Steel Design Test 2

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Steel Design Test 2

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Weekly Quiz Competition

Question 1
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The IS 800:2007 says that the effective slenderness ratio of the laced built-up column is to be increased by a factor X to that of the maximum slenderness ratio of the column. Let Y be the reason for doing this. Which of the following is correct?

A
X = 1.1 and Y = to enhance bending resistance.

B
X = 1.05 and Y = to enhance bending resistance.

C
X = 1.1 and Y = to account for shear deformations

D
X = 1.05 and Y = to account for shear deformations.

Solution

Concept
As per IS 800:2007, the effective slenderness ratio of the laced built-up column shall be taken as 1.05 times maximum slenderness ratio of the column. This is done to account for the shear deformations.
∴ X = 1.05 and Y = to account for shear deformations.
Note:
The same value in the case of a battened built-up column system is 1.1
Question 2
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Consider the following two statements:
Statement I: A semi-compact section cannot reach yield stress.
Statement II: A semi-compact section cannot develop plastic moment of resistance.
Which of the following is CORRECT?

A
Statement I is true, but Statement II is false

B
Both Statement I and Statement II are false

C
Both Statement I and Statement II are true

D
Statement I is false, but Statement II is true

Solution

Concept:
A plate may buckle locally due to compressive stresses. This local buckling can be avoided before the limit state is reached, by limiting the width to thickness ratio (b/t) of each element of the cross-section. On this basis 4 classes of sections are defined as follows:
1) Plastic section (Class 1): Those sections which can develop plastic hinges and has the rotational capacity required for the failure of the structure by plastic mechanism formation.
2) Compact section (Class 2): Those sections which can develop plastic hinge but have inadequate rotational capacity for the formation of plastic mechanism due to local buckling.
3) Semi-compact section (class 3): Those section in which extreme fibers can reach yield stress (f_y) but can not develop the plastic moment of resistive due to local buckling.
4) Slender-section (Class 4): Those section in which local buckling occurs before yield stress is reached are called slender section.
Hence statement I is false and statement II is true for a semi-compact section.
Question 3
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A beam column for a non-sway column in a building frame is subjected to a factored axial load of 500 kN, factored moment at bottom of column of 45 kNm. For ISHB 200, the values are A = 4750 mm², γ_y = 45.1, h = 200 mm, b = 200 mm, b_f = 9 mm and the effective length is 0.8 L. Its buckling load will be

A
910 kN

B
930 kN

C
950 kN

D
980 kN

Solution

Concept:
As per yura,
Bucking load of column, \(p_e\; =p + \frac{{2M}}{d}\)
Where, p = Factored axial load
M = Factored axial moment
d = depth of beam section
Calculation:
Given, p = 500 kN, M = 45 kNm and d = 200 mm = 0.2 m
∴ \({p_e} = 500 + \frac{{2 \times 45}}{{0.2}} = 950\;kN\)
Question 4
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An I-section beam is fabricated with plates having an elastic section modulus of 200 × 10^-5m³. If the laterally unsupported steel beam is classified as a semi-compact section, then the design plastic moment capacity of the beam considering Fe 410 grade of steel and the partial safety factor of material as 1.1 will be ____ kNm. (Use Bending stress reduction factor (X_LT= 1.0)

A
453-456

B
4534-4561

C
4537-4566

D
4539-4562

Solution

Concept:
The bending strength or the plastic moment capacity of a laterally unsupported beam is given by:
Md = β_b Z_p f_bd
Where
β_b = 1 → for plastic and concept section
\(= \frac{{{{\rm{z}}_{\rm{e}}}}}{{{{\rm{z}}_{\rm{p}}}}} \to {\rm{for\;semi}} - {\rm{compact\;section}}.\)
Z_e = Elastic section modulus
Z_p = Plastic section modulus
f_bd = design bending compressive strength
\(= {{\rm{X}}_{{\rm{LT}}}}\frac{{{{\rm{f}}_{\rm{y}}}}}{{{{\rm{y}}_{{\rm{mo}}}}}}\)
y_mo = Partial safety factor for material = 1.1
X_LT = bending stress reduction factor to account for lateral torsional buckling
X_LT ≤ 1.0
Calculation:
For a semi-compact section of a laterally unsupported steel bean plastic moment capacity
\({{\rm{M}}_{\rm{d}}} = \frac{{{\rm{ze}}}}{{{\rm{zp}}}} \times {\rm{zp}} \times {{\rm{X}}_{{\rm{LT}}}}\frac{{{\rm{fy}}}}{{{{\rm{y}}_{{\rm{mo}}}}}}{\rm{\;}}\)
X_LT = 1.0
γ_mo = 1.1
f_y = 250 Mpa for grade fe 410
\(\therefore {\rm{\;Md\;}} = {\rm{\;}}200{\rm{\;}} \times {\rm{\;}}{10^{ - 5}}{{\rm{m}}^3} \times {\rm{\;}}1{\rm{\;}} \times \frac{{250}}{{1.1}}{\rm{N}}/{\rm{m}}{{\rm{m}}^2}\)
\({\rm{Md\;}} = {\rm{\;}}200{\rm{\;}} \times {\rm{\;}}{10^{ - 5}}{{\rm{m}}^3}{\rm{\;}} \times \frac{{250}}{{1.1}} \times \frac{1}{{{{10}^{ - 6}}}} \times {10^{ - 3}}{\rm{KN}}/{{\rm{m}}^2}\) = 454.54 kNm