CE 562 – Structural Design I Name:____________________________ Midterm No. 1 – Closed Book Portion (30 / 100 pts)

1. [2 pts / 30] A very slender column with perfectly pinned ends is subjected to an ever-increasing compressive axial force. The column has an effective slenderness ratio much greater than

y4.71 E F about all axes. The column is perfectly ideal (perfectly homogenous, perfectly isotropic, perfectly prismatic, perfectly straight, and loaded such that absolutely no eccentricity is generated). Will the perfect column buckle, or will it crush / yield?

Circle one: Buckle Yield

2. [2 pts / 30] A column is in a one-story unbraced frame, and is rigidly fixed at its base. The column is connected through a hinge to a girder at its top end. Is a value of k = 1.0 conservative when determining the effective length of this column?

Circle one: Yes No

3. [2 pts / 30] Some cross-section A is comprised only of stiffened elements with identical lengths,

b, and thicknesses, t. Another cross-section B, has is comprised only of unstiffened elements of identical lengths, b, and thicknesses, t. If cross-sections A and B are subjected to the same uniform compression, which cross-section will be more susceptible to local buckling?

Circle one: Cross-section A Cross-section B

CE 562 – Fall 2007 Midterm No.1 Page 1 of 3

4. [4 pts / 30] What is the reliability index, β, and why is it important to load and resistance factor design?

Answer: The reliability index is how engineers ensure structures remain safe. β>1. Specifically, β is the number of standard deviations the probability of failure is away from the statistical mean.

5. [4 pts / 30] A tension member connected to a gusset plate with six bolts (A, B, C, D, E, and F) fails in net section fracture on failure plane AC. What percentage of the applied force, P, was that failure plane subjected to?

Answer: 5/6

A

B

C

D

E

FT

6. [5 pts / 30] What are residual stresses, and why are they important with relation to column design?

Answer: Residual stresses are stresses that remain in a member after it has been formed into a finished product. They are the result of plastic deformations, which can be caused by different sources, such as:

• Uneven cooling of hot-rolled shapes • Cold-bending or cambering during fabrication • Punching of holes or other cutting operations • Welding

They are important to column design because their presence in a steel section means that some of the resistance capacity has already been “used up”. Therefore, under a given loading, a steel section with residual stresses may reach its capacity before expected, if residual stresses were unaccounted for.

CE 562 – Fall 2007 Midterm No.1 Page 2 of 3

7. [6 pts / 30] Two channels are connected back-to-back, and used as a tension member. The tension member is connected to a column as shown. When calculating the net section fracture resistance of the tension member, should a shear lag reduction factor, U, be included? Why or why not?

Answer: Yes, the shear lag reduction factor should be accounted for because the web of the bending member is not attached to the column. Therefore, the exchange of stress from the beam to column will not occur uniformly throughout the connection.

8. [5 pts / 30] One of the underlying assumptions behind the development of the alignment charts for (framed) column effective length factors is that behavior of the columns is purely elastic. Is this a reasonable assumption? How does this assumption affect the design of real columns?

Answer: It is not a reasonable assumption, unless it is accounted for by using τA, the inelastic reduction factor. Most columns fall within the inelastic buckling portion of the column design curve. While assuming all columns will be controlled by elastic buckling is conservative, this can result in high levels of over-conservatism and is not generally considered good practice. If a column is designed with a “k” from the alignment charts, and is found to fall within the inelastic buckling portion of the column design curve, then he inelastic reduction factor can be used to decrease the “k” value to return a more realistic (and less overly-conservative) column capacity.

CE 562 – Fall 2007 Midterm No.1 Page 3 of 3

CE 562 – Structural Design I Name:__C. Bennett________________ Midterm No. 1 – Open Book Portion (70 / 100 pts) 1. [35 pts / 70 pts] Given the following four-story steel-framed structure, determine the design strength of column HI. All steel is A36. Assume that all buckling and bending takes place about all sections’ strong axes. All joints are rigid unless otherwise noted on the drawing. DO NOT USE THE COLUMN DESIGN TABLES AS YOUR SOLUTION. Account for the stiffness reduction factor if appropriate (tabular value may be used).

( )( )

( ) ( )( )( ) ( )( )1 12 2

881 88120 20 1.459

2070 207040 30

columnH

girder

EILG

EIL

+= =

+

∑∑

=

( )( )

( ) ( )( )( )12

881 88120 20 3.405

207040

columnI

girder

EILG

EIL

+= = =∑∑

Using the alignment chart for unbraced frames, k = 1.66

CE 562 – Fall 2007 Midterm No.1 Page 1 of 11

Therefore, the effective length for column HI is ( ) ( )( )1.66 20' 33.2'xkL = =

( )33.2' 12"/' 29,00065.85 4.71 133.696.05" 36x

kLr

⎛ ⎞⎛ ⎞ = = < =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

therefore inelastic buckling controls.

n crP F A= g

y

e

FF

cr yF 0.658 F⎡ ⎤⎢ ⎥=⎢ ⎥⎣ ⎦

where

( )( )

22ksi

e 2 2

x

29,000EF 665.85KL

r

ππ= = =⎛ ⎞⎜ ⎟⎝ ⎠

6.01

CE 562 – Fall 2007 Midterm No.1 Page 2 of 11

( )36

ksi ksi66.01crF 0.658 36 28.65

⎡ ⎤= =⎢ ⎥⎣ ⎦

( )( )ksi 2 kipsnP 28.65 24.0in 687.66= =

( )kips kipsnP 0.90 687.66 618.89φ = = Since the column was controlled by inelastic buckling and the alignment charts were used, the value for nPφ is expected to be over conservative. Therefore, τA should be applied to the effective length of the column to obtain a more realistic column capacity. Referencing Table 4-21:

2

kipsksiu n

ing g

P P 618.89Let 25.79A A 24.0

φ= = =

Interpolating,

Aτ =0.4944

( )( )

( )( ) ( )( )( )( ) ( )( )1 12 2

881 8810.4944 0.494420 20 0.7212070 2070

40 30

columnH

girder

EILG

EIL

+= = =

+

∑∑

( )( )

( )( ) ( )( )( )( )12

881 8810.4944 0.494420 20 1.6832070

40

columnI

girder

EILG

EIL

+= = =∑∑

CE 562 – Fall 2007 Midterm No.1 Page 3 of 11

Using the alignment chart for unbraced frames, k = 1.37

Therefore, ( )( )x

1.37 20' 12"/'kL 29,00054.35 4.71 133.69r 6.05" 36

⎛ ⎞⎛ ⎞ = =

( )36

ksi ksi96.89crF 0.658 36 30.82

⎡ ⎤= =⎢ ⎥⎣ ⎦

( )( )ksi 2 kipsnP 30.82 24.0in 739.57= =

( )kips kipsnP 0.90 739.57 665.6φ = = Conclusion: The design strength of column HI for buckling about its strong axis is 665.6kips. Check Local Buckling: Flanges:

ff l

f

b 5.922t

λ = = (1-23)

r,fly

E 29,0000.56 0.56 15.89F 36

λ = = = (16.1-16)

Since , local buckling will not occur in the flanges. f l rλ < λ

Web:

webw

h 22.4t

λ = = (1-23)

r,weby

E 29,0001.49 1.49 42.29F 36

λ = = = (16.1-16)

Since , local buckling will not occur in the web. web rλ < λ

CE 562 – Fall 2007 Midterm No.1 Page 5 of 11

2. [35 pts / 70 pts] The tension member shown below is being used to support service dead loads of 35kips, service live loads of 100kips, and service wind loads of +/-40kips. Is the member sufficient to resist the given loads? The tension member is 35’ long. Investigate all pertinent block shear connection (tension member and gusset plate) failure modes in your analysis, in addition to all other required checks.

T

8 spaces @ 3" = 2'-0"2 ½” 4 ½”

4"

2"

4"

2"

2 C12 x 25 back-to-back, A588 Gr. 50 Steel

5/8" thick gusset plate, A572 Gr. 50 Steel

All holes are for 1" diameter bolts.

Load Combinations: 1. (1.4u i i )R Q Dγ= = +∑ F = 1.4(35) = 49 kips 2. ( ) ( ) ( )1.2 1.6 0.5u i i rR Q D F T L H L or S orγ= = + + + + +∑ R = 1.2(35) + 1.6(100) = 202 kips 3. ( ) ( )1.2 1.6 0.8u i i r rR Q D L or S or R L or Wγ= = + +∑ = 1.2(35) + 0.5(100) or 0.8(40) = 92 kips 4. ( )1.2 1.6 0.5u i i rR Q D W L L or S orγ= = + + +∑ R = 5. 1.2 1.0 0.2u i iR Q D E Lγ= = + + +∑ S = … 6. 0.9 1.6 1.6u i iR Q D Wγ= = + +∑ H = … 7. 0.9 1.0 1.6u i iR Q D Eγ= = + +∑ H = … Gross Section Yielding of the Tension Member:

( ) ( )( )250 2 7.34 734.0ksi in kipsn y gR F A= = × = ( )( )0.90 734 660.6kips kipsnRφ = =

Net Section Fracture of the Tension Member:

By inspection, the rightmost line of bolts will control NSF. Not only are they subject to 100% of the force (T), there are three hole diameters without any staggers.

CE 562 – Fall 2007 Midterm No.1 Page 6 of 11

n u eR F A= e netA UA=

( )( ) ( )( )( )( )2 21 116 162 7.34 3 0.387" 2 1" " " 12.07in innetA holes thick channels= − + + = 1 xU

l= − (Case 2 from Table D3.1 applies)

0.674"1 .97224"

U = − =

( )( )2 2.972 12.07 11.73in ineA = = ( )( )270 12.73 821.0ksi in kipsnR = = ( )( )0.75 821.0 615.8kips kipsnRφ = =

Block Shear Failure of the Tension Member:

Tension Member Block Shear Mode 1:

0.6 0.6n u nv bs u nt y gv bs u ntR F A U F A F A U F A= + ≤ + Gross area in tension, Agt

( )( )( ) 24" 2 0.387" 3.096ingtA channels thick= =

Net area subject to tension, Ant

( )( )( )2 21 1 1 12 2 16 163.096 1 1" " " 2 0.387" 1.355

nt gt holes

in innt

A A A

A

= −

= − + + + + × =

Gross area in shear, Agv

( )( ) 22 26.5" 2 0.387" 41.022ingvA = × =

Net area subject to shear, Anv

( )( )( )2 21 1 12 16 1641.022 2 1 1 1 1 1" " " 2 0.387" 33.185

nv gv holes

in innv

A A A

A

= −

= − + + + + + + × =

0.6 0.6n u nv bs u nt y gv bs u ntR F A U F A F A U F A= + ≤ +

CE 562 – Fall 2007 Midterm No.1 Page 7 of 11

Ubs = 1.0 as per Commentary to Ch. J, pg. 16.1-352 ( )( )20.6 0.6 70 33.185 1393.78ksi in kipsu nvF A = = ( )( )21.0 70 1.355 94.815ksi in kipsbs u ntU F A = = ( )( )20.6 0.6 50 41.022 1230.66ksi in kipsy gvF A = = 0.6 1393.78 94.82 1488.6kips kips kipsu nv bs u ntF A U F A+ = + = 0.6 1230.66 94.82 1325.5kips kips kipsy gv bs u ntF A U F A+ = + = 1488.6 1325.5 1325.5kips kips kipsn nR R= ≤ ∴ = ( )0.75 1325.5 994.1kips kipsnRφ = = Block Shear Failure of the Gusset Plate:

T T

0.6 0.6n u nv bs u nt y gv bs u ntR F A U F A F A U F A= + ≤ + Gross area in tension, Agt

( )( ) 24" 0.625" 2.50ingtA thick= =

Net area subject to tension, Ant

( )( )( )2 21 1 1 12 2 16 162.50 1 1" " " 0.625" 1.094

nt gt holes

in innt

A A A

A

= −

= − + + + + =

Gross area in shear, Agv

( )( ) 22 28.5" 0.625" 35.625ingvA = =

Net area subject to shear, Anv

( )( )( )2 21 1 12 16 1635.625 2 1 1 1 1 1" " " 0.625" 29.297

nv gv holes

in innv

A A A

A

= −

= − + + + + + + =

0.6 0.6n u nv bs u nt y gv bs u ntR F A U F A F A U F A= + ≤ + Ubs = 1.0 as per Commentary to Ch. J, pg. 16.1-352

CE 562 – Fall 2007 Midterm No.1 Page 8 of 11

( )( )20.6 0.6 65 29.297 1142.58ksi in kipsu nvF A = = ( )( )21.0 65 1.094 71.094ksi in kipsbs u ntU F A = = ( )( )20.6 0.6 50 35.625 1068.75= =ksi in kipsy gvF A 0.6 1142.58 71.094 1213.67kips kips kipsu nv bs u ntF A U F A+ = + = 0.6 1068.75 71.094 1139.84+ = + =kips kips kipsy gv bs u ntF A U F A 1213.67 1139.84 1139.84= ≤ ∴ =kips kips kipsn nR R ( )0.75 1184.38 854.9= =kips kipsnRφ

Tension Member Block Shear Mode 2:

You could also have investigated the following mode of tension member failure, however, you would have to account for the area of the flange in net tension.

T T

Serviceability:

min

300Lr

≤

( )min

35' 12"/'300

r≤

Turning to Table 1-16 for 2C shapes, note that there are no tabulated values for a separation of 5/8”. Linearly interpolating between the values for 3/8” and 3/4” of separation,

rmin, 3/8” = 1.16” rmin, 3/4” = 1.31”

Therefore, rmin, 5/8” ≈ 1.26”

( )35' 12"/' 333.33 300

1.26"= ≤

CE 562 – Fall 2007 Midterm No.1 Page 9 of 11

Conclusion: Net section fracture of the tension member controls, and 616kipsnRφ = . However, the tension member does not satisfy serviceability, and should be resized

such that min

300Lr

≤ .

CE 562 – Fall 2007 Midterm No.1 Page 10 of 11

[3 pts. Extra credit] Almost inevitably, instructors fail to ask you in an exam all those things for which you so carefully prepared. As it happens, generating good questions is almost as difficult as generating adequate answers. Thus, to give you your turn on the pitcher's mound, if you have the time and the inclination, write an original exam question pertinent to material that was fair-game to appear on this exam. You will receive between 0 and 3 points depending upon the quality of your question. JUST THE QUESTION PLEASE - DON'T SUPPLY THE ANSWER.

CE 562 – Fall 2007 Midterm No.1 Page 11 of 11