Final Exam Solution

Dilla University, Department of Construction Technology and Management
Steel and Timer Structures Final Exam Examination Time Allowed 25min.
Name____________________________________ID_____________Section____________
Instruction:-
 Show all the necessary steps clearly for what you are asked only
 Unclear works has no mark values
 If any data is missing assume it reasonably
Part II Open Book: - (Time allowed 2:00 hours.)

1. The simply supported steel beam of span 6.4m free to rotate on plan at the ends shown below is laterally
restrained along its length and has bearing lengths of 50 mm at the unstiffened supports and 75 mm
under the mid span factored point load P=200KN. Neglecting self weight of the beam and using
Fe430 steel grade, Check the adequacy of UB 457x191x67 for the following.
 Resistance for Bending moment
 Resistance for Shear
 Shear buckling
 flange induced buckling and
 Resistance of web to crushing at mid span point load.
2. The connection shown in the Figure below is subjected to a design tensile force of 300KN. The
steel Grade is Fe430, the bolt Grade 7.8 and its diameter is 22 mm. Check that the connection is
adequate.
Steel and Timer Structures Final Exam:- Open Book Time Allowed 2:00 hours
Solution Q1:
Step 1: Design moment and Design shear
𝑃𝑑 𝐿 200 ∗ 6.4
𝑀𝑠𝑑 = = = 320𝐾𝑁𝑚
4 4
𝑃𝑑 200
𝑉𝑠𝑑 = = = 100𝐾𝑁𝑚
2 2
Step 2: Selection of the profile and section classification
The relevant section properties are:
Material strength
𝐹𝑜𝑟 𝐹𝑒430 𝑎𝑛𝑑 𝑡 ≤ 40𝑚𝑚,
𝑓𝑦 = 275𝑁/𝑚𝑚2 𝑎𝑛𝑑 𝑓𝑢 = 430𝑁/𝑚𝑚2
235 235
ε=√ =√ = 0.92
fy 275
Out stand element subjected to compression

c implies
= 7.48 ≤ 8.5ε = 7.82 ⇒ flange is class 1
tf
Web subjected to bending only
d implies
= 48 ≤ 79ε = 72.68 ⇒ Web is class 1
tw
Thus the overall section is class 1.
Step 3: Moment Resistance. (For class 1 section).
𝑊𝑝𝑙,𝑦 𝑓𝑦 1471 ∗ 103 ∗ 275
𝑀𝑐,𝑅𝑑 = = ∗ 10−6 = 367.75𝐾𝑁𝑚 ≥ 𝑀𝑠𝑑 = 320𝐾𝑁𝑚 … . 𝑜𝑘!
𝛾𝑚𝑜 1.1
Step 4: Check for shear Resistance.
Maximum shear force, VSd = 100 KN.
𝐴𝑣 = 4090𝑚𝑚2 𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒 𝑜𝑟 𝐴𝑣 ≅ 1.04ℎ𝑡𝑤 = 1.04 ∗ 453.4 ∗ 8.5 = 4008.06
𝐴𝑣 𝑓𝑦 /√3 4090 ∗ 275/√3

𝑉𝑝𝑙,𝑅𝑑 = = ∗ 10−3 = 590.34𝐾𝑁 > 𝑉𝑠𝑑 = 100𝐾𝑁 … . 𝑜𝑘!
𝛾𝑚𝑜 1.1
Vsd 100
Vpl,Rd
= 590.34 = 0.17 < 0.5
𝑖𝑚𝑝𝑙𝑖𝑒𝑠
⇒ 𝑛𝑜 𝑛𝑒𝑒𝑑 𝑓𝑜𝑟 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑑𝑒𝑠𝑖𝑔𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑.
Step 5: Check for web crushing at mid span point load:
(𝑆𝑠 + 𝑆𝑦 )𝑡𝑤 𝑓𝑦𝑤

𝑅𝑦,𝑅𝑑 = 𝑖𝑛 𝑤ℎ𝑖𝑐ℎ 𝑆𝑠 = 90𝑚𝑚 (𝑔𝑖𝑣𝑒𝑛)𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑖𝑓𝑓 𝑏𝑒𝑎𝑟𝑖𝑛𝑔
𝛾𝑚1
2
𝑏𝑓 𝑓𝑦𝑓 𝜎𝑓,𝐸𝐷
𝑆𝑦 = 2𝑡𝑓 √( ) ( ) [1 − ( )]
𝑡𝑤 𝑓𝑦𝑤 𝑓𝑦𝑓
𝑏𝑓 = 189.9𝑚𝑚 ≤ 25 𝑡𝑓 = 25 ∗ 12.7 = 317.5 … … 𝑜𝑘!
𝑡𝑤 = 8.5𝑚𝑚, 𝑎𝑛𝑑 𝑓𝑦𝑓 = 𝑓𝑦𝑤 = 275𝑁/𝑚𝑚2
6 453.4
𝑀𝐶 320 ∗ 10 ∗ 2
𝜎𝑓,𝐸𝐷 = = = 246.92𝑁/𝑚𝑚2
𝐼𝑦 29380 ∗ 104
2
𝑏𝑓 𝑓𝑦𝑓 𝜎𝑓,𝐸𝐷
𝑆𝑦 = 2𝑡𝑓 √( ) ( ) [1 − ( )]
𝑡𝑤 𝑓𝑦𝑤 𝑓𝑦𝑓
189.9 275 246.92 2

√
𝑆𝑦 = 2 ∗ 12.7 ( )( ) [1 − ( )] = 12.26𝑚𝑚
8.5 275 275
(𝑆𝑠 + 𝑆𝑦 )𝑡𝑤 𝑓𝑦𝑤 (90 + 12.26)8.5 ∗ 275

𝑅𝑦,𝑅𝑑 = = 𝑅𝑦, 𝑅𝑑 = ∗ 10−3 = 217.3𝐾𝑁
𝛾𝑚1 1.1
𝑅𝑦,𝑅𝑑 = 217.3𝐾𝑁 ≥ 𝑃𝑑 = 200𝐾𝑁 … … . 𝑜𝑘‼!
Step 6: Check for Shear buckling resistance

Webs with d/tw greater than 69 for an un stiffened web, shall be checked for resistance to shear buckling.
𝑑 407.6
= = 47.95 ≤ 69𝜀 = 69 ∗ 0.92 = 63.48 … … 𝑜𝑘‼!
𝑡𝑤 8.5
∴ 𝑛𝑜 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐h𝑒𝑐𝑘 𝑓𝑜𝑟 𝑠h𝑒𝑎𝑟 𝑏𝑢𝑐𝑘𝑙𝑖𝑛𝑔
Step 7: Check for flange induced buckling
To prevent the possibility of the compression flange buckling in a plane of the web, the ratio d/tw of the web
shall satisfy the following criterion: (𝑑/𝑡𝑤 ) ≤ 𝑘(𝐸/𝑓𝑦𝑓 )√𝐴𝑤 /𝐴𝑓𝑐
𝑖𝑚𝑝𝑙𝑖𝑒𝑠
𝑓𝑙𝑎𝑛𝑔𝑒 𝑖𝑠 𝑐𝑙𝑎𝑠𝑠 1 ⇒ 𝑘 = 0.3
𝐴𝑤 = 𝑑𝑡𝑤 = 407.6 ∗ 8.5 = 3464.6𝑚𝑚2 𝑎𝑛𝑑 𝐴𝑓𝑐 = 𝑏𝑓 𝑡𝑓 = 189.9 ∗ 12.7 = 2411.73𝑚𝑚2
𝑑 407.6 𝐸 𝐴𝑤 2.1 ∗ 105 3464.6

= = 47.95 ≤ 𝑘 ( ) √ = 0.3 ∗ √ = 274.58 … … 𝑜𝑘‼
𝑡𝑤 8.5 𝑓𝑦𝑓 𝐴𝑓𝑐 275 2411.73
Because the beam is laterally restrained there is no need to check for lateral torsional buckling, and hence the
section UB 457x191x67 is safe.
Solution Q2:
Check for the Geometry and material:
Fe430  fy = 275MPa and fu = 430MPa
Bolts: grade 7.8  fyb = 560MPa, fub = 700MPa,
Diameter of the holes: the hole diameter shal be do =d+2= 22+2=24mm,
Assume Area at the bottom of thread: As = 303mm2
Min. edge distance, e1or e2 = 1.25do = 1.25 * 24 = 30mm≤30mm Ok
Minimum hole distance, P1 = 2.5do = 2.5*24 = 60mm< 50mmnot Ok
Max.edge distance, e1or e2 = 12t = 12 * 8= 96mm>35mmOk
Max hole distance, P1 = 14t = 14*8 = 112mm>50mmOk
Shear capacity of bolts:
Assumption threads are in shear plane i.e. As = 303mm2
The bolts are subjected to double shear (two shear planes each at the interface b/n the inner and outer gusset plate)
0.6𝑓𝑢𝑏 𝐴𝑠 0.87𝑓𝑦𝑏 𝐴𝑠
𝐹𝑣,𝑅𝑑 = ≤
𝛾𝑚𝑏 𝛾𝑚𝑏
0.6 ∗ 700 ∗ 303 ∗ 10−3 0.87 ∗ 560 ∗ 303 ∗ 10−3
𝐹𝑣,𝑅𝑑 = = 101.82𝐾𝑁 ≤ = 118.1𝐾𝑁
1.25 1.25
Because there are four bolts and one bolt is in double shear
300
𝐹𝑣,𝑅𝑑 = 101.82𝐾𝑁 ≥ = 37.5𝐾𝑁 … … . 𝑜𝑘!
2∗4
The bearing capacity of the bolt
The thicker plate with t=10mm is the critical
𝑑𝑡{0.9(𝑓𝑢𝑏 + 𝑓𝑦𝑏 )} 22 ∗ 10 ∗ 0.9(700 + 560) 300

𝐹𝑏𝑏,𝑅𝑑 = = ∗ 10−3 = 199.58𝐾𝑁 ≥ = 75𝐾𝑁 − − − 𝑜𝑘‼
𝛾𝑚𝑏 1.25 4
The thinner plate with t=8mm
𝑑𝑡{0.9(𝑓𝑢𝑏 + 𝑓𝑦𝑏 )} 22 ∗ 8 ∗ 0.9(700 + 560) 300

𝐹𝑏𝑏,𝑅𝑑 = = ∗ 10−3 = 159.67𝐾𝑁 ≥ = 37.5𝐾𝑁 − − − 𝑜𝑘‼
𝛾𝑚𝑏 1.25 2∗4
Bearing capacity of the connected members (gusset plate)
The inner(thicker) gusset plate is the critical member
Since t = 10mm < 2*8 = 16mm
0.8(𝑓𝑢 + 𝑓𝑦 )𝑑𝑡 1 (𝑓𝑢 + 𝑓𝑦 )

𝐹𝑏𝑝,𝑅𝑑 = ≤ 𝑒1 𝑡 ∗ 0.8 ∗ 𝑤𝑖𝑡ℎ 𝑑 = 22𝑚𝑚, 𝑡 = 10𝑚𝑚, 𝑒1 = 30𝑚𝑚
𝛾𝑚𝑏 2 𝛾𝑚𝑏
0.8(430 + 275) ∗ 22 ∗ 10
𝐹𝑏𝑝,𝑅𝑑 = ∗ 10−3 = 99.26𝐾𝑁
1.25
1 (430 + 275)
≤ ∗ 30 ∗ 10 ∗ 0.8 ∗ ∗ 10−3 = 67.68𝐾𝑁
2 1.25
300
𝐹𝑏𝑝,𝑅𝑑 = 67.68𝐾𝑁 ≥ = 75𝐾𝑁 − − − 𝑛𝑜𝑡 𝑜𝑘‼
4
The outer (thinner) gusset plate
0.8(𝑓𝑢 + 𝑓𝑦 )𝑑𝑡 1 (𝑓𝑢 + 𝑓𝑦 )

𝐹𝑏𝑝,𝑅𝑑 = ≤ 𝑒1 𝑡 ∗ 0.8 ∗ 𝑤𝑖𝑡ℎ 𝑑 = 22𝑚𝑚, 𝑡 = 8𝑚𝑚, 𝑒1 = 30𝑚𝑚
𝛾𝑚𝑏 2 𝛾𝑚𝑏
0.8(430 + 275) ∗ 22 ∗ 8
𝐹𝑏𝑝,𝑅𝑑 = ∗ 10−3 = 79.41𝐾𝑁
1.25
1 (430 + 275)
≤ ∗ 30 ∗ 8 ∗ 0.8 ∗ ∗ 10−3 = 54.14𝐾𝑁
2 1.25
300
𝐹𝑏𝑝,𝑅𝑑 = 54.14𝐾𝑁 ≥ = 37.5𝐾𝑁 − − − 𝑜𝑘‼
2∗4
Therefore the connection is not adequate to resist the design tensile load.
Short answer:
1. b/c of the high content of silicate acid

2. True
3. False
4. because the insects are less active and hence will reduce attack by beetles.
5. Method of preservation includes
 Smoking
 Heating
 Immersion
 Impregnating coatings
6. yes

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Dilla University, Department of Construction Technology and Management

Part II Open Book: - (Time allowed 2:00 hours.)

Out stand element subjected to compression

𝐴𝑣 = 4090𝑚𝑚2 𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒 𝑜𝑟 𝐴𝑣 ≅ 1.04ℎ𝑡𝑤 = 1.04 ∗ 453.4 ∗ 8.5 = 4008.06

𝐴𝑣 𝑓𝑦 /√3 4090 ∗ 275/√3

(𝑆𝑠 + 𝑆𝑦 )𝑡𝑤 𝑓𝑦𝑤

𝑏𝑓 = 189.9𝑚𝑚 ≤ 25 𝑡𝑓 = 25 ∗ 12.7 = 317.5 … … 𝑜𝑘!

𝑡𝑤 = 8.5𝑚𝑚, 𝑎𝑛𝑑 𝑓𝑦𝑓 = 𝑓𝑦𝑤 = 275𝑁/𝑚𝑚2

189.9 275 246.92 2

(𝑆𝑠 + 𝑆𝑦 )𝑡𝑤 𝑓𝑦𝑤 (90 + 12.26)8.5 ∗ 275

Step 6: Check for Shear buckling resistance

𝐴𝑤 = 𝑑𝑡𝑤 = 407.6 ∗ 8.5 = 3464.6𝑚𝑚2 𝑎𝑛𝑑 𝐴𝑓𝑐 = 𝑏𝑓 𝑡𝑓 = 189.9 ∗ 12.7 = 2411.73𝑚𝑚2

𝑑 407.6 𝐸 𝐴𝑤 2.1 ∗ 105 3464.6

Fe430  fy = 275MPa and fu = 430MPa

Bolts: grade 7.8  fyb = 560MPa, fub = 700MPa,

Diameter of the holes: the hole diameter shal be do =d+2= 22+2=24mm,

Assume Area at the bottom of thread: As = 303mm2

Min. edge distance, e1or e2 = 1.25do = 1.25 * 24 = 30mm≤30mm Ok

Minimum hole distance, P1 = 2.5do = 2.5*24 = 60mm< 50mmnot Ok

Max.edge distance, e1or e2 = 12t = 12 * 8= 96mm>35mmOk

Max hole distance, P1 = 14t = 14*8 = 112mm>50mmOk

Shear capacity of bolts:

Assumption threads are in shear plane i.e. As = 303mm2

The bearing capacity of the bolt

The thicker plate with t=10mm is the critical

𝑑𝑡{0.9(𝑓𝑢𝑏 + 𝑓𝑦𝑏 )} 22 ∗ 10 ∗ 0.9(700 + 560) 300

The thinner plate with t=8mm

𝑑𝑡{0.9(𝑓𝑢𝑏 + 𝑓𝑦𝑏 )} 22 ∗ 8 ∗ 0.9(700 + 560) 300

Bearing capacity of the connected members (gusset plate)

The inner(thicker) gusset plate is the critical member

Since t = 10mm < 2*8 = 16mm

0.8(𝑓𝑢 + 𝑓𝑦 )𝑑𝑡 1 (𝑓𝑢 + 𝑓𝑦 )

0.8(𝑓𝑢 + 𝑓𝑦 )𝑑𝑡 1 (𝑓𝑢 + 𝑓𝑦 )

1. b/c of the high content of silicate acid

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