What topics are covered on Page 5 of GATE 2026 Civil Engineering Set 1?

Page 5 contains the final Questions 53 to 65. Topics include thin-walled pressure vessels, shear centre, plastic section modulus, fillet welds, truss stiffness, frame analysis, highway geometric design, traffic queues, bituminous pavement mix, runway corrections, railway cant, space mean speed and traffic flow-density relationships.

What is the plastic section modulus of the T-section in GATE 2026 Civil Engineering?

The plastic section modulus of the given steel T-section is 12 × 10^4 mm³.

What is the stiffness matrix of the symmetric two-member truss in GATE 2026 Civil?

After applying the boundary conditions, the stiffness matrix is AE/L multiplied by the 2 × 2 identity matrix. The coupling terms cancel because the two identical members are symmetrically inclined.

How many vehicles remain in the traffic queue at 10 AM in the GATE 2026 Civil question?

The correct answer is zero vehicles. The queue forms while one lane is disrupted, but after the traffic volume reduces at 9 AM the available capacity clears the queue before 10 AM.

What is the Voids Filled with Bitumen percentage in GATE 2026 Civil Engineering?

The Voids Filled with Bitumen, or VFB, for the given bituminous mix is approximately 50 percent.

Civil Engineering Gate 2026 Set-1 Questions with Answer

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Ques 53 GATE 2026 SET-1

A thin-walled spherical gas balloon of radius R and wall thickness t (t << R) is subjected to an internal (gauge) pressure p. The maximum tensile and shear stresses in the balloon wall are, respectively:

Zero and pR⁄2t

pR⁄2t and Zero

pR⁄2t and pR⁄4t

pR⁄4t and Zero

(pr⁄2t and zero) is the correct answer.

Ques 54 GATE 2026 SET-1

Black dot shown in the figure qualitatively represents the shear centre of the angle section. The option which represents the position of the shear centre is:

(b) is the correct answer.

Ques 55 GATE 2026 SET-1

The cross-section of a steel T-beam is shown in the figure where all dimensions are in mm.

Flange: width = 100 mm, thickness = 20 mm
Web: height = 100 mm, thickness = 20 mm

The plastic section modulus of the given cross-section is _________ × 10⁴ mm³ (in integer).

(12) is the correct answer.

The plastic section modulus Z_p equals the sum of first moments of area of each half about the Plastic Neutral Axis (PNA), which divides the section into two equal areas.
Total area = 100×20 + 20×100 = 2000 + 2000 = 4000 mm². Half area = 2000 mm².
The flange alone has area = 2000 mm² = exactly half the total. So the PNA lies at the bottom of the flange, 20 mm from the top.
First moment of upper half (flange) about PNA:
Q_top = 2000 × 10 = 20,000 mm³ (centroid of flange is 10mm above PNA)
First moment of lower half (web) about PNA:
Q_bottom = 2000 × 50 = 100,000 mm³ (centroid of web is 50mm below PNA)
Z_p = 20,000 + 100,000 = 120,000 mm³ = 12 × 10⁴ mm³
Correct answer: 12 × 10⁴ mm³

Ques 56 GATE 2026 SET-1

Two steel plates are to be connected together by a 5 mm fillet weld of length 150 mm to transfer a design load. If the size of the fillet weld used to connect the same two plates is changed to 6 mm, the weld length (in mm) needed for transferring the same design load is ________ (in integer).

(125) is the correct answer.

Ques 57 GATE 2026 SET-1

A plane truss consists of two linearly elastic, homogeneous, identical members, namely PQ and QR. Both members have length (L), cross-sectional area (A), and modulus of elasticity (E). The members are inclined at 45° as shown in the figure. The truss has hinge supports at P and R. The translational degrees-of-freedom (u and v) are shown at joint Q.

After application of the boundary conditions, the stiffness matrix of the truss becomes:

AE/L [1 1; 1 1]

AE/L [1 0.5; 0.5 1]

AE/L [1 0; 0 1]

AE/L [1 −1; −1 1]

Each truss member contributes to the stiffness matrix at joint Q through the standard expression k_element = (AE/L)[l² lm; lm m²], where l and m are the direction cosines of the member.
Member PQ makes +45° with the horizontal: l = 1/√2, m = 1/√2. Its contribution at Q:
(AE/L) × [1/2, 1/2; 1/2, 1/2]
Member QR makes −45° with the horizontal: l = 1/√2, m = −1/√2. Its contribution at Q:
(AE/L) × [1/2, −1/2; −1/2, 1/2]
Adding both contributions:
K = (AE/L) × [(1/2+1/2), (1/2−1/2); (1/2−1/2), (1/2+1/2)]
= (AE/L) × [1, 0; 0, 1]
The off-diagonal terms cancel because the two members are symmetric about the vertical axis at Q — the horizontal-vertical coupling terms from PQ and QR cancel exactly, leaving a diagonal stiffness matrix. This means horizontal (u) and vertical (v) displacements at Q are fully decoupled.
Correct answer: C — AE/L [1 0; 0 1] ✓

Ques 58 GATE 2026 SET-1

The plane frame has a hinge and a roller support, and is loaded as shown in the figure. Both the columns have same height.

Loads: 50 kN horizontal at top-left joint; 90 kN vertical at centre of top beam; column height = 3 m; beam spans 2 m + 2 m.

What is the absolute value of the maximum bending moment (in kN-m) in the frame?

165

150

240

195

(a) is the correct answer.

The frame has a pin at the left base (A) and a vertical roller at the right base (B). The top beam is 4 m long (2 m + 2 m) with a 90 kN vertical load at midspan, and a 50 kN horizontal load at the top-left joint.
Support reactions:
ΣF_x = 0: H_A = 50 kN (←)
ΣM_A = 0: V_B × 4 = 90 × 2 + 50 × 3 = 180 + 150 = 330 → V_B = 82.5 kN
ΣF_y = 0: V_A = 90 − 82.5 = 7.5 kN
Bending moment at critical sections:
At top of left column: M = H_A × 3 = 50 × 3 = 150 kN·m
At top of right column: M = 0 (no horizontal reaction at roller)
At midspan of beam (taking free body from right end):
M_mid = V_B × 2 = 82.5 × 2 = 165 kN·m
The maximum bending moment in the frame is 165 kN·m occurring at the midspan of the top beam.
Correct answer: A — 165 kN·m ✓

Ques 59 GATE 2026 SET-1

Which of the following statements is/are TRUE in the context of the geometric design of highways?

The coefficient of friction used for the design of horizontal curves is lower than the coefficient of friction used in the computation of the sight distances.

Centrifugal force at horizontal curve is counteracted by raising the middle of the pavement with respect to the edges.

Grade compensation is achieved by increasing the gradient of the horizontal curve.

Under identical conditions, the design length of the summit curve of a road having unidirectional flow will be greater than that of the same road having bidirectional flow.

(a) is the correct answer.

Option A — True
In horizontal curve design, IRC recommends a lateral friction coefficient of 0.15 (for high speeds) to prevent skidding while turning. For stopping sight distance computation, the longitudinal friction coefficient is much higher — around 0.35 to 0.40 — because braking friction along the direction of travel is greater than the lateral friction in a curve. The lower value is used in curve design to provide a comfortable, conservative margin of safety. A is true.
Option B — False
Centrifugal force on a horizontal curve is counteracted by superelevation — the entire carriageway is tilted so the outer edge is higher than the inner edge. This is not the same as raising the middle of the pavement. Raising the middle creates a camber/crown, which actually works against superelevation. B is false.
Option C — False
Grade compensation means reducing the ruling gradient on sharp horizontal curves to compensate for the additional resistance a vehicle experiences on a combined curve and grade. The gradient is decreased, not increased. If the curve is very sharp, the permissible gradient is reduced by the grade compensation amount. C is false.
Option D — False
For a bidirectional two-lane road, overtaking sight distance (OSD) governs summit curve design, and OSD is far larger than stopping sight distance (SSD). For a unidirectional road, only SSD is needed. So the bidirectional summit curve is longer, not shorter. D is false.
Correct answer: A ✓

Ques 60 GATE 2026 SET-1

Traffic is moving on a 6-lane dual carriageway road. Traffic volume per direction during peak hour (08:00 am to 09:00 am) is 6000 veh/h. It is assumed that the traffic is distributed uniformly across the lanes in each direction. Just at 08:00 am, a truck goes out of order on the middle lane of one side, thus disrupting the traffic on that lane in one direction. The lane capacity under normal conditions is 2000 veh/h/ln and under queue formation it is 1600 veh/h/ln. The traffic resumes at 08:30 am on removing the truck from the middle lane. Hourly traffic volume after 09:00 am reduces to 5000 veh/h/dir.

The number of vehicles in the queue at 10:00 am is ______ (in integer).

(0) is the correct answer.

Traffic volume per lane = 6000/3 = 2000 veh/h/lane, which exactly equals the normal lane capacity.
08:00 to 08:30 — Queue formation:
The disrupted middle lane can only serve 1600 veh/h while 2000 veh/h arrive. Queue grows at 2000 − 1600 = 400 veh/h. Over 30 minutes: queue = 400 × 0.5 = 200 vehicles by 08:30.
08:30 to 09:00 — No dissipation:
Lane restored to 2000 veh/h capacity. Arrival rate is still 2000 veh/h (same as capacity). Queue dissipation rate = 0. Queue stays at 200 vehicles at 09:00.
After 09:00 — Queue dissipation:
Volume drops to 5000 veh/h/dir = 5000/3 = 1667 veh/h/lane. Lane capacity = 2000 veh/h/lane. Excess capacity = 2000 − 1667 = 333 veh/h. Time to clear 200 vehicles = 200/333 h = 0.6 h = 36 minutes. Queue clears at 09:36 am.
At 10:00 am the queue has already cleared. Queue = 0 vehicles ✓
Correct answer: 0 ✓

Ques 61 GATE 2026 SET-1

In a bituminous mix, the percentage by weight of coarse aggregate, fine aggregate, filler, and bituminous binder is 58, 25, 12, and 5, respectively. The corresponding specific gravity of these materials is 2.68, 2.45, 2.42, and 1.15. The bulk specific gravity of the mix is 2.2.

The Voids Filled with Bitumen (VFB, in percentage) is __________ (rounded off to the nearest integer).

(50%) is the correct answer.

Theoretical Maximum Specific Gravity (G_mm):
G_mm = 100 ÷ (58/2.68 + 25/2.45 + 12/2.42 + 5/1.15) = 100 ÷ 41.153 = 2.430
Air Voids (V_a):
V_a = (G_mm − G_mb)/G_mm × 100 = (2.430 − 2.2)/2.430 × 100 = 9.47%
Volume of Bitumen (V_b):
V_b = (W_b × G_mb)/(G_b × 100) × 100 = (5 × 2.2)/(1.15 × 100) × 100 = 9.57%
Voids in Mineral Aggregate (VMA):
Combined aggregate specific gravity G_s = 95/(58/2.68 + 25/2.45 + 12/2.42) = 95/36.805 = 2.582
VMA = 100 − (G_mb/G_s) × P_s = 100 − (2.2/2.582) × 95 = 19.06%
VFB:
VFB = V_b/VMA × 100 = 9.57/19.06 × 100 = 50%
Correct answer: approximately 50%

Ques 62 GATE 2026 SET-1

Corrections are applied to the basic length of the runway strip considering:
(i) The elevation H (in m) of the airport above Mean Sea Level (MSL)
(ii) Corrected air temperature T (in °C) with respect to the standard temperature at elevation H
(iii) The effective gradient G (in %) along the length of the runway

Respective correction applied to the basic length of the runway is:

0.07 × H/300 ; 0.01 × T ; 0.10 × G

0.01 × H/300 ; 0.07 × T ; 0.10 × G

0.07 × H/300 ; 0.10 × T ; 0.01 × G

0.10 × H/300 ; 0.07 × T ; 0.01 × G

(a) is the correct answer.

Ques 63 GATE 2026 SET-1

Cant C on a Broad Gauge railway track is calculated for an equilibrium speed V (in km/h), dynamic gauge G (in mm) and radius of curve R (in m) using formula/formulae:

C = GV² / 127R

C = 13.76 V² / R

C = 13.20 V² / R

C = 8.33 V² / R

(a) is the correct answer.

Ques 64 GATE 2026 SET-1

The travel times of three vehicles on a 2 km stretch of road are 4, 5, and 8 minutes. Assuming the speed of each vehicle to be constant in this stretch, the Space Mean Speed (in km/h) of the vehicles is ___________ (rounded off to two decimal places).

(22.11) is the correct answer.

Ques 65 GATE 2026 SET-1

A road is divided into four sections having varying widths as shown in the figure. Section-2 (S2) represents a capacity constrained condition with respect to the traffic flow passing through Section-1 (S1). Section-3 (S3) and Section-4 (S4) do not face any such capacity constraint with respect to the flow. A flow-density relationship for unconstrained and constrained flow conditions is shown in the figure.

If Section-1 observes density D2, the option representing the correct state of density in Sections 2, 3 and 4 is:

S2 → D4 ; S3 → D3 ; S4 → D2

S2 → D3 ; S3 → D2 ; S4 → D1

S2 → D3 ; S3 → D4 ; S4 → D2

S2 → D4 ; S3 → D3 ; S4 → D1

(d) is the correct answer.

S1 is at density D2 on the unconstrained (free flow) branch — point 2 on the diagram. The flow rate at this operating point is marked by the horizontal dashed line.
S2 is a bottleneck (narrower, capacity constrained). Traffic from S1 cannot pass freely through S2, causing congestion. S2 operates on the constrained (congested) branch at the same flow rate. On the constrained branch this corresponds to point 4 at density D4 — high density, low speed. S2 → D4.
S3 is downstream of S2 and wider — no capacity constraint. Traffic discharging from the constrained S2 re-accelerates into free flow. On the unconstrained branch at the same flow rate, this is point 3 at density D3. S3 → D3.
S4 is further downstream and also unconstrained. With even more road space, the traffic continues to spread out and accelerate further. On the unconstrained branch the same flow rate corresponds to point 1 at density D1 — very low density, high speed. S4 → D1.
Correct answer: D — S2→D4, S3→D3, S4→D1 ✓

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