Design of Steel Structures Multiple Choice Questions 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’p^{2}t/8g

c) ndt – n’p^{2}t/4g

d) nd + n’p^{2}t/4g

Answer: c

Clarification: The net area of flat with staggered hole is given by : A = (b – ndh + n’p^{2}/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 cm^{2}

b) 20 cm^{2}

c) 240 mm^{2}

d) 480 mm^{2}

Answer: a

Clarification: b = 40cm = 400mm, t = 10mm, d_{h} = 18mm

Net section area = 400×10 – 16×10 = 3820mm^{2} = 38.2 cm^{2}.

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

Answer: d

Clarification: 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) 2796mm^{2}

b) 2681mm^{2}

c) 2861mm^{2}

d) 3055mm^{2}

Answer: b

Clarification: B = 300mm, t = 10mm, d_{h} = 18+2 =20mm, n = 3, n’ = 1, p = 75mm, g = 50mm

Effective net area = (B-nd_{h}+n’p^{2}/4g)t = (300 – 3×20 + 1×75^{2}/4×50)x10 = 2681.25 mm^{2}.

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 mm^{2}

b) 756 mm^{2}

c) 800 mm^{2}

d) 640 mm^{2}

Answer: d

Clarification: In case of drilled hole, d_{h} = 20mm

Net Area A_{n} = A_{g} – d_{h}t = 100 x 8 – 20 x 8 = 640mm^{2}.

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

Answer: c

Clarification: Net area of connected leg, A_{nc} = (100 – 12/2) x 12 = 1128 mm^{2}

Net area of outstanding leg, A_{go} = (75 – 12/2) x 12 = 828 mm^{2}

Total net area = 1128 + 828 = 1956 mm^{2}.

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

Answer: a

Clarification: 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

Answer: c

Clarification: 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 mm^{2}

Tensile strength due to gross section yielding, T_{dg }= A_{g}f_{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

Answer: d

Clarification: 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 mm^{2}, A_{g0} = (75 – 10/2) x 10 = 700 mm^{2}

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_{u}A_{nc} /γ_{m1} + βA_{g0}f_{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

Answer: b

Clarification: f_{u} = 410 MPa, f_{y} = 250 MPa, γ_{m0 }= 1.1, γ_{m1 }= 1.25

A_{vg} = ( 1×100 + 50 ) x 8 = 1200 mm^{2}

A_{vn} = (1×100 + 50 – (2 – 1/2)x20 ) x 8 = 1440 mm^{2}

A_{tg} = 35 x 8 = 280 mm^{2}

A_{tn} = (35 – 1/2 x 20) x 8 = 200 mm^{2}

T_{db1} = (A_{vg}f_{y}/√3 γ_{m0})+(0.9A_{tn}f_{u}/γ_{m1}) = [(1200×250/ 1.1x√3) + (0.9x200x410 / 1.25) ] x 10-3 = 216.49 kN

T_{db2} = (A_{tg}f_{y}/ γ_{m0})+(0.9A_{vn}f_{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.