identifier: ------­

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Architecture 314

Fall2018

Name: _1'---<~~=-----..~.y_A __ GSI: 6/V'I,ID

ARCHITECTURAL STRUCTURES I Test 1

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Write your name and GSI on right top of each page plus your personal device on the top left of the first page (to identify your paper). Enter your answers in the boxes provided. Show work for problems directly on the page. If you need more space, work on the back side of the same page as the problem.

10 pts. 1.

10 pts. 2.

10 pts. 3.

10pts. 4.

10 pts. 5.

10 pts. 6.

10 pts. 7.

What is the standard material weight of steel?

A) 250 PCF B) 940 PCF C) 490 PCF D) 150 PCF

What are two sources of Dead Load in a building?

What are two sources of Live Load in a building?

What is the dominant stress condition in the outer ring?

outer ring

What type of force system is illustrated

A) coplanar, Nonconcurrent B) Noncoplanar, concurrent C) Noncoplanar, Nonconcurrent

)H7 on

D) coplanar, concurrent

Find the horizontal and vertical components of the force shown at right. K S" s-6 e. . 4~ -·;..;;---Honz.: J.f ' At> o

3oGK Vert.:---------

; . ,. .. -·-3 'foo Make a sketch showing the graphic vector addition of the three forces shown below. Show the resultant and approximate its value.

k '

1.-,,. k : .... .......... K. .... l.) • L ? .. r-1-f-.._L lF ... ~ ~

l k ~ I'" ...... "' '~ lL' .. II /

I~ r_f..

4..-.------ "X

~

Page 2 of 5 Arch 314 Test 1 Fall2018 Name: {(£Y A GSI: (/~VoJJ

1 o pts. 8. Replace the non-uniform distributed load shown with an equivalent single point load.

Magnitude of point load '2 l K ! 1( Z 0 -;: z. { I<.

t FT 'Z,.

zlt./~F ~~ I I . <.JI'

Location from left end

w p 10 pts. 9. Using the lever principle, what is the force P, needed

to balance the weight W, if W = 60 LBS?

~oit( lj 1 ): f~( 15') ~ ~ I

10 pts. 10.

10 pts. 11.

10 pts. 12.

10 pts. 13.

10 pts. 14.

10 pts. 15.

PinLBS S ~

i::: ~(; l? :: zo ~ ~ 5' ----+---15' ____j

"The Principle of the Lever'' used in solving parallel force systems is attributed to whom?

A) Archimedes B) Newton C) Stevin D) Socrates

What is an ideal compression arch?

A) an arch that acts only in compression B) an arch that acts only in tension C) an arch that acts only in flexure D) an arch with the peak hinge exactly at the center line

Where within a cable system is the moment the greatest?

A) the ends B) nowhere, because the moment is zero everywhere C) at the center line D) the inside nodes

The internal tensile force in the cable shown will be of the same magnitude at all points along the cable. (True or False)

l I i

I I 1920 1

2880 lb

What is the shape of a suspended cable loaded by its own weight called?

4k

I Solve for the direction and magnitude of the reactions (R1 and Rz). (Draw the direction arrows on the illustration.)

192011b i-+

I

'' : i

i' . I

I I

I I i I

' I. I I I I i "19 I' I I I I I I J I l

I'

I l

I I

I I I

1100#

-t

I I I

I 20 I

I II

I I I I

I I

28801b

100#

-t·

lb l

0 R2~ crt;

12'~ z.t<

Page 3 of 5 Arch 314 Test 1 Fall2018 Name: Kccy tl, GSI: UJJ

70 pts. 16. Calculate the dead weight load of the concrete beam shown. Use a density of 150 LBS/FT3.

Determine both end reactions based on the dead load of the beam. ) vJ 12.' 'z'

16' f 8' Rl R2

13"

4"

Weight of beam in PLF ~ !""\ TO~ w"-'C. to~ r -;. F<..f (_ L. ')

::: ?1.11 ( zLl') ~J]oo ,_ I 'i1.n LBS/FT

TotaT~h~o~:m I LBS

Location of the total load vector from R1

I 12, I FT

Vertical reaction at R1

1 32$"" I LBS

Direction of vertical reaction at R1

11' I) p I (up or down)

Vertical reaction at R2

11?5" I LBS

Direction of vertical reaction at R2

I 1 U f" I (up or down)

&<e r<t ~ o :: -11"<> -?( 4) + R, C 1'')

~~ ::: 32~ t ~

Page 4 of 5 Arch 314 Test 1 Fall2018 Name: __._K.::::..=r:.-'-+-y__,_t\......_____

GSI: _qA _ _,.___t-=tJ'-L.~ __

so pts. 17. Determine the cable force in each segment of the hanging cable system shown. The nodes are spaced evenly at an 8 FT spacing. Find the vertical dimensions a and b of the cable nodes for the given center sag.

Vertical reaction Ey

} t-5 LBS

Horizontal reaction Ex

I "3-s-s. ~ I LBS

Vertical dimension b

I I'~ FT

Vertical dimension a

I 4.z FT

Cable force in seqment DE

LBS

Z:.Fu =o :.Cx-C.;r. £:x = C:..x-:. ~ ~~ +-

2.Fv = :...C y -tt:Jtt> + 1~ C.y=15J,

l IS"o*

G.( 0

, "

Horizontal component of force in segment CD 2: M@ C "o" i~'{p') -11~t~ )+~ (c; ')

E:)( G, ;:: 2 Ot:>O

t-1<::. 3> 'S. 'S ~ +1~ -J) 51~

LBS

Vertical component of force in segment CD

LBS 7s 513 -Total cable force in segment CD • 0

<:.1 ::::- ~ , Z I 34l.t LBS

1o -:: ~ - 4. 2 ~ I . t;

l· b I

;: 3 ft,, 'i '4-

-- ::. s4l. 1 ~