Technological Institute of the Philippines

1338 Arlegui St., Quiapo, Manila

College of Engineering and Architecture

Department of Civil Engineering

Integration Course 3
CE 412
CE43S4

Assignment No. 3

Steel Design

Score

Submitted by: Submitted to:

Lasac, Claudine P. Engr. Rolls Grant B. Vasquez

1911936

Date Submitted:
July 9, 2023
Problem 1 Design of Connections 1

Two plates, each with a thickness of 12mm, are bolted together with 6-22mm diameter bolts forming a lap connection. The
hole diameters are 3mm larger than the bolt diameters. The plates used are A36 steel with Fy=248 MPa and Fu=400 MPa
The following are the allowable stresses
Allowable shear stress of bolts = 120 MPa
Allowable bearing stress of plates = 1.20 Fu
Calculate the following:
a.) Value of P (ASD) in kN based on the shear of bolts
𝑃
𝜏=
𝐴
𝑃𝑛
120 = 6𝜋
(22)2
4

𝑃𝑛 = 273.695552 𝑘𝑁

𝑃𝑛 273.695
Ω
= 2.0

𝑃𝑎 = 136.85 𝑘𝑁

b.) Value of P (ASD) in kN based on bearing of plates

𝑃
𝜎 = 𝐴 ; 𝑃 = 𝜎𝐴

𝑃𝑎 = 𝐹𝑛 𝐴𝑏
(12(22)(6)(1.2)(400)
𝑃𝑎 = 103

𝑃𝑎 = 760.32 𝑘𝑁

𝑃𝑎 760.32
Ω
= 2.0

𝑃𝑎 = 380.16 𝑘𝑁

c.) Value of P (ASD) based on block shear

Case 1:
𝐴𝑔𝑣 = (200)(12)(2) = 4800 𝑚𝑚2

𝐴𝑛𝑣 = (200 − 2.5(25))(12)(2) = 3300 𝑚𝑚2

𝐴𝑛𝑡 = (100 − 1(25))(12) = 900 𝑚𝑚2

𝑃𝑛 = 0.6𝐹𝑢 𝐴𝑛𝑣 + 𝑈𝑏𝑠 𝐹𝑢 𝐴𝑛𝑡 ≤ 0.6𝐹𝑦 𝐴𝑔𝑣 + 𝑈𝑏𝑠 𝐹𝑢 𝐴𝑛𝑡

𝑃𝑛 = 0.6(400)(3300) + 400(900) ≤ 0.6(248)(4800) + 400(900)

 𝑃𝑛 = 1152 𝑘𝑁 ≤ 1074.24 𝑘𝑁 ∴ 𝑈𝑠𝑒 𝑈𝑝𝑝𝑒𝑟𝑙𝑖𝑚𝑖𝑡

𝑃𝑛 1074.24
𝑃𝑎 = Ω
= 2.00

𝑃𝑎 = 537.12 𝑘𝑁

Case 2:
𝐴𝑔𝑣 = (200)(12) = 2400 𝑚𝑚2

𝐴𝑛𝑣 = (200 − (2.5)(25))(12) = 1650 𝑚𝑚2

𝐴𝑛𝑡 = ((140) − 1.5(25))(12) = 1230 𝑚𝑚2

𝑃𝑛 = .6𝐹𝑢 𝐴𝑛𝑣 + 𝑈𝑏𝑠 𝐹𝑢 𝐴𝑛𝑡 ≤ .6𝐹𝑦 𝐴𝑔𝑣 + 𝑈𝑏𝑠 𝐹𝑢

𝑃𝑛 = 0.6(400)(1650) + 400(1230) ≤ 0.6(248)(2400) + 400(1230)

𝑃𝑛 = 888 𝑘𝑁 ≤ 849.12 𝑘𝑁 ∴ 𝑈𝑠𝑒 𝑈𝑝𝑝𝑒𝑟𝑙𝑖𝑚𝑖𝑡

𝑃𝑛 849.12
𝑃𝑎 = Ω
= 2

𝑃𝑎 = 424.56 𝑘𝑁
Problem 2 Design of Connections 2

An angle L125x100x8 is connected to a gusset plate shown. The effective hole diameter is 25mm. Use Fy=248 MPa and
Fu = 400 MPa

a.) Calculate the value of P (kN) based on the yielding of the angle. Use ASD
𝑃𝑛 = 𝐹𝑦 𝐴𝑔

𝑃𝑛 = 248((125 ∗ 8) + 9100 − 8)(8)

𝑃𝑛 = 430.528 𝑘𝑁

𝑃𝑛 430.528
Ω
= 1.67
= 257.8011976

𝑃𝑎 = 257.80 𝑘𝑁
b.) Calculate the value of P (kN) based on the tearing of the angle. Use ASD
Case 1:
𝑃𝑛 = 𝐹𝑢 𝐴𝑒

𝐴𝑛 = 𝐴𝑔 − 𝑛𝑑𝑡 = 1735 − 1(25)(8) = 1536 𝑚𝑚2

𝑃𝑛 = 400(1536) = 614.400 𝑘𝑁
𝑃𝑛 614.400
𝑃𝑎 = Ω
= 2.00

𝑃𝑎 = 307.20 𝑘𝑁

Case 2:
𝑠2 402
𝐴𝑛 = 𝐴𝑔 − 𝑛𝑑𝑡 + 𝑡 = 1736 − 2(25)(8) + )(8) = 1411.29 𝑚𝑚^2
4𝑔 4(42.50)

𝑃𝑛 = 400(1411.29) = 564.52 𝑘𝑁
𝑃𝑛 564.52
𝑃𝑎 = Ω
= 2.00

𝑃𝑎 = 282.26 𝑘𝑁

c.) Calculate the value of P (kN) based on block shear. Use ASD

𝐴𝑔𝑣 = 240(8) + 280(8) = 4160 𝑚𝑚2

𝐴𝑛𝑣 = [240 − 2.5(25)](8) + [280 − 3.5(25)](8) = 2960 𝑚𝑚2

402
𝐴𝑛𝑡 = [42.5 − (1)25 + 4(42.50)] (8) = 215.2941176 𝑚𝑚2

𝑃𝑛 = 0.6𝐹𝑢 𝐴𝑛𝑣 + 𝑈𝑏𝑠 𝐹𝑢 𝐴𝑛𝑡 ≤ 0.6𝐹𝑦 𝐴𝑔𝑣 + 𝑈𝑏𝑠 𝐹𝑢 𝐴𝑛𝑡

𝑃𝑛 = 0.6(400)(2960) + 400(215.2941176) ≤ 0.6(248)(4160) + 400(215.2941176)

𝑃𝑛 = 796.518 𝑘𝑁 ≤ 705.126 𝑘𝑁 ∴ 𝑈𝑠𝑒 𝑈𝑝𝑝𝑒𝑟𝑙𝑖𝑚𝑖𝑡

𝑃𝑛 705.126 𝑘𝑛
𝑃𝑎 = =
Ω 2

𝑃𝑎 = 352.563 𝑘𝑁

𝐴𝑔𝑣 = 240(8) = 1920 𝑚𝑚2

𝐴𝑛𝑣 = 240(80) − 2.5(28)(8) = 1420 𝑚𝑚2
402
𝐴𝑛𝑡 = [75(8) − (1.5)(25)(8) + 4(42.50) (8)] = 375.294 𝑚𝑚2

𝑃𝑛 = 0.6𝐹𝑢 𝐴𝑛𝑣 + 𝑈𝑏𝑠 𝐹𝑢 𝐴𝑛𝑡 ≤ 0.6𝐹𝑦 𝐴𝑔𝑣 + 𝑈𝑏𝑠 𝐹𝑢 𝐴𝑛𝑡

 𝑃𝑛 = 0.6(400)(1420) + 1.0(400)(375.294) ≤ 0.6(248)(1920) + 400(375.294)

𝑃𝑛 = 490.92 𝑘𝑁 ≤ 435.814 𝑘𝑁 ∴ 𝑈𝑠𝑒 𝑈𝑝𝑝𝑒𝑟𝑙𝑖𝑚𝑖𝑡

𝑃𝑛 435.814
𝑃𝑎 = Ω
= 2.00

𝑃𝑎 = 217.91 𝑘𝑁

Problem 3 Design of Compression Members

To reinforce a column in an existing structure, two channels are welded to the W 12 x 50 column section as shown in the
figure. The effective length of the column with respect to each axis is 4.8 m.
Kx = Ky = 1.0
Properties of W12 x 50
A = 9484 mm2
rx = 131.57 mm
ry = 49.78 mm
Ix = 164 x 106 mm4
Iy = 23 x 106 mm4
Fy = 345 MPa
d = 309.63 mm
Properties of C 6 x 13
A = 2471 mm2
Ix = 7.24 x 106 mm4
Iy = 0.44 x 106 mm4
bf = 54.79 mm
x = 13.06 mm (centroid from the outer side of the web)
Applied concentrated loads on the column
Dead Load = 480 kN
Live Load = 870 kN
a.) Determine the critical slenderness ratio.
𝐴𝑻 = 9484 + 2(2471) = 14426 𝑚𝑚2
309.63
𝑥̄ = 2
+ 54.79 = 209.605 𝑚𝑚

𝑥2 = 309.63 + 2(54.79) − 13.06 = 406.15 𝑚𝑚

𝑥3 = 13.06 𝑚𝑚

𝐼𝑥 = 164 𝑥 106 + 2(7.24 𝑥 106 ) = 178.48𝑥106 𝑚𝑚4

𝐼𝑦 = 23 𝑥 106 + 2(0.44 𝑥 106 ) + 9848(0) + 2471(406.15 − 209.60)2 + (209.60 − 13.06)2

𝐼𝑦 = 119.37 𝑚𝑚4

𝐼 178.48𝑥 106
𝑟𝑥 = √ 𝑥 = √ = 111.2299686 𝑚𝑚
𝐴 14426

𝐼 119.37𝑥 106
𝑟𝑦 = √ 𝐴𝑦 = √ 14426
= 90.96628565 𝑚𝑚
 𝐾𝐿 1.0(4800)
( 𝑟 )𝑥 = 111.229
= 43.15383758
𝐾𝐿 1.0(4800)
( )𝑦 = = 52.76680218 ≈ 52.77 ∴ 𝐺𝑜𝑣𝑒𝑟𝑛𝑠
𝑟 90.966

b.) Determine the design strength of the column section

𝐸 200000
𝐶𝑐 = 4.71√𝐹 = 4.71√ 345
= 113.4034736
𝑦

𝐾𝐿
𝑟
< 𝐶𝑐 ∴ 𝐼𝑛𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝐿𝑇𝐵

𝜋2 𝐸 𝜋2 (200000)
𝐹𝑒 = 𝐾𝐿 = (52.77)2
= 708.9378929
( )2
𝑟

345
𝐹𝑐𝑟 = 0.658708.937 = 281.4232099
𝑃𝑛 = 𝐹𝑐𝑟 𝐴𝑔 = 281.423(14426)

𝑃𝑛 = 4059.81125 𝑘𝑁

𝑃𝑢 = ∅𝑃𝑛 = 0.90(4059.811)
𝑃𝑢 = 3653.83 𝑘𝑁

c.) Determine the allowable strength of the column section.

𝑃𝑛 4059.911
𝑃𝑎 = 𝛺
= 2.00

𝑃𝑎 = 2431.02 𝑘𝑁

d.) Determine the Demand Capacity Ratio in LRFD. Indicate if safe or failed
𝑃 1.2 (480)+1.6 (870)
𝐷𝐶𝑅 = ∅𝑃𝑢 = 3653.83
𝑛

𝐷𝐶𝑅 = 0.5386 < 1.0 ∴ 𝑆𝐴𝐹𝐸

e.) Determine the Demand Capacity Ratio in ASD. Indicate if safe or failed
𝑃𝑎 480+870
𝐷𝐶𝑅 = 𝑃 =
𝑛 /𝛺 241.02

𝐷𝐶𝑅 = 0.5553 < 1.0 ∴ 𝑆𝐴𝐹𝐸

Problem 4 Design of Flexural Members

A W 16 x 26 is used for beam AB as shown in the figure. Service dead loads consists of 125 mm thick reinforced concrete
slab. Assume that the floor slab provides continuous lateral support.
Partition load (movable, live) = 0.96 kPa, Suspended ceiling and mechanical equipment = 0.48 kPa. Service live load = 3.12
kPa and unit weight of concrete = 24 kN/m3.
Properties of W 16 X 26
Wt. = 39 kg/m
A = 4955 mm2
d = 398.53 mm
bf = 139.70 mm
tf = 8.76 mm
tw = 6.35 mm
Sx = 629 x 103 mm3
Sy = 57 x 103 mm3
Zx = 724 x 103 mm3
Zy = 90 x 103 mm3
Fy = 345.6 MPa
Applied uniform loads on the beam
Dead Load = 12 kN/m (including self-weight)
Live Load = 10 kN/m
a.) Determine the design flexural strength capacity of the beam. Ø =0.90 (kN-m)
𝑏𝑓 139.70
𝜆= = = 7.973744292
2𝑡𝑓 2(8.76)

𝐸 200000
𝜆𝑝 = 0.38√𝑓𝑦 = 0.38√ 345.6
= 9.141379262

𝜆 < 𝜆𝑝 ∴ 𝐶𝑜𝑚𝑝𝑎𝑐𝑡

𝐿𝑏 = 9000 𝑚𝑚
𝑀𝑝 = 𝐹𝑦 𝑍𝑥 = 345.6(724 ∗ 103 ) = 250.2144 𝑘𝑁 ∗ 𝑚
𝑀𝑢 = 0.9(250.21)
𝑀𝑢 = 225.19 𝑘𝑁 ∗ 𝑚
b.) Determine the design shear strength capacity of the beam. (kN)
𝑉𝐶𝐴𝑃𝐴𝐶𝐼𝑇𝑌 = 0.6𝐹𝑦 𝐴𝑛 𝐶𝑣 = 0.6(345.6)(398.53 ∗ 6.35)(1)

𝑉𝐶𝐴𝑃𝐴𝐶𝐼𝑇𝑌 = 524.7587981
𝑉𝑢 = 1.0(524.758)
𝑉𝑢 = 524.76 𝑘𝑁
c.) Determine the deflection of the beam if it has a flexural rigidity of 23800 kN-m2 based on live load only.
𝑘𝑁
𝑤𝑢 𝐿 = 4.080(1.8) = 7.344 𝑚

5𝑊𝐿4 5(7.344)(9)4
∆𝑀𝐴𝑋 = 384𝐸𝐼 = 354(23,800)
∗ 103 = 26.36116071

∆𝑀𝐴𝑋 = 26.36 𝑚𝑚
 d.) Is the beam adequate for live load deflection limit as per code?
𝐿 9000
∆= 360 = 360

∆= 25 𝑚𝑚 < 26.36 𝑚𝑚 ∴ 𝐼𝑛𝑎𝑑𝑒𝑞𝑢𝑎𝑡𝑒

e.) Determine the Demand Capacity Ratio for LRFD. Indicate if failed or safe.
𝑤𝑢 𝐿2 1.2(12)+1.6(10)(9)2
𝑀𝑢 = 8
= 8
= 307.8 𝑘𝑁 ∗ 𝑚
𝑀 307.8
𝐷𝐶𝑅 = ∅𝑀𝑢 = 225.192
𝑛

𝐷𝐶𝑅 = 1.366857802 > 1.0 ∴ 𝐹𝐴𝐼𝐿, 𝑁𝑂𝑇 𝑆𝐴𝐹𝐸

f.) Determine the Demand Capacity Ratio for ASD. Indicate if failed or safe.
𝑤𝑢 𝐿2 (12+10)(9)2
𝑀𝑎 = = = 222.75 𝑘𝑁 ∗ 𝑚
8 8
𝑀𝑛 250.2144
Ω
= 1.67
= 149.828982 𝑘𝑁 ∗ 𝑚
𝑀𝑎 222.75
𝐷𝐶𝑅 = 𝑀𝑛 = 149.828
Ω

𝐷𝐶𝑅 = 1.48869501 > 1.0 ∴ 𝐹𝐴𝐼𝐿, 𝑁𝑂𝑇 𝑆𝐴𝐹𝐸

Problem 5 Design of Roof Purlins

An A36 channel section is used as a purlin in a roof system with spacing of trusses at 4.5 m. apart. Purlins are to be placed
at the joints and at the midpoints of each top chord member. Sag rods are placed at the center of each purlin. The total
gravity load consists of a dead load of 1.2 kPa of roof surface (including wt. of roofing and purlins) and a live load of 1.4
kPa of the roof surface. Assume the steel section and that all loads are applied at the top flange of the purlins. Use LRFD
Method.

Properties of channel section

Sx = 221.28 x 103 mm3
Sy = 19.01 x 103 mm3
Zx = 258.92 x 103 mm3
Zy = 38.51 x 103 mm3
Lp = 0.90
Lr = 3.34 m
Cb = 1.30
Fy = 248 MPa
U = 1.2DL + 1.6LL
a.) Determine the flexural strength of the purlins about the x-axis.

∴ 𝐿𝑝 < 𝐿𝑏 < 𝐿𝑟 , 𝐼𝑛𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝐿𝑇𝐵

𝑀𝑝 = 𝐹𝑦 𝑍𝑥 = 248(258.9 ∗ 103 ) = 64.212160 𝑘𝑁 ∗ 𝑚

𝐿 −𝐿
𝑀𝑛 = 𝐶𝑏 (𝑀𝑝 − (𝑀𝑝 − 0.7𝐹𝑦 𝑆𝑥 ) (𝐿𝑏−𝐿𝑝 )) ≤ 𝑀𝑝
𝑟 𝑝

((0.7(248)(221.23∗103 ) 2.25−0.9
𝑀𝑛 = 1.3 (64.212 − (64.212 − 106
) (3.34−0.9)) ≤ 𝑀𝑝
 𝑀𝑛 = 64.920313 𝑘𝑁 ∗ 𝑚 > 64.212 𝑘𝑁 ∗ 𝑚 ∴ 𝑁𝑂𝑇 𝑂𝐾

𝑀𝑛 = 64.21 𝑘𝑁 ∗ 𝑚

b.) Determine the flexural strength of the purlins about the y-axis.

𝑀𝑝𝑦 = 𝐹𝑦 𝑍𝑦 ≤ 1.6𝐹𝑦 𝑆𝑦

248(38.51 ∗ 103 ) 1.6(248)(19.01 ∗ 103 )

𝑀𝑝𝑦 = ≤
106 106
𝑀𝑝𝑦 = 9.55 ≤ 7.54 ∴ 𝑈𝑝𝑝𝑒𝑟𝑙𝑖𝑚𝑖𝑡 𝑔𝑜𝑣𝑒𝑟𝑛𝑠

𝑀𝑝𝑦 = 7.54 𝑘𝑁 ∗ 𝑚

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