# GATE CE 2013

 Question 1
There is no value of x that can simultaneously satisfy both the given equations. Therefore, find the 'least squares error' solution to the two equations, i.e., find the value of x that minimizes the sum of squares of the errors in the two equations. __________

2x= 3
4x= 1
 A 0.5 B 1.5 C 2 D 4
Engineering Mathematics   Linear Algebra
 Question 2
What is the minimum number of multiplications involved in computing the matrix product PQR?
MatrixP has 4 rows and 2 columns, matrixQ has 2 rows and 4 columns, and matrixR has 4 rows and 1 column. __________
 A 256 B 32 C 16 D 128
Engineering Mathematics   Linear Algebra
Question 2 Explanation:
If we multiply QR first then,
$Q_{2\times 4}\times R_{\left ( 2\times 1 \right )}$ having multiplication number 8.
Therefore $P_{\left ( 4\times 2 \right )}QR_{\left ( 2\times 1 \right )}$ will have minimum number of multiplication =(8+8)=16.
 Question 3
A 1-h rainfall of 10 cm magnitude at a station has a return period of 50 years. The probability that a 1-h rainfall of magnitude 10 cm or more will occur in each of two successive years is:
 A 0.04 B 0.2 C 0.02 D 0.0004
Engineering Hydrology   Flood Routing and Flood Control
Question 3 Explanation:
Return period of rainfall,
 T&=50 years
$\therefore$ probability of occurrence once in 50 years,
p&=$\frac{1}{50}$&=0.02
Probability of occurrence in each of 2 successive year $=p^{2}=\left ( 0.02 \right )^{2}=0.0004$
 Question 4
Maximum possible value of Compacting Factor for fresh (green) concrete is:
 A 0.5 B 1 C 1.5 D 2
RCC Structures   Working Stress and Limit State Method
 Question 5
As per IS 800:2007, the cross-section in which the extreme fiber can reach the yield stress, but cannot develop the plastic moment of resistance due to failure by local buckling is classified as
 A plastic section B compact section C semi-compact section D slender section
Design of Steel Structures   Plastic Analysis
Question 5 Explanation:
As per clause 3.7.2 of IS 800:2007
 Question 6
The creep strains are
Solid Mechanics   Properties of Metals, Stress and Strain
Question 6 Explanation:
Creep strains occur with time at a constant level of stress. So they occur due to permanent load i.e., Dead Load.
 Question 7
As per IS 456:2000 for M20 grade concrete and plain barsin tension the design bond stress $\tau _{bd}=1.2 MPa$.Further, IS 456:2000 permits this design bond stress value to be increased by 60% for HSD bars. The stress in the HSD reinforcing steel barsin tension, $\sigma _{s}=360MPa$. Find the required development length, $L_{d}$, for HSD barsin terms of the bar diameter, $\phi$. __________
 A 34.54 B 46.87 C 89.25 D 93.47
RCC Structures   Shear, Torsion, Bond, Anchorage and Development Length
Question 7 Explanation:
\begin{aligned} L_{d} &=\frac{\phi \sigma_{s t}}{4 \tau_{b d}}=\frac{\phi \times 360}{4 \times 1.2 \times 1.60} \\ &=46.875 \phi \end{aligned}
 Question 8
The 'plane section remains plane' assumption in bending theory implies:
 A strain profile is linear B stress profile is linear C both strain and stress profiles are linear D shear deformations are neglected
Solid Mechanics   Bending and Shear Stresses
 Question 9
Two steel columns P (length L and yield strength $f_{y}$= 250 MPa) and Q (length 2L and yield strength $f_{y}$= 500 MPa) have the same cross-sections and end-conditions. The ratio of buckling load of column P to that of column Q is:
 A 0.5 B 1 C 2 D 4
Solid Mechanics   Theory of Columns and Shear Centre
Question 9 Explanation:
\begin{aligned} P &=\frac{\pi^{2} E \mid}{\left(l_{e f f}\right)^{2}} \\ \therefore \quad P_{p} &=\frac{\pi^{2} E \mid}{(L)^{2}} ; P_{Q}=\frac{\pi^{2} E \mid}{(2 L)^{2}} \\ \therefore \quad \frac{P_{P}}{P_{a}} &=4 \end{aligned}
 Question 10
The pin-jointed 2-D truss is loaded with a horizontal force of 15 kN at joint S and another 15 kN vertical force at joint U, as shown. Find the force in member RS (in kN) and report your answer taking tension as positive and compression as negative. __________
 A 1 B 2 C 4 D 0
Structural Analysis   Trusses
Question 10 Explanation:
\begin{aligned} \Sigma M_{T}&=0 \\ \Rightarrow R_{v} \times 8-15 \times 4+15 \times 4 & =0 \\ \therefore \quad R_{V} & =0\\ \text{Take moment about V, } M_{V}&=0 \\ \Rightarrow \quad R_{H} \times 4&=0 \\ \therefore \quad R_{H}&=0 \end{aligned}

We know that if two members meet at a joint which are not collinear and also there is no external forces acting on that joint then both members will carry zero forces.
$\therefore F_{O V}=F_{Q R}=0$
Now, consider joint R ,
$\Sigma F_{y}=0$

\begin{aligned} \Rightarrow F_{R U} \sin 45^{\circ}&=0 \\ \Rightarrow F_{R U}&=0\\ \text{Now},\quad \Sigma F_{x}&=0 \\ \Rightarrow \quad F_{R S}+F_{R U} \cos 45^{\circ}&=0 \\ \therefore F_{R S}&=0 \end{aligned}
So, force is member is "RS" will be zero.
There are 10 questions to complete.