1. The design strength of tension member corresponding to gross section yielding is given by :
a) γm0 fyAg
b) γm0fy/Ag
c) fy/Ag γm0
d) fyAg/ γm0
Explanation: The design strength of tension member corresponding to gross section yielding is given by Tdg = fyAg/ γm0, where fy = yield strength of material in MPa, Ag = gross cross-sectional area in mm2, γm0 = partial safety factor for failure in tension by yielding = 1.10.
2. Which of the following relation is correct?
a) Net area = Gross area x deductions
b) Net area = Gross area + deductions
c) Net area = Gross area – deductions
d) Net area = Gross area / deductions
Explanation: Net area = Gross area – deductions, that is net area of tensile members is calculated by deducting areal of holes from the gross area.
3. The design strength of tension member corresponding to net section rupture is given by :
a) Anfyγm1
b) 0.9Anfyγm1
c) 0.9An/fyγm1
d) 0.9Anfy/γm1
Explanation: The design strength of tension member corresponding to net section rupture is given by Tdn = 0.9Anfy/γm1, where An = net effective area of cross section in mm2, fy = ultimate strength of material in MPa, γm1 = partial safety factor for failure due to rupture of cross section = 1.25.
4.The block shear strength at an end connection for shear yield and tension fracture is given by :
a) (Avgfy/√3 γm0)+(0.9Atnfu/γm1)
b) (Atgfy/√3 γm0)+(0.9Avnfu/γm1)
c) (0.9Atgfy/√3 γm0)+( Avnfu/γm1)
d) (0.9Avgfy/√3 γm0)+(Atnfu/γm1)
Explanation: The block shear strength at an end connection for shear yield and tension fracture is given by Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfu/γm1), where Avg = minimum gross area in shear along line of action of force, Atn = minimum net area of cross section in tension from hole to toe of angle or last row of bolts in plates perpendicular to line of force, fy and fu are yield and ultimate stress of material respectively, γm1 = 1.25, γm0 = 1.10.
5. The block shear strength at an end connection for shear fracture and tension yield is given by :
a) (Avgfy/√3 γm0)+(0.9Atnfu/γm1)
b) (Atgfy/ γm0)+(0.9Avnfu/√3 γm1)
c) (0.9Avgfy/√3 γm0)+(Atnfu/γm1)
d) (0.9Atgfy/√3 γm0)+( Avnfu/γm1)
Explanation: The block shear strength at an end connection for shear fracture and tension yield is given by Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γm1), where Avn = minimum net area in shear along line of action of force, Atg = minimum gross area in tension from hole to toe of angle or last row of bolts in plates perpendicular to line of force, fy and fu are yield and ultimate stress of material respectively, γm1 = 1.25, γm0 = 1.10.
6. The block shear strength of connection is ________
a) block shear strength at an end connection for shear fracture and tension yield
b) block shear strength at an end connection for shear yield and tension fracture
c) larger of block shear strength at an end connection for (shear fracture, tension yield) and (shear yield, tension fracture)
d) smaller of block shear strength at an end connection for (shear fracture, tension yield) and (shear yield, tension fracture)
Explanation: The block shear strength of connection is smaller of block shear strength at an end connection for shear yield, tension fracture Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfu/γm1) and block shear strength at an end connection for shear fracture, tension yield Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γvm1).
7. The design tensile strength of tensile member is
a) minimum of strength due to gross yielding, net section rupture, block shear
b) maximum of strength due to gross yielding, net section rupture, block shear
c) strength due to gross yielding
d) strength due to block shear
Explanation: The design tensile strength of tensile member is taken as the minimum of strength due to gross yielding (Tdg=fyAg/1.1), net section rupture(Tdn=0.9Anfy/γm1), block shear (Tdb1=(Avgfy/√3 γm0)+(0.9Atnfu/γm1), Tdb2=(Atgfy/ γm0)+(0.9Avnfu/√3 γm1)).
8. Which of the following is not true for angles as tension members?
a) Angles if axially loaded through centroid can be designed as plates
b) Angles connected to gusset plates by welding or bolting only through one of the two legs results in eccentric loading
c) When load is applied by connecting only one leg of member, there is shear lag at the end connection
d) When angles are connected to gusset plates by welding or bolting only through one of the two legs resulting in eccentric loading, there is a uniform stress distribution over cross section.
Explanation: Angles if axially loaded through centroid can be designed as plates. Angles connected to gusset plates by welding or bolting only through one of the two legs results in eccentric loading, causing non-uniform stress distribution over cross section. When load is applied by connecting only one leg of member, there is shear lag at the end connection.
9. Which of the following is true statement?
a) thickness of angle has no significant influence on member strength
b) net section efficiency is lower when long leg of angle is connected rather than short leg
c) when length of connection decreases, the tensile strength increases
d) effect of gusset plate thickness on ultimate tensile strength is significant
Explanation: (i) The effect of gusset plate thickness on ultimate tensile strength is not significant, (ii) the thickness of angle has no significant influence on member strength, (iii) the net section efficiency is higher(7-10%) when long leg of angle is connected rather than short leg, (iv) when length of connection increases, the tensile strength increases upto four bolts and effect of any further increase in number of bolts on tensile strength of member is not significant.
10. The additional factor to be added for angles for design strength of tension member corresponding to net section rupture is given by :
a) βAg0fyγm0
b) βAg0fy/γm0
c) βAg0γm0
d) βAg0/fyγm0
Explanation: The design strength of angle section governed by tearing at net section is given by Tdn = (0.9Ancfu/γm1) + βAg0fy/γm0 , where Anc = net area of connected leg, Ag0 = gross area of outstanding leg, fu = ultimate strength of material, γm1 = partial safety factor for failure due to rupture of cross section = 1.25, γm0 = partial safety factor for failure in tension by yielding = 1.10.