Tension Members Design Questions and Answers

This set of Design of Steel Structures Multiple Choice Questions & Answers (MCQs) focuses on “Design of Tension Members”.

1. For the calculation of net area of flat with staggered bolts, the area to be deducted from gross area is :
a) nd
b) n’p²t/8g
c) ndt – n’p²t/4g
d) nd + n’p²t/4g
View Answer

Answer: c
Explanation: The net area of flat with staggered hole is given by : A = (b – ndh + n’p²/4g)t, where b = width of plate, n = number of holes in zig-zag line, n’ = number of staggered pitches, p = pitch distance, g= gauge distance, t = thickness of flat.

2. What is the net section area of steel plate 40cm wide and 10mm thick with one bolt if diameter of bolt hole is 18mm?
a) 38.2 cm²
b) 20 cm²
c) 240 mm²
d) 480 mm²
View Answer

Answer: a
Explanation: b = 40cm = 400mm, t = 10mm, d_h = 18mm
Net section area = 400×10 – 16×10 = 3820mm² = 38.2 cm².

3. Which section to be considered in the design for the net area of flat?

a) 1-5-6-3
b) 2-7-4
c) 1-5-7-4
d) 1-5-7-6-3
View Answer

Answer: d
Explanation: The section giving minimum area of plate is considered for design. So, section 1-5-7-6-3 is used for net area of flat.

4. Calculate the minimum effective net area for the given section (300mm width, 10mm thick) connected to a 10 mm thick gusset plate by 18mm diameter bolts.

a) 2796mm²
b) 2681mm²
c) 2861mm²
d) 3055mm²
View Answer

Answer: b
Explanation: B = 300mm, t = 10mm, d_h = 18+2 =20mm, n = 3, n’ = 1, p = 75mm, g = 50mm
Effective net area = (B-nd_h+n’p²/4g)t = (300 – 3×20 + 1×75²/4×50)x10 = 2681.25 mm².

5. What is the net area for the plate 100 x 8 mm bolted with a single bolt of 20mm diameter in case of drilled hole ?
a) 624 mm²
b) 756 mm²
c) 800 mm²
d) 640 mm²
View Answer

Answer: d
Explanation: In case of drilled hole, d_h = 20mm
Net Area A_n = A_g – d_ht = 100 x 8 – 20 x 8 = 640mm².

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6. Determine the effective net area for angle section ISA 100 x 75 x 12 mm, when 100mm leg is connected to a gusset plate using weld of length 140mm.
a) 1795 mm2
b) 1812 mm2
c) 1956 mm2
d) 2100 mm2
View Answer

Answer: c
Explanation: Net area of connected leg, A_nc = (100 – 12/2) x 12 = 1128 mm²
Net area of outstanding leg, A_go = (75 – 12/2) x 12 = 828 mm²
Total net area = 1128 + 828 = 1956 mm².

7. Calculate the value of β for the given angle section ISA 150x115x8mm of Fe410 grade of steel connected with gusset plate : Length of weld = 150mm
a) 0.89
b) 0.75
c) 0.5
d) 1
View Answer

Answer: a
Explanation: w=115mm, t=8mm, b=115mm, L_c=150mm, f_y=250MPa, f_u=410MPa
β = 1.4 – [0.076 (w/t)(f_y/f_u)(b_s/L_c)] = 1.4 – [ 0.076 x (115/8) x (250/410) x (115/150)] = 0.89 (>0.7) .

8. Calculate the tensile strength due to gross section yielding of an angle section 125 x 75 x 10mm of Fe410 grade of steel connected with a gusset plate.
a) 780 kN
b) 586.95 kN
c) 432.27 kN
d) 225.36 kN
View Answer

Answer: c
Explanation: f_u = 410MPa, f_y = 250MPa, γ_m0 = 1.1, γ_m1= 1.25,
For ISA 125 x 75 x 10mm : gross area A_g = 1902 mm²
Tensile strength due to gross section yielding, T_dg= A_gf_y/γ_m0 = 1902 x 250 x 10-3 / 1.1 = 432.27 kN.

9. A single unequal angle 100 x 75 x 10 of Fe410 grade of steel is connected to a 10mm thick gusset plate at the ends with six 16mm diameter bolts with pitch of 40mm to transfer tension. Find the tensile strength due to net section rupture if gusset is connected to 100mm leg.
a) 526.83 kN
b) 385.74 kN
c) 450.98 kN
d) 416.62 kN
View Answer

Answer: d
Explanation: d_h=18 mm, f_u = 410MPa, f_y = 250MPa, γ_m0 = 1.1, γ_m1 = 1.25
A_nc= (100 – 10/2 – 18) x 10 = 770 mm², A_g0 = (75 – 10/2) x 10 = 700 mm²
β = 1.4 – 0.076(w/t)(f_y/f_u)(b_s/L_c)
= 1.4 – 0.076 [(75-5)/10] [250/410] [{(75-5)+(100-40)}/{40×5}] = 1.19 > 0.7 and < 1.44[(410/250)(1.1/1.25)] (=2.07)
T_dn = 0.9f_uA_nc /γ_m1 + βA_g0f_y/γ_m0
= [0.9x410x770/1.25 + 1.19x700x250/1.1] x 10-3 = 416.62 kN.

10. Determine block shear strength of tension member shown in figure if grade of steel is Fe410.

a) 309.06 kN
b) 216.49 kN
c) 258.49 kN
d) 326.54 kN
View Answer

Answer: b
Explanation: f_u = 410 MPa, f_y = 250 MPa, γ_m0= 1.1, γ_m1= 1.25
A_vg = ( 1×100 + 50 ) x 8 = 1200 mm²
A_vn = (1×100 + 50 – (2 – 1/2)x20 ) x 8 = 1440 mm²
A_tg = 35 x 8 = 280 mm²
A_tn = (35 – 1/2 x 20) x 8 = 200 mm²
T_db1 = (A_vgf_y/√3 γ_m0)+(0.9A_tnf_u/γ_m1) = [(1200×250/ 1.1x√3) + (0.9x200x410 / 1.25) ] x 10-3 = 216.49 kN
T_db2 = (A_tgf_y/ γ_m0)+(0.9A_vnf_u/√3 γ_m1) = [ (0.9x1440x410 / 1.25x√3) + (280×250/1.1) ] x 10-3 = 309.06 kN
Block shear strength of tension member is 216.49 kN.

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