Compression Members - Architectural Structures - Exam, Exams of Structural Design and Architecture

Compression Members, On Programmable, Conceptual Questions, Formula Provided, Distributed Loads, Support Uniformly, Unbraced Length, Flexural Strength, Dead Load Determines, Live Load Moment. This is past exam of Architectural Structures. Key points are given above.

Typology: Exams

2011/2012

Uploaded on 12/22/2012

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bg1
32 ft
A
B
wL = 1150 lb/ft
8 ft
13 ft
wD = 980 lb/ft
wselt wt
Note: No aids are allowed for part 1. One side of a letter sized paper with notes is allowed
during part 2, along with a silent, non-programmable calculator. There are reference charts for
part 2, shown on pages 2-6.
Clearly show your work and answer.
Part 1) Worth 5 points (conceptual questions)
Part 2) Worth 45 points
(NOTE: The loading type [ex, live, dead, wind...] and sizes can
and will be changed for the quiz with respect to the beam
diagrams and formula provided. The support condition,
section, and bracing for the column can and will be changed.)
A wide flange beam of A992 steel (Fy = 50 ksi, E = 30 x 103 ksi) is
needed to span 32 ft and support uniformly distributed loads of 980 lb/ft of dead load (from materials),
the self weight, and 1150 lb/ft of live load over a length of 11 feet as shown. The beam is simply
supported with a maximum unbraced length of 15 ft.
a) Select the most economical beam adequate for flexural strength using LRFD design and the chart
provided (including self weight). Assume that the dead load determines the location of the maximum
moment and superimpose the live load moment there.
b) Determine the minimum moment of inertia required such that the dead load deflection does not
exceed 1.25 inches assuming a self weight of 60 lb/ft. [or live load deflection using
max = wl4/(152EI) because there is no equation does not exceed 0.8 in; or total deflection
assuming that the dead load determines the location of the maximum moment using
x=5wx(l3-3lx2+2x3)/(6EIl) because there is no equation does not exceed 1.75 in.]
A W 250 x 49 metric column is 5.75 m tall of A36 steel (Fy = 250 MPa, E = 200 x 103 MPa). The base is
fixed and the top is pinned in the weak axis, while the strong axis is considered pinned at the top and
bottom (no picture and approximated conditions). The section properties are:
A = 6260 mm2, Ix = 70.7 x 106 mm4, rx = 106 mm, Iy = 15.2 x 106 mm4, ry = 49.3 mm
c) If the column is to support 200 kN of dead load and 600 kN of live load, is it adequate for design
using LRFD?
Answers Not provided on actual quiz!
a) Mu = 279.4 k-ft, use W18x55 (Mu* < 294 k-ft)
b) Ireq’d = 654.3 in4 (dead only) [Ireq’d-li ve only = 571.2 in4; Ireq’d-total = 725.8 in4]
c) Pn = 886 kN Not OK (weak axis governs because Pn-strong = 1368 kN)
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A 32 ft B

wL = 1150 lb/ft

13 ft 8 ft

wD = 980 lb/ft

wselt wt

Note: No aids are allowed for part 1. One side of a letter sized paper with notes is allowed

during part 2, along with a silent, non-programmable calculator. There are reference charts for

part 2, shown on pages 2-6.

Clearly show your work and answer.

Part 1) Worth 5 points (conceptual questions)

Part 2) Worth 45 point s

(NOTE: The loading type [ex, live, dead, wind...] and sizes can

and will be changed for the quiz with respect to the beam

diagrams and formula provided. The support condition,

section, and bracing for the column can and will be changed.)

A wide flange beam of A992 steel (Fy = 50 ksi, E = 30 x 10^3 ksi) is

needed to span 32 ft and support uniformly distributed loads of 980 lb/ft of dead load (from materials),

the self weight, and 1150 lb/ft of live load over a length of 11 feet as shown. The beam is simply

supported with a maximum unbraced length of 15 ft.

a) Select the most economical beam adequate for flexural strength using LRFD design and the chart

provided (including self weight). Assume that the dead load determines the location of the maximum

moment and superimpose the live load moment there.

b) Determine the minimum moment of inertia required such that the dead load deflection does not

exceed 1.25 inches assuming a self weight of 60 lb/ft. [ or live load deflection– using

 max = wl^4 /(152EI) because there is no equation – does not exceed 0.8 in; or total deflection

assuming that the dead load determines the location of the maximum moment – using

 x=5wx(l^3 -3lx^2 +2x^3 )/(6EIl) because there is no equation – does not exceed 1.75 in.]

A W 250 x 49 metric column is 5.75 m tall of A36 steel (Fy = 250 MPa, E = 200 x 10^3 MPa). The base is

fixed and the top is pinned in the weak axis, while the strong axis is considered pinned at the top and

bottom (no picture and approximated conditions). The section properties are:

A = 6260 mm^2 , Ix = 70.7 x 10^6 mm^4 , rx = 106 mm, Iy = 15.2 x 10^6 mm^4 , ry = 49.3 mm

c) If the column is to support 200 kN of dead load and 600 kN of live load, is it adequate for design

using LRFD?

Answers – Not provided on actual quiz!

a) Mu = 279.4 k-ft, use W18x55 (Mu* < 294 k-ft) b) Ireq’d = 654.3 in^4 (dead only) [Ireq’d-live only = 57 1.2 in^4 ; Ireq’d-total = 72 5.8 in^4 ] c) Pn = 886 kN Not OK (weak axis governs because Pn-strong = 1368 kN)

ARCH 331 F2012abn

REFERENCE CHARTS FOR QUIZ 5

W =

wl^2 W =

wl^2

W =

wl^2