What topics are covered on Page 1 of GATE 2026 Civil Engineering Set 1?

Page 1 contains Questions 1 to 13 of GATE 2026 Civil Engineering Set 1. Topics include concrete creep, reinforcement spacing, prestressed concrete, Newton-Raphson method, polynomial interpolation, matrices and eigenvalues, probability, Cholesky decomposition, integration, differential equations and engineering mechanics involving friction.

What spacing is required when 12 mm bars replace 10 mm bars in the GATE 2026 Civil slab question?

The required centre-to-centre spacing is 216 mm. The reinforcement area per unit width must remain equivalent when the 10 mm diameter bars at 150 mm spacing are replaced by 12 mm diameter bars of the same grade.

What is the coefficient of static friction in the GATE 2026 Civil Engineering mechanics question?

For the two identical blocks connected by a rigid rod with impending sliding at 45 degrees, the coefficient of static friction is approximately 0.41.

Civil Engineering Gate 2026 Set-1 Questions with Answer

Q: What determines long-term deformation of concrete under sustained loading in GATE 2026 Civil Set 1?

The correct answer is creep. Creep represents the gradual increase in strain or deformation of concrete when it remains subjected to sustained stress over a long period.

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Ques 1 GATE 2026 SET-1

Which one of the following is utilized to determine the long-term deformation of concrete under sustained loading?

Creep

Shrinkage

Modulus of rupture

Split tensile strength

(a) is the correct answer.

Ques 2 GATE 2026 SET-1

The required centre-to-centre spacing of 10 mm diameter bars is 150 mm to resist the design moment in a concrete slab. Instead of 10 mm diameter bars, if 12 mm diameter bars of the same grade are used, the required centre-to-centre spacing (in mm) to resist the same design moment becomes ________ (rounded off to the nearest integer).

(216) is the correct answer.

Ques 3 GATE 2026 SET-1

Select ALL CORRECT option(s) which can be considered to check whether the flexural stresses in a prestressed concrete beam are within the allowable stresses at the transfer and the service stages.

Limiting zone for prestressing

Magnel’s graph

Hoyer effect

Load balancing method

(A,B,D) is the correct answer.

The question asks which methods can be used to verify that flexural stresses in a prestressed concrete beam stay within allowable limits at both transfer and service stages.
Option A — Limiting zone — True
The limiting zone defines the region of eccentricity within which the prestressing cable must lie to ensure all four stress conditions (top and bottom fibre stresses at transfer and service) are satisfied simultaneously. It is a direct and standard method for checking stress limits.
Option B — Magnel''s graph — True
Magnel''s diagram plots 1/P versus eccentricity e. Each of the four stress conditions produces a linear boundary, and the feasible region enclosed by all four lines gives all valid (P, e) combinations. It directly checks whether stresses are within allowable limits and is one of the most powerful design tools in prestressed concrete.
Option C — Hoyer effect — False
The Hoyer effect relates to the bond mechanism in pre-tensioned beams — when the prestressing wire is cut, it expands laterally due to Poisson''s effect, creating a wedge action that transfers prestress into the concrete. This is a bond/anchorage concept, not a method for checking flexural stress limits.
Option D — Load balancing method — True
T.Y. Lin''s load balancing method treats the prestress as an equivalent transverse load. The net stress at any section is computed as P/A ± the stress from any unbalanced load. This can be used to verify stresses against allowable limits at both transfer and service stages.
Correct answer: A, B, D

Ques 4 GATE 2026 SET-1

Starting with the first approximation as x = 0.5, the second approximation for the root of the following function by the Newton-Raphson method is ______ (rounded off to two decimal places).

f(x) = e^−x − x

(0.57) is the correct answer.

f(x) = e^−x − x and f′(x) = −e^−x − 1.
The Newton-Raphson update formula is x₁ = x₀ − f(x₀)/f′(x₀).
Starting with x₀ = 0.5:
f(0.5) = e^−0.5 − 0.5 = 0.6065 − 0.5 = 0.1065
f′(0.5) = −e^−0.5 − 1 = −0.6065 − 1 = −1.6065
x₁ = 0.5 − (0.1065)/(−1.6065) = 0.5 + 0.0663 = 0.5663 ≈ 0.57
Correct answer: 0.57 ✓

Ques 5 GATE 2026 SET-1

Values of y for different values of x are tabulated below.

If a second-degree interpolating polynomial P₂(x) is used to represent y, the value of P₂(0) is ___________ (rounded off to the nearest integer).

(2) is the correct answer.

Using the three data points (−2, 28), (1, 4), (2, 16), a second-degree polynomial P₂(x) = ax² + bx + c is fitted.
Setting up the three equations:
4a − 2b + c = 28 ... (i)
a + b + c = 4 ... (ii)
4a + 2b + c = 16 ... (iii)
Subtracting (ii) from (iii): 3a + b = 12 ... (iv)
Subtracting (ii) from (i): 3a − 3b = 24 → a − b = 8 ... (v)
Adding (iv) and (v): 4a = 20 → a = 5
From (v): b = 5 − 8 = −3
From (ii): c = 4 − 5 + 3 = 2
So P₂(x) = 5x² − 3x + 2
P₂(0) = 5(0)² − 3(0) + 2 = 2
Correct answer: 2

Ques 6 GATE 2026 SET-1

Matrix P is given as

The TRUE option is

Trace of P is equal to the sum of the Eigen values of P.

P^TP is an identity matrix.

P is a skew-symmetric matrix.

Absolute magnitude of each Eigen value is 1.

(a) is the correct answer.

Ques 7 GATE 2026 SET-1

Given:

The above system of equations represents a

plane

line

volume

point

(line) is the correct answer.

Ques 8 GATE 2026 SET-1

Bag I contains 4 white and 6 black balls.
Bag II contains 4 white and 3 black balls.
One ball is drawn at random from any one of the two bags and it is found to be a black ball. The probability that the black ball was drawn from Bag I is ______ (rounded off to two decimal places).

(0.56) is the correct answer.

Ques 9 GATE 2026 SET-1

A matrix is given as:

[9 15]
[15 50]

By performing Cholesky decomposition, |l₂₂| of the lower triangular matrix is ______ (in integer).

(5) is the correct answer.

Ques 10 GATE 2026 SET-1

It is given that x and y are integers in the following equation:

(x + y − 7)² + (y + 3x − 13)² = 0

The value of (x³ + y³) is ________ (in integer).

(152) is the correct answer.

Ques 11 GATE 2026 SET-1

Let f(x) be a continuous function defined in [0,2] → ℝ and satisfying the equation

∫₀² f(x)[x − f(x)]dx = 2/3

The value of f(1) is

1/2

The given condition is ∫₀² f(x)[x − f(x)]dx = 2/3, which expands to ∫₀² [xf(x) − f(x)²]dx = 2/3.
The key trick is to consider the integral of the perfect square [f(x) − x/2]²:
∫₀² [f(x) − x/2]² dx = ∫₀² f(x)²dx − ∫₀² xf(x)dx + ∫₀² (x²/4)dx
From the given condition, ∫₀² xf(x)dx − ∫₀² f(x)²dx = 2/3, so ∫₀² f(x)²dx − ∫₀² xf(x)dx = −2/3.
Computing the remaining integral: ∫₀² (x²/4)dx = [x³/12]₀² = 8/12 = 2/3.
Adding everything: ∫₀² [f(x) − x/2]² dx = −2/3 + 2/3 = 0.
Since [f(x) − x/2]² is non-negative everywhere on [0,2] and continuous, and its integral equals zero, the function itself must be identically zero throughout the interval. This forces f(x) = x/2 for all x in [0,2].
At x = 1: f(1) = 1/2.
Correct answer: C) 1/2.

Ques 12 GATE 2026 SET-1

An ordinary differential equation is given below.

x² d²y/dx² = 6y

Considering a and b as arbitrary constants, the general solution of the equation is

y(x) = ax³ + b/x²

y(x) = ax² + b/x³

y(x) = ax² + b ln x

y(x) = ax³ + b ln x

(a) is the correct answer.

I have the full question. This is a Cauchy-Euler (equidimensional) equation. Let me solve it.
Given: x²y″ = 6y, i.e., x²y″ − 6y = 0
This is a Cauchy-Euler equation of the form x²y″ + pxy′ + qy = 0, with p = 0, q = −6.
Assume solution y = x^m:
y′ = mx^m−1, y″ = m(m−1)x^m−2
Substituting:
x² × m(m−1)x^m−2 − 6x^m = 0
m(m−1)x^m − 6x^m = 0
m(m−1) − 6 = 0
m² − m − 6 = 0
(m−3)(m+2) = 0
m = 3 or m = −2
General solution:
y(x) = ax³ + bx⁻² = ax³ + b/x²
Your page''s answer of A is correct ✓ 1. EXPLANATION: The equation x²y″ = 6y, rewritten as x²y″ − 6y = 0, is a Cauchy-Euler (equidimensional) equation. For such equations, assume a trial solution y = x^m.
Substituting y = x^m, y′ = mx^m−1, y″ = m(m−1)x^m−2 into the equation:
x² × m(m−1)x^m−2 − 6x^m = 0
m(m−1) − 6 = 0
m² − m − 6 = 0
(m−3)(m+2) = 0
This gives two roots: m = 3 and m = −2. Since the roots are real and distinct, the general solution is a linear combination of x³ and x⁻²:
y(x) = ax³ + b/x²
Correct answer: A) y(x) = ax³ + b/x² ✓

Ques 13 GATE 2026 SET-1

Two identical blocks A and B are connected by a rigid rod. The blocks rest against vertical and horizontal planes, as shown in the figure. The coefficient of static friction at the vertical and horizontal planes is the same. If the sliding impends when θ = 45°, the value of the coefficient of static friction is ______ (rounded off to two decimal places).

(0.41) is the correct answer.

Block A rests against the vertical wall and block B rests on the horizontal floor. Both are connected by a rigid inclined rod at θ = 45°. Let T be the compressive force in the rod, W the weight of each block, N_A and N_B the normal reactions, and μ the coefficient of friction.
Equilibrium of Block A:
Horizontal: N_A = T cos45° = T/√2
Vertical: μN_A + T sin45° = W → (T/√2)(1 + μ) = W ... (i)
Equilibrium of Block B:
Vertical: N_B = W + T sin45° = W + T/√2
Horizontal: T cos45° = μN_B → T/√2 = μ(W + T/√2)
Rearranging: T/√2 = μW/(1 − μ) ... (ii)
Substituting (ii) into (i):
[μW/(1 − μ)] × (1 + μ) = W
μ(1 + μ) = 1 − μ
μ² + μ = 1 − μ
μ² + 2μ − 1 = 0
Solving the quadratic:
μ = (−2 + √8)/2 = √2 − 1 = 0.4142 ≈ 0.41
Correct answer: 0.41 ✓

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