Abstract

    This Staircase Force Analysis project is a requirement of the ENGR 20 -
Engineering Mechanics I course.  The subject matter of this course is centered around
analyzing static forces on stationary objects, which include internal, external, frictional,
distributed, and reactant forces.  Moments due to these forces are also considered.  The
professor for this course is Kurt Schulz.  Professor Schulz made this assignment and also
designated the groups.  The team for this project consists of Matt Ortiz and Nick Dean.
This assignment was given on October 28, 2002.
    The assignment is to model, analyze and present a static loading situation.  An
object was chosen to analyze, the staircase on the back side of the Omega Phi Alpha
Fraternity building.  Background information must be researched on the object including
manufacturers’ specifications, materials used, manufacturing methods used, and alternate
designs, if any.  Any loads on the object must be measured or estimated.  All tension or
compression in beams or cables will be calculated, as well as the support reactions such
that all forces acting on the staircase will be known or measured.  Forces to be analyze
include the applied load of the stairs themselves acting downward on the support truss,
support reactions at the anchor points of the beams, and all internal forces of the beams.
This will be done by drawing free body diagrams of the structure.  The support reactions
will be found by using the equations of equilibrium.  Once the supporting forces are found,
the internal forces of the beams can be analyzed by using the method of sections.  The
staircase foundation is concrete blocks and the steps, rails and support beams are steel.
The support beams appear to be welded together and bolted to the steps and foundation.
Steel was used for its strength and durability.  Cost and aesthetics may have been
considered in choosing the materials to be used.  Concrete or wood also could have been
used.  One advantage of this design is that triangles are used in the truss which makes it
very sturdy and stable.  Also, a plus would be the low maintenance of steel.  One
disadvantage may be that the staircase is only supported in two places.  It may have been
better to bind it to the wall of the building as well.  Another disadvantage could be rain
water pooling on the individual steps causing them to be slippery and hazardous.  Perhaps
the pooling of water could be avoided by making the steps out of steel gridwork instead of
solid panels.
    The majority of past relevant experience was gained in the ENGR 5 - Introduction
to Engineering course.  In that course two major projects were required.  The first one
was to design a boat made entirely out of cardboard and duct tape.  Some force analysis
was required for that project in order to discern how to keep the boat afloat.  Also with
this assignment a written report was required.  The second one was to design a parabolic
hackey-sack launcher.  Essentially, this was a catapult.  Some force analysis was required
for this in order to calculate the distance that the hackey-sack would launch.  An oral
presentation was included in this assignment.

Background

    The purpose of the staircase is to provide access from downstairs to the upstairs balcony and backdoor entrance in the back of the Omega Phi Alpha house.  Mild steel materials were used in the fabrication process of the staircase, the steps have rigid surfaces (for traction) and bent pans, which require shearing and press breaking in order to form the staircase structure.  The anchor bolts are rods that are threaded on one end and connect to the bottom of the staircase and supports.  A press breaking machine was also used to create the bent “L” shape, which is cemented down into the base of the ground.  The steps and handrails were constructed by a welding method called dual shield process, this process consists of a hollow welding wire with flux tips.  For the staircase, the blueprints have been acquired of which there has been a slight change in the overall plan.  The original plan had truss beams only on the smaller support beams, however the final product of the staircase includes two more trusses on the long support beams.  The original staircase was made of wood with a brick supporting base, which was constructed by (at that time) active Omega Phi Alpha bros.  The American Metal Fabricators Agency constructed the materials and WMB architects produced the blueprints for the staircase.  The fabrication design is certified by a registered professional engineer, Carl R. Ballantyne, practicing in the State of California.  The architectural blueprints are stamped with approval by licensed architect of the State of California, Thomas E. Bowe.  The previous staircase was old, frail, and had a couple of broken steps; and so UOP Physical Plant stepped in and made sure that a new staircase be fabricated and constructed using more durable, stronger, and safer steel methods.  The given information is from personal knowledge and first hand experience, while fabrication and manufacturing methods were attained from American Metal Fabricators Agency and WMB Architects.

Materials And Methods

Steps:  Mild steel steps welded together by a dual shield process with the handrails and support beams.  The steps are bent pans by press breaking.
Rails:  Mild steel handrails are dual shield welded to steps and, at specific spots, to the support beams.  A separate handrail is bolted to only the house.
Support Beams:  Mild steel truss beams were dual shield welded to the steps and anchor bolted in cement foundation.
Foundation:  Cement blocks are bolted to the bottom of the support beams and are set at different depths to ensure proper stability.
Anchor Bolts:  A rod threaded on one end and “L” shaped on the other end by press breaking.

Calculations

    With the above calculations, we found the total distributed load from the upper set of stairs to be 503 lb. The total distributed load on the landing was found to be 149.5 lb. plus the 200 lb. from the man we have positioned in the center of the landing. The total distributed load from the lower set of stairs is 802.4 lb. The force of the weight of each of the support beams was also calculated. Theses forces were calculated using force distributions we were given in either pounds per foot or pounds per square foot.

    When we began we had 43 unknown support reactions. This is far too many to find using the conventional equations of equilibrium, so we began to search for assumptions we could make. We recognized that all applied loads were in the negative z direction so there should be no x or y support reactions. So we assumed:
Fax= 0 lb.    Fay= 0 lb.    Fbx= 0 lb.    Fby= 0 lb.    Fcx= 0 lb.    Fcy= 0 lb.    Fex= 0 lb.    Fey= 0 lb.
Ffx= 0 lb.    Ffy= 0 lb.    Fgx= 0 lb.    Fgy= 0 lb.    Fhx= 0 lb.    Fhy= 0 lb.
Since there are no x or y forces in our system, there can be no z moments so we can assume:
Maz= 0 lb*ft    Mbz= 0 lb*ft    Mcz= 0 lb*ft    Mez= 0 lb*ft    Mfz= 0 lb*ft    Mgz= 0 lb*ft    Mhz= 0 lb*ft
    This leaves us still with 22 unknown support reactions. This leaves our system statically indeterminate unless we make some estimations. This will cause our calculated reactions to be different from the actual reactions, but they will hopefully be near to the real reactions. Our first non-realistic assumption is to state that all of the load of the landing is supported by the trusses directly beneath it. We will call these the front and rear low trusses.
    In section 6.5-Space Trusses in the text book, it is stated that, "In cases where the weight of a member is to be included in the analysis, it is generally satisfactory to apply it as a vertical force, half of its magnitude applied at each end of the member." So we divided the wieght of each stair set among their respective supports at each end, so the normal reaction at the foot of the lower stair set, Fdz, can be found with the following equation:
SFz=0=Fdz-(1/2)(802.4)        Fdz= 401.2 lb.
    Next we move to the upper stair set and use the section 6.5 statement again.since there is the 'tall truss' next to the supporting wall at the top of the upper stair set, we assume each will absorb 1/4 of the load of the stairs. We can now find the reaction Fcz on the wall.
SFz=0= Fcz-(1/4)(503)        Fcz= 125.8 lb.
    Now we consider the supports at the bases of the tall truss. We will call these points A and G. On the above FBD labeled 'East Elevation', it can be seen that there is a symmetry of forces on the tall truss so Faz=Fgz. Sum the forces in the z direction on the tall truss:
SFz=0= Faz+Fgz-125.5-58.9-19.8-(43.1)(2)-(117)(2)
2Faz= 524.4            Faz= 262.2 lb.        Fgz= 262.2 lb.
When moments are summed about AG, it can be seen that:
Max= 0 lb*ft        Mgx= 0 lb*ft
Sum the moments about point G in the y direction:
SMgy=0= ((43.1lb)(16.5")(2)+19.8lb(16.5")+58.9lb(16.5")+125.5lb(16.5")+117lb(33")-Faz(33")+Mgy)*(1ft/12")
Mgy= 0 lb*ft
Summing the moments about point A in the Y direction will give us the same equation with all of the applied moments being opposite sense (negative). So:
May= 0 lb*ft
    Now we consider the rear lower truss. We assume the load from the landing and the man is distributed evenly to the front and rear lower trusses, and half of the load of the upper stair set is distributed evenly to the top of the corners of the front and rear trusses. Since the load of the upper stair set acts directly down on member H, we will assume Fhz counteracts it as well as half of the load applied from the landing:
Fhz=125.8lb+((1/2)(200lb+149.5lb))(1/2)        Fhz= 213.2 lb.
Ffz only counteracts half of the landing load:
Ffz=((1/2)(200lb+149.5lb))(1/2)        Ffz= 87.4 lb.
When moments are summed about the y-axis, it is discovered that no y moments are created, so:
Mfy= 0 lb*ft        Mhy= 0 lb*ft
Sum moments about point H in the x direction:
SMhx=0=((-1/2)(200lb+149.5lb)(25.5")+Ffz(51")+Mhx)*(1ft/12")
Mhx= -0.1 lb*ft
Sum moments about point F in the x direction:
SMfx=0=((1/2)(200lb+149.5lb)(25.5")+(125.8lb)(51")-Fhz(51")+Mfx)*(1ft/12")
Mfx= 0.1 lb*ft
We treat the front lower truss just the same as the rear lower truss, except there is another distributed load accross the top of it from the top of the lower stair set:
Fbz=125.8lb+((1/2)(200lb+149.5lb)+401.2lb)(1/2)         Fbz= 413.8 lb.
Fez=((1/2)(200lb+149.5lb)+401.2lb)(1/2)        Fez= 288 lb.
When moments are summed about the y-axis, it is discovered that no y moments are created, so:
Mbx= -0.1 lb*ft        Mey= 0 lb*ft
Sum moments about point B in the x direction:
SMbx=0=((-1/2)(200lb+149.5lb)(25.5")+Fez(51")-(401.2lb)(25.5")+Mbx)*(1ft/12")
Mbx= -0.1 lb*ft
Sum moments about point E in the x direction:
SMex=0=(((1/2)(200lb+149.5lb)+401.2lb)(25.5")+(125.8lb)(51")-Fbz(51")+Mex)*(1ft/12")
Mex= 0.1 lb*ft
Now that we only have two unknown reactions left, Mcx and Mcy, we can sum moments on the entire structure in the x and y directions to find them. First we will show the summation about the y at point C:
SMcy=0=(Mcy+May+Mgy+Mby+Mhy+Mey+Mfy-Faz(15.25")+Fgz(15.25")-Fbz(15.25")+Fhz(15.25")-Fez(15.25")
+Ffz(15.25")-Fdz(145")+802.4lb(96"))*(1ft/12")
Mcy= -1065 lb*ft
Now we will sum the moments about point C in the x direction:
SMcx=0=(Mcx+Max+Mgx+Mbx+Mhx+Mex+Mfx-117lb(12")(2)+Faz(12")+Fgz(12")-43.1lb(12")(2)-19.8lb(12")
-58.9lb(12")-(149.5lb+200lb)(120.5")+Fbz(76")+Fhz(76")-63.7lb(76")(2)-94.5lb(101.5")(2)-51.3lb(101.5")(2)
-63.7lb(127")(2)+Fez(127")+Ffz(127")+Fdz(120.5")-802.4lb(120.5")-503lb(49.5"))*(1ft/12")
Mcx= 6165.3 lb*ft
Mcx and Mcy seem too large to be realistic so they are either off due to a math error or because of our assumptions.
Calculated support reactions:
Fdz= 401.2 lb.        Fcz= 125.8 lb.        Mcx= 6165.3 lb*ft        Mcy= -1065 lb*ft
Faz= 262.2 lb.        Max= 0 lb*ft           May= 0 lb*ft
Fgz= 262.2 lb.        Mgx= 0 lb*ft           Mgy= 0 lb*ft
Fhz= 213.2 lb.        Mhx= -0.1 lb*ft      Mhy= 0 lb*ft
Ffz= 87.4 lb.           Mfx= 0.1 lb*ft        Mfy= 0 lb*ft
Fbz= 413.8 lb.        Mbx= -0.1 lb*ft      Mbx= -0.1 lb*ft
Fez= 288 lb.           Mex= 0.1 lb*ft        Mey= 0 lb*ft

Discussion

    The staircase design is a definite improvement compared to the old wooden model.  For the new staircase a mild steel material was used, which is much more sturdy and durable than the old model.  The steel staircase will have a much longer life expectancy than the wooden staircase had.  Steel does not require too much maintenance whereas wood would require a proper finish and occasional repair to steps and handrails.  The steel staircase design includes handrails on both sides of the staircase and is wide and long, virtues that the wooden staircase did not have.  The steel staircase is also treaded, which helps to add traction and avoid slipping.  The quality of the steel stairs will not suffer from weathering in the way that wood will suffer.  However, the steel staircase does have some negative aspects from weather in that it becomes slippery when, it is a heat absorber and will become excessively hot in the summer, and it could potentially rust in certain areas if the paint finish were to flake or be scratched.  Improvements could be made on the final design, as far as improving the slip resistance when wet.  This could possibly be done by integrating some slip pads onto the steps.  The slip pads could be easily found and, if added in the correct manner, would not be too expensive.

Conclusions

    The steel staircase design is a significant improvement compared to the wooden staircase.  The steel staircase is more safe, stable, and durable; it is expected to be in good condition as long as the house is standing.  We modeled and analyzed a static loading situation from adding a 200 pound man standing on the landing section of the staircase.  We gained experience in examining a tangible object and finding all the acting forces on the staircase and made assumptions to rid of any negligible forces and/or moments.  Finally, we learned to cooperate with a partner and communicate with certified engineers to complete an objective.

References

American Remote Vision Co., http://www.xmlsearch.gator.com.

DeGarmo, E. Paul.  Materials and Processes in Manufacturing / E. Paul DeGarmo, J.
Temple Black, Ronald, A. Kohser.-8th Ed.

Guenther, Clifford. Project Manager:  American Metal Fabricators; 2955 Farrar Ave;   Modesto, CA 95354-4118.  Phone (209) 571-6004, Fax (209) 577-3735.

Hibbeler, R.C. Engineering Mechanics.  Statics / R.C. Hibbeler.-9th Ed.

Schulz, Kurt. Mechanical Engineering Professor:  UOP; 3601 Pacific Ave; Stockton, CA  95211.

United States Patent Organization. http://www.uspto.gov.

Wennell, L.; Mattheis, T., Bowe, T.; WMB.  Architecture Planning Interiors:  246 E.
Main St; Stockton, CA 95202.  Phone (209) 944-9110, Fax (209) 944-5711.

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