relative error
π 331 psi 400 psi 17.25%
ππ 151 4.2 3,495%
The large relative error for factors of safety shown above is most likely due to using an incorrect
formula for the factor of safety in my theoretical calculation. I wasnβt 100% sure on how to
properly calculate the factor of safety of the weld since we didnβt go over it in class in detail.
Another possibility is that instead of creating fillets as welds for the simulation, edge weld
connections were used. The somewhat large relative error of the shear stresses could be due to
the way that the solid works simulation was performed. Figures 15 and 16 below illustrate the
edge weld anaylsis for welds c and h using linear geometry. This appears to be a more precise
way of calculating the shear stress of a weld since welds are based off of linear geometry.
Figure 15
Evaluation
The point of this section is to see whether or not our staircase met specs and is safe. Since
all the factors of safety are greater than the target value of 1 (theoretical and SW) and that
we are within the building code, we can conclude that our design is safe.
References
[1] Budynas and Nisbett, 2011, SHIGLEYβS MECHANICAL ENGINEERING DESIGN, 9th ed,
McGraw-Hill, NewYork.
[2] M. Griffis, 2014, EML3005 Fall 2015 Lectures.
[3] www.mcnichols.com, Schedule 40 thickness pipe dimension table
Figure 16