Sowa Market Hall Design of Components Avery Watterworth Lauren Mitchell Vivian Nguyen
Column
Truss (Compression Members)
Beams (horizontal Columns)
63’
20’
44’
35’ Lateral bracing moved above mezzanine to reduce load on knife plate joint
20’ Stair moved between structural bays to act as column bracing
Pin connection at the base of the columns in order to reduce stress in the knife plate joint
Changes The two major changes we made were shifting the stairs from floating in the middle of the mezzanine span to in between the column bays. This will provide extra lateral support in the longitudinal direction as well as bracing for the columns. We had a lot of suggestions about our stair placement during the review and we tried some of those solutions. However, this one was the most helpful because it allows us to utilize the stair in a structural way. The next change was raising the lateral support in the cross-sectional direction so that there would not be as much force going into the joint where the knife plates were originally. Shifting this up would also change the effective length of the columns by providing bracing at a different interval. Again, the joint with the knife plates would receive less force as a result of lateral forces, helping to keep the columns reasonably sized. We also changed the rigid connection at the base of the columns to a pin connection. This would allow for movement and reduce stress at the connection of the bracing and the columns. With a rigid connection, the column would bend in a way that would create a lot of stress within the knife plate joint, which is also rigid. This will allow the column to rotate when lateral stresses are applied.
Truss Compressive Member Calculations (Lauren) Assumed 6” by 4” Compression Member for Truss in Part 2. L = 12’ A = 19.25 in2 Fc = 1000 psi E = 1.6 x610 Le/Dmin = (12 x 12)/4 = 36 < 50 (max slenderness ratio) FcE = 0.822 (1.6 x 10 6 ) = 1,014.8 psi 36 2 Fc* = Fc x Cd x Ct x Cf = 1000 (1.15)(0.9)(0.9)(1.15) = 1,071.22 psi FcE 1014.8 psi 0.95 = = Fc* 1071.22 psi
6,900 lb
Cp = 1 + 0.95 1.6
12’
2
1 + 0.95 - 0.95 1.6 0.8
Cp = 1.2187 - 0.5457 = 0.673 R (Ground)
fc = P/A < Fc (Cd)(Ct)(Cm)(Cf)(Cp) fc = 6900 lb/ 19.25 in2< 1000(1.15)(0.9)(0.9)(1.15)(0.673) fc = 358.4 psi < 720.93 psi
If we use a 4” by 4” fc = P/A < Fc (Cd)(Ct)(Cm)(Cf)(Cp) fc = 6900 lb/ 12.25 in2 < 1000 (1.15)(0.9)(0.9)(1.15)(0.673) fc = 563.3 psi < 720.93 psi If we use a 2” by 4” Le/dmin = (12x12)/2 = 72 > 50 ( max slenderness ratio) Column to slender to use in trusse design. We would design our longest truss compression member from 4” by 4” lumber, as well as the rest of the compression and tension members.
ix= bh^3/12 iy=ix i min= 833.3 in^4 Cross Sectional Area=100 in^2 Radius of Gyration=sqrt. i/A = sqrt(833.3/100) = 2.88
Glulam Column (10”x10”)
Avery 59,000 lb
F= P/A (Fc) (Cd) (Cm) (Ct) (Cp) (Fc) Compression (Parallel to Grain) Glulam Column = 1650 psi (E) Elasticity Glulam Column = 1.8 x 10^6 psi (C) Constant Glulam Column = 0.418 (Cd) Load Duration Factor = 1.15 (Cm) Wet Service Factor = 0.9 (Cp) Column Stability Factor = 0.403 Effective Length/Minimum Depth = 23’x12”/10” = 27.6” Fce = c (E)/L^2 Fce = 0.418(1.8x10^6 psi)/(27.6”)^2 = 987.7 psi
23’
Fc* = Fc (Cd) = Fc* (Cp) Fc* = 1650 psi (1.15) = 1898 (0.403) = 765 psi Pa = Fc* (A) Pa = 765 psi (100 in^2) = 76,500 psi Fc = 59,000lb/100in^2 = 590 psi < 76,500 psi
Main Column Design
R (Ground)
TRY: 8 3/4” x 10 1/2” A = 92 in^2 Pa = Fc* (A) Pa = 765 psi (92 in^2) = 70,380 psi Fc = 59,000 lb/92 in^2 = 641.3 psi < 70,380 psi 8 3/4” x 10 1/2” would be more efficient than a 10” x 10”, both still have lots of capacity.
Beam as Column Vivian
The calculations shows that our intuition for the beam size was very off and the actual size for the beam doesn’t need to be very big to carry the weight that’s being put onto it. I had a little problem thinking of a way to figure out the size of the beam because we designed the cross-section of it and there is no chart like a typical column would have. I took a guess on what size the make this “column” and did the calculations for that.
Reflections Avery
Lauren
I learned that sizing is a process. While precedents and experience can make us feel like we have a grasp on how big elements need to be, the design really takes multiple tries to get the right look and strength. This combination does not come easily or by accident. Making changes to a design is part of the process, and I think it is exciting to have the ability to affect that process one way or the other. I have learned that going back and really examining your work and your intentionality are important and making sure that things will function is only one part or the considerations that designers make. There are many other factors that go in to designing components than just strength, but there are few as important.
In doing the third part of our SoWA design competition project, I learned an awful lot about how to size a column component in a building. At first, all the variables where confusing, especially when attempting to figure out how to find each one. After some research and example problems it made much more sense. The processes of sizing member is much easier than it seems or looks from the complicated equations and multitude of variables. I feel confident that i could size wood members from doing this exercise. Overall, I enjoyed the SoWa Market design project. It not only helped me master concepts we had previously learned in class through application, but it also was interesting and fun to design. I definitely enjoyed the competition part of the project, partly because I am motivated by competition, but it was also a fun activity and something that I had not experienced in another class.
This project really helped me to make design decisions based on structure as well as aesthetics. It is really important to be able to change the way your building looks and functions in conjunction with one another. The structure always makes the biggest impression on a user, so this was a great project to practice that.
Vivian The calculations shows that our intuition for the beam size was very off and the actual size for the beam doesn’t need to be very big to carry the weight that’s being put onto it. I had a little problem thinking of a way to figure out the size of the beam because we designed the cross-section of it and there is no chart like a typical column would have. I took a guess on what size the make this “column” and did the calculations for that. The size I chose for the calculation was an approximate size that fits the structure better than the old size but is still not the best one. I could’ve done a more exact sizing of the member if I have a chance to test out all the different sizes from the cross section and find the moment of inertia for different sizes to see which one fits best. Overall, the project has been pretty successful and satisfying since I finally understand how structure works and was able to design a whole project with my teammates and size every member of the structure.