Tension Member MCQ Quiz - Objective Question with Answer for Tension Member - Download Free PDF
Last updated on Jun 13, 2025
Latest Tension Member MCQ Objective Questions
Tension Member Question 1:
The shear lag effect in beam flanges are disregarded when the outstand of the beam flange is less than or equal to
Answer (Detailed Solution Below)
Tension Member Question 1 Detailed Solution
Explanation:
IS 800(Page number 53)
The shear lag effects in flanges may be disregarded provided:
a) For outstand elements (supported along one edge), bO ≤ LO/20;
and b) For internal elements (supported along two edges), bi ≤ LO/ 10.
where LO = length between points of zero moment (inflection) in the span, bO = width of the flange with outstand, and bi = width of the flange as an internal element..
Additional InformationShear Lag
Shear lag is a structural effect where not all parts of a cross-section effectively resist stress, especially in tension or bending. This leads to non-uniform stress distribution, typically in wide flanges or tension plates.
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It occurs because stresses don’t distribute instantly across the entire section.
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Areas far from the point of load application (e.g., edges of wide flanges) respond with a delay.
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Shear lag is accounted for by reducing effective width or using effective area in calculations.
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Helps prevent overestimation of capacity, especially in critical structures.
Tension Member Question 2:
The best tension member section will be a
Answer (Detailed Solution Below)
Tension Member Question 2 Detailed Solution
Explanation:
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Double angle sections on opposite sides of a gusset plate provide better symmetry, load distribution, and stability under tension.
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They reduce eccentricity and bending effects compared to single angles.
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Bolted or welded single angles and channel sections are less efficient due to eccentricity and potential twisting.
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Double angles connected back-to-back with a gusset plate are commonly used as tension members in trusses and bracing.
Additional Information
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Tension members are structural elements designed to carry axial tensile forces.
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Important considerations include cross-sectional area, connection type, and load transfer efficiency.
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IS 800 provides guidelines on design and selection of tension members in steel structures.
Tension Member Question 3:
For a steel member in tension, the permissible stress in axial tension is given by
Answer (Detailed Solution Below)
Tension Member Question 3 Detailed Solution
Concept:
The permissible stresses for a steel member are given in the below table:
|
Steel member |
Permissible stress |
|
In axial tension |
0.6 fy |
| In axial compression | 0.6 fy |
|
In bending |
0.66 fy |
|
In bearing |
0.75 fy |
|
In average shear |
0.4 fy |
| In maximum shear | 0.45 fy |
Tension Member Question 4:
Match List-I with List-II and select correct answer using the codes given below the list:
|
List I (Member) |
List II (Max. effective slenderness ratio) |
|
A. A member carrying compressive loads resulting from dead loads and imposed loads |
P. 400 |
|
B. Compression flange of a beam against lateral torsional buckling |
Q. 350 |
|
C. Member always under tension |
R. 300 |
|
D. A member normally acting as a tie in a roof truss |
S. 180 |
Answer (Detailed Solution Below)
Tension Member Question 4 Detailed Solution
Explanation:
As per IS: 800 - 2007
|
Type of member |
Maximum slenderness ratio |
|
A member carrying compressive loads resulting from the dead load and the imposed load |
180 |
|
A tension member in which a reversal of direct stress occurs due to loads other than wind and seismic forces |
180 |
|
A member subjected to compression force resulting only from combination with wind/earthquake actions, provided the deformation of such members does not adversely affect the stress in any part of the structure |
250 |
|
Compression flange of a beam against lateral torsional buckling |
300 |
|
A member normally acting as a tie in a roof truss or a bracing system not considered effective when subjected to a possible reversal of stress into compression resulting from the action of wind earthquake forces. |
350 |
|
Member always under tension (other than pre-tensioned members) |
400 |
Tension Member Question 5:
Find the design tension strength due to yielding of the given single angle (ISA 60 × 60 × 8 in the figure) tension member.
(Gross area of the angle = 896 mm2, fy = 250 N/mm2 and fu = 410 N/mm2)
Answer (Detailed Solution Below)
Tension Member Question 5 Detailed Solution
Concept:
Design strength due to yielding of the gross section is calculated using the formula:
\( T_{dg} = \frac{A_g \cdot f_y}{\gamma_{mo}} \)
Where:
•
•
•
Given:
• Gross area,
• Yield stress,
•
Calculation:
\( T_{dg} = \frac{896 \times 250}{1.10} = \frac{224000}{1.10} = 203636.36 \, \text{N} \)
Convert to kN: \( T_{dg} = 203.64 \, \text{kN} \)
Top Tension Member MCQ Objective Questions
A steel rod of 20 mm diameter is used as a tie member in the roof bracing system and may be subjected to possible reversal of stress due to wind load. What is the maximum permissible length of the member?
Answer (Detailed Solution Below)
Tension Member Question 6 Detailed Solution
Download Solution PDFAs per IS 800:2007, The maximum slenderness ratio for a member normally acting as a tie in a roof truss or a bracing system but subjected to possible reversal of stresses resulting from the action of wind or earthquake forces will be 350.
Slenderness ratio “λ” is given by:
\({\rm{\lambda }} = \frac{{{{\rm{l}}_{{\rm{eff}}}}}}{{{{\rm{r}}_{{\rm{min}}}}}}\)
Where,
rmin = minimum radius of gyration of the member, and ℓeff = effective length of the member
\({{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{\rm{I}}}{{\rm{A}}}} = \sqrt {\frac{{\frac{{\rm{\pi }}}{{64}}{{\rm{D}}^4}}}{{\frac{{\rm{\pi }}}{4}{{\rm{D}}^2}}}} = \frac{{\rm{D}}}{4} = \frac{{20}}{4} = 5{\rm{\;mm}}\)
Where,
A = Area of the member and I = Moment of the inertia of the member about its center of gravity
λ = 350
∴ ℓeff = 350 x 5 = 1750 mmA tie member consists of ISA 100 × 75 × 8 longer leg is connected to a gusset plate. The gross cross sectional area of connected and unconnected leg are _______ mm2 respectively.
Answer (Detailed Solution Below)
Tension Member Question 7 Detailed Solution
Download Solution PDFConcept:
Case i) Tensile strength due to gross yielding
\(\therefore {{\text{T}}_{1}}=\text{Ag}\times \frac{{{\text{f}}_{\text{y}}}}{{{\text{ }\!\!\gamma\!\!\text{ }}_{\text{m}0}}}\)
Case ii) Tensile strength due to net section fracture
\({{\text{T}}_{2}}=0.9\frac{{{\text{A}}_{\text{nc}}}\times {{\text{f}}_{\text{u}}}}{{{\text{ }\!\!\gamma\!\!\text{ }}_{{{\text{m}}_{1}}}}}+\text{ }\!\!\beta\!\!\text{ }\times \frac{{{\text{A}}_{{{\text{g}}_{\text{o}}}}}\times {{\text{f}}_{\text{y}}}}{{{\text{ }\!\!\gamma\!\!\text{ }}_{{{\text{m}}_{0}}}}}\)
Area of connected leg (Anc) \(=\left( a-\frac{t}{2} \right)\times t\)
Area of outstanding leg (Ago) \(=\left( b-\frac{t}{2} \right)\times t\)
Calculation:
Gross cross sectional area of connected leg (Anc) \(=\left( 100-\frac{8}{2} \right)\times 8=768~m{{m}^{2}}\)
Gross cross sectional area of outstanding leg (Ago) \(=\left( 75-\frac{8}{2} \right)\times 8=568~m{{m}^{2}}\)
As per IS 800:2007, the maximum effective slenderness ratio of a member normally carrying tension but subjected to reversal of stress due to wind or earthquake forces should be
Answer (Detailed Solution Below)
Tension Member Question 8 Detailed Solution
Download Solution PDFExplanation:
The maximum effective slenderness ratio of a member normally carrying tension but subjected to reversal of stress due to wind or earthquake forces should be 350
Maximum slenderness ratio for tension members:
|
Type of Member |
λ |
|
A tension member in which a reversal of direct stress occurs due to loads other than wind or seismic forces. |
180 |
|
A member normally acts as a tie-in roof truss or a bracing system but is subjected to a possible reversal of stresses resulting from the action of the wind or earthquake forces. |
350 |
|
Member always under tension |
400 |
Additional InformationMaximum slenderness ratio for compression members:
|
Type of Member |
λ |
|
A tension member in which a reversal of direct stress occurs due to loads other than wind or seismic forces. |
180 |
|
A member carrying compressive loads resulting from dead loads and imposed loads. |
180 |
|
A member subjected to compressive forces resulting only from combination with wind/earthquake actions, provided the deformation of such members does not adversely affect the stress in any part of the structure. |
250 |
|
Compression flange of a beam restrained against lateral-torsional buckling. |
300 |
|
A member normally acting as a tie in a roof truss or a bracing system is not considered effective when subjected to a possible reversal of stresses resulting from the action of wind or earthquake forces. |
350 |
The design strength of a tension member is given by a minimum of
i) Block shear strength of end region
ii) Rupture of critical section
iii) Yielding of net section
Which of the above statements(s) is/are correct?Answer (Detailed Solution Below)
Tension Member Question 9 Detailed Solution
Download Solution PDFAs per IS 800: 2007 codal provisions Design strength of tension member is the minimum of the following:
1. Design strength due to yielding of Gross section.
2. Design strength due to Rupture of Net Section.
3. Block Shear Strength
Hence, Statement 3 is false as such Design strength of tension member is based on the yielding of Gross section and not net section.
Important Point:
Design strength due to Rupture of Net Section for plates is based on net effective area at critical section and for angle section is based on shear lag effect.
Maximum slenderness ratio for tension members shall not exceed
Answer (Detailed Solution Below)
Tension Member Question 10 Detailed Solution
Download Solution PDFAs per IS: 800 - 2007
|
Type of member |
Maximum slenderness ratio |
|
A member carrying compressive loads resulting from the dead load and the imposed load |
180 |
|
A tension member in which a reversal of direct stress occurs due to loads other than wind and seismic forces |
180 |
|
A member subjected to compression force resulting only from combination with wind/earthquake actions, provided the deformation of such members does not adversely affect the stress in any part of the structure |
250 |
|
Compression flange of a beam against lateral torsional buckling |
300 |
|
A member normally acting as a tie in a roof truss or a bracing system not considered effective when subjected to a possible reversal of stress into compression resulting from the action of wind earthquake forces. |
350 |
|
Member always under tension (other than pre-tensioned members) |
400 |
For preliminary sizing, the rupture strength of net section may be approximately taken by using the below equation as per IS ∶ 800-2007,
Tdn = αAnfu/γml
What is the "α" value for one or two bolts along the length in the end connection or equivalent weld length.
Answer (Detailed Solution Below)
Tension Member Question 11 Detailed Solution
Download Solution PDFConcept:
As per IS 800 : 2007, Clause 6.3.3
For preliminary sizing, the rupture strength of net section may be approximately taken as:
\(T_{dn}={α A_nf_u\over\gamma_{ml}}\)
Where
α = 0.6 for one or two bolts, 0.7 for three bolts, and 0.8 for four or more bolts among the length in the end connection or equivalent weld length.
An = Net area of the total cross-section;
Anc = Net area of the connecting leg; and
\(\gamma_{ml}\) = Partial safety factor for failure at ultimate stress.
Hence option (4) is correct.
The slenderness ratio in tension member as per BIS Code where reversal of stress is due to loads other than wind or seismic shall not exceed:
Answer (Detailed Solution Below)
Tension Member Question 12 Detailed Solution
Download Solution PDFExplanation:
As per IS 800: 2007 limit of slenderness ratio for tension member where reversal of stress is due to loads other than wind or seismic shall not exceed 180.
Refer table 3 of IS800: 2007, we get following values of maximum slenderness ratios for tension members:
|
Maximum Slenderness Ratios for tension members |
|
|
A tension member in which reversal of stress due to loads other than wind or seismic loads. |
180 |
|
A member normally acting as a tie in roof truss or a bracing system but subjected to reversal of stresses resulting from the action of wind or earthquake forces. |
350 |
|
Tension Members (other than Pre-tensioned members) |
400 |
The maximum effective slenderness ratio for a tension member subjected to reversal of stress due to others (other than wind and earthquake) is ______.
Answer (Detailed Solution Below)
Tension Member Question 13 Detailed Solution
Download Solution PDFExplanation:
Maximum values of Effective Slenderness Ratios
| Sl. No. | Member | Maximum Effective Slenderness Ratio (KL/r) |
| (i) | A member carrying compressive loads resulting from dead loads and imposed loads | 180 |
| (ii) | A tension member in which a reversal of direct stress occurs due to loads other than wind or seismic forces | 180 |
| (iii) | A member subjected to compression forces resulting only from combination with wind/earthquake actions provided the deformation of such member does not adversely affect the stress in any part of the structure. | 250 |
| (iv) | Compression flange of a beam against lateral-torsional buckling. | 300 |
| (v) | A member normally acts as a tie in a roof truss or a bracing system not considered effective when subject to possible reversal of stress into compression resulting from the action of wind or earthquake forces. | 350 |
| (vi) | Members always under tension (other than pre-tensioned members) | 400 |
A steel plate is 300 mm wide and 40 mm thick is connected using 24 mm dia bolts of 3 numbers. The net sectional area of the plate is
Answer (Detailed Solution Below)
Tension Member Question 14 Detailed Solution
Download Solution PDFConcept:
Tensile strength of a section, (Tdn)
\({{\rm{T}}_{{\rm{dn}}}} = \frac{{0.9{{\rm{A}}_{\rm{n}}}{{\rm{f}}_{\rm{u}}}}}{{{{\rm{\gamma }}_{{\rm{m}}1}}}}\)
\({{\rm{A}}_{\rm{n}}} = \left[ {{\rm{b}} - {\rm{n}}{{\rm{d}}_{\rm{h}}} + \sum \frac{{{\rm{P}}_{{\rm{si}}}^2}}{{4{{\rm{g}}_1}}}} \right]{\rm{\;}}\)× t
Where,
An = net section area (mm2),
fu = ultimate strength of the material,
γm1 = partial safety factor,
g = gauge length between the bolt holes,
Ps = Staggered pitch length between lines of bolt holes (Zero for no staggering),
dh = Diameter of the bolt hole, and
n = Number of bolt holes in the critical section.
Calculation:
Diameter of bolt hole, dh = 24 + 2 = 26 mm
Width of steel plate, b = 300 mm
Thickness of steel plate, t = 40 mm
No. of bolts, n = 3
∴ Net sectional area, An = [300 – 3 × 26] × 40 = 8880 mm2
For a steel member in tension, the permissible stress in axial tension is given by
Answer (Detailed Solution Below)
Tension Member Question 15 Detailed Solution
Download Solution PDFConcept:
The permissible stresses for a steel member are given in the below table:
|
Steel member |
Permissible stress |
|
In axial tension |
0.6 fy |
| In axial compression | 0.6 fy |
|
In bending |
0.66 fy |
|
In bearing |
0.75 fy |
|
In average shear |
0.4 fy |
| In maximum shear | 0.45 fy |