Question 1 |
Consider the horizontal axis passing through the centroid of the steel beam cross-section shown in the figure. What is the shape factor (rounded off to one decimal place) for the cross-section ?
1.5 | |
1.7 | |
1.3 | |
2 |
Question 1 Explanation:
\begin{aligned} Z_{P} & =\frac{A}{2}\left(\bar{y}_{1}+\bar{y}_{2}\right) \\ & =\left(3 b \times \frac{b}{2}+b \times b\right)\left(2 \times \frac{11}{20} b\right) \\ & =\frac{11}{4} b^{3} \end{aligned}
z_{e}=\frac{1}{y_{max }}=\frac{\frac{b \times(3 b)^{3}}{12}+\frac{2 b(b)^{3}}{12}}{\frac{3 b}{2}}
z_{e}=\frac{29}{18} b^{3}
Shape factor (S)=\frac{Z_{P}}{Z_{e}}=\frac{\frac{11}{4} b^{3}}{\frac{29}{18} b^{3}}=1.7
Question 2 |
The flange and web plates of the doubly symmetric built-up section are connected by
continuous 10 mm thick fillet welds as shown in the figure (not drawn to the scale). The
moment of inertia of the section about its principal axis X-X is 7.73 \times 10^{6} mm^4. The
permissible shear stress in the fillet welds is 100 N/ mm^2. The design shear strength
of the section is governed by the capacity of the fillet welds.
The maximum shear force (in kN, round off to one decimal place) that can be carried by the section, is _______.
The maximum shear force (in kN, round off to one decimal place) that can be carried by the section, is _______.
393.5 | |
125.4 | |
256.3 | |
485.2 |
Question 2 Explanation:
q = Shear stress at the level mn in the weld = 100 MPa =\frac{SA\bar{y}}{Ib}
Shear force at the given section
A = Area of the cross-section above the level mn=100 \times 10 mm^2
\bar{y}=C.G. of shaded area above the level mn = 60-5 = 55 m
I=7.73 \times 10^6 mm^4
b = Width of weld at mn (4 welds) = 4 x t = 4 x 7 = 28 mm
t = Throat thickness
\begin{aligned} &=0.7 \times s=0.7 \times 10 \times 4=28 mm \\ 100&= \frac{F \times (100 \times 10) \times 55}{7.73 \times 10^6 \times 28}\\ F&=\frac{100 \times 7.73 \times 10^6 \times 28}{1000\times 55} \\ &=393.527kN \end{aligned}
Question 3 |
A rolled I-section beam is supported on a 75 mm wide bearing plate as shown in the figure. Thicknesses of flange and web of the I-section are 20 mm and 8 mm, respectively. Root radius of the I-section is 10 mm. Assume: material yield stress, f_y=250 MPa and partial safety factor for material, \gamma _{mo}=1.10.
As per IS: 800-2007, the web bearing strength (in kN, round off to 2 decimal places) of the beam is ________
As per IS: 800-2007, the web bearing strength (in kN, round off to 2 decimal places) of the beam is ________
272.73 | |
185.64 | |
286.92 | |
312.44 |
Question 3 Explanation:
Web bearing strength
\begin{aligned} &=[b+2.5(t_f+R)]\times t_w \times \frac{f_y}{\gamma _{mo}} \\ &= [75+2.5(20+10)] \times 8 \times \frac{250}{1.1}\\ &= 272.73kN \end{aligned}
Question 4 |
Assuming that there is no possibility of shear buckling in the web, the maximum reduction permitted by IS 800-2007 in the (low-shear) design bending strength of a semi-compact steel section due to high shear is
Zero | |
25% | |
50% | |
governed by the area of the flange |
Question 4 Explanation:
As per IS 800 : 2007
For semi compact section
(i) In low shear case (V \leq 0.6 V_d)
M_d = Z_ef_y/\gamma _{mo}
(ii) In high shear case (V \gt 0.6 V_d)
M_d = Z_ef_y/\gamma _{mo}
So reduction is zero.
For semi compact section
(i) In low shear case (V \leq 0.6 V_d)
M_d = Z_ef_y/\gamma _{mo}
(ii) In high shear case (V \gt 0.6 V_d)
M_d = Z_ef_y/\gamma _{mo}
So reduction is zero.
Question 5 |
A steel column of ISHB 350 @72.4 kg/m is subjected to a factored axial compressive load of 2000 kN. The load is transferred to a concrete pedestal of grade M20 through a square base plate. Consider bearing strength of concrete as 0.45f_{ck}, where f_{ck} is the characteristic strength of concrete. Using limit state method and neglecting the self weight of base plate and steel column, the length of a side of the base plate to be provided is
39 cm | |
42 cm | |
45 cm | |
48 cm |
Question 5 Explanation:
\begin{aligned}
&\text{Area required for base plate}\\ &=\frac{\text{Factored load}}{\text{Bearing capacity of concrete}}\\ &=\frac{2000\times 10^{3}}{0.45\times 20}=222222.222mm^{2}
\end{aligned}
So, side of base plate =\sqrt{\text{Area}}=471.4 mm= 47.14 cm
Since, provided area must be more than required
So, answer should be 48 cm.
So, side of base plate =\sqrt{\text{Area}}=471.4 mm= 47.14 cm
Since, provided area must be more than required
So, answer should be 48 cm.
There are 5 questions to complete.