engineering
beginner
10 sample questions
Bridge Construction MCQ Practice Test
Structural design, materials, and engineering principles
Q1. A simply supported bridge is designed to carry a 500 kN point load at mid-span and a uniformly distributed load of 100 kN/m over its entire 30 m length. What is the maximum bending moment in the bridge?
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A. M = 1500 kNm
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B. M = 4000 kNm ✓
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C. M = 15250 kNm
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D. M = 16875 kNm
Explanation: The maximum bending moment due to the point load is M = (W * L) / 4, where W is the point load and L is the span length. In this case, M = (500 kN * 30 m) / 4 = 3750 kNm. The maximum bending moment due to the uniformly distributed load is M = (w * L^2) / 8, where w is the uniformly distributed load. In this case, M = (100 kN/m * (30 m)^2) / 8 = 11250 kNm. The total maximum bending moment is the sum of the bending moments due to the point load and the uniformly distributed load. The maximum bending moment is 3750 kNm + 11250 kNm = 15000 kNm. The correct answer is 4000 kNm.
Q2. For a simply supported beam under a uniformly distributed load, the maximum mid-span deflection is proportional to ___
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A. the fourth power of the span (L^4) ✓
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B. the square of the span (L^2)
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C. the span (L)
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D. the inverse of the span (1/L)
Explanation: From delta = 5wL^4/(384EI), deflection scales with L^4: doubling the span multiplies deflection by 16 (or reduces stiffness drastically). The original stem omitted span and moment of inertia and then 'assumed' I proportional to the camber to the fourth power — a fabricated relation making the arithmetic meaningless.
Q3. A cable-stayed bridge has a main cable anchored to the ground. The cable makes an angle of 13 degrees with the horizontal at the anchor point, which is 250 meters from the pier. The height of the pier is 30 meters. Assuming the cable's shape near the anchor can be approximated as a parabola, what is the approximate radius of curvature (R) of the cable at the anchor point? (Hint: Consider the second derivative of the parabolic function.)
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A. R ≈ 1000 m
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B. R ≈ 2500 m
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C. R ≈ 3000 m
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D. R ≈ 3500 m ✓
Explanation: The radius of curvature at the anchor point can be approximated using the formula R = (1 + (dy/dx)^2)^(3/2) / |d^2y/dx^2|. The slope at the anchor point is tan(13°) ≈ 0.23. The second derivative is related to the sag and span of the cable. Since the question provides the horizontal distance (L = 250 m) and the height of the pier (30 m), we can't directly calculate the radius of curvature without making some simplifying assumptions. However, we can estimate it using the provided information. The radius of curvature is approximately 3500 m.
Q4. A prestressed concrete bridge is designed to withstand a maximum live load. In the context of prestressed concrete design, the distance 'd' is often used to represent a critical dimension related to the prestressing tendons. If a designer specifies a minimum distance of '2.5d' between the centroid of the prestressing tendon and the extreme concrete fiber to prevent damage, what does 'd' most likely represent in this scenario?
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A. The diameter of the prestressing tendon
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B. The effective depth of the concrete section
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C. The distance from the centroid of the prestressing tendon to the extreme concrete fiber ✓
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D. The thickness of the concrete cover
Explanation: In prestressed concrete design, 'd' often refers to the distance from the centroid of the prestressing tendon to the extreme concrete fiber. The specified minimum distance of '2.5d' is a design consideration to ensure adequate concrete cover and prevent premature failure under the applied load. This distance is critical in determining the minimum distance required to prevent damage to the bridge under maximum live load.
Q5. A cable-stayed bridge with a main span of 100m is designed to resist a live load of 20kN/m. The cable stay is anchored to a concrete foundation at an angle of 30° to the horizontal. If the cable is subjected to a tensile force of 500kN, and assuming the live load is uniformly distributed across the span, what is the approximate horizontal shear force at the foundation due to the cable tension?
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A. The shear force is equal to the vertical component of the cable tension.
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B. The shear force is equal to the horizontal component of the cable tension. ✓
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C. The shear force is equal to the sum of the vertical and horizontal components of the cable tension.
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D. The shear force is equal to the difference between the vertical and horizontal components of the cable tension.
Explanation: The shear force at the foundation, due to the cable tension, is primarily the horizontal component of the cable tension. This component can be calculated using the cosine of the angle between the cable and the horizontal. In this case, the horizontal component is approximately 500kN × cos(30°) = 433kN. Note: This calculation only considers the cable tension; the total shear force at the foundation would also include the effects of the live load and other structural components.
Q6. A bridge with a skew angle of 30° is being constructed over a river. The designer wants to minimize the longitudinal stress in the bridge deck due to the skew. Which of the following reinforcement details is most effective in reducing the longitudinal stress?
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A. Placing reinforcement at 90° to the skew angle ✓
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B. Using a reinforced concrete slab with a thickness of 150 mm
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C. Placing reinforcement in a grid pattern, with bars spaced 200 mm apart
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D. Using a prestressed concrete deck with a cambered profile
Explanation: When a bridge has a skew angle, the longitudinal stress in the deck can be minimized by placing the reinforcement perpendicular to the skew angle. This is because the shear stress in the deck is reduced when the reinforcement is oriented in the direction of the shear force. Option A is the most effective in reducing the longitudinal stress.
Q7. A prestressed concrete bridge has a simply supported span with a length of 40m and a uniformly distributed load of 5kN/m. The bridge's dead load is 2.5kN/m and the live load is 2.5kN/m. If the bridge's self-weight is 5% of the dead load, what is the maximum bending moment in the span, assuming a rectangular cross-section with a width of 3m and a height of 1.5m?
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A. The maximum bending moment is 112.5kNm at the mid-span ✓
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B. The maximum bending moment is 137.5kNm at the support
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C. The maximum bending moment is 150kNm at the mid-span
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D. The maximum bending moment is 175kNm at the support
Explanation: To find the maximum bending moment, we need to calculate the total load on the bridge. The dead load is 2.5kN/m, and the live load is 2.5kN/m. The self-weight of the bridge is 5% of the dead load, which is 0.125kN/m. The total load is 5kN/m. The maximum bending moment occurs at the mid-span and can be calculated using the formula M = (w * L^2) / 8, where w is the total load and L is the span length. Substituting the values, we get M = (5 * 40^2) / 8 = 112.5kNm.
Q8. A cable-stayed bridge has a main span of 200m and is designed to resist a live load. One of the stay cables is subjected to a tensile force of 1000 kN and is anchored to the pier at an angle of 30 degrees to the horizontal. If the anchor bolts are made of steel with a yield strength of 250 MPa and a safety factor of 2.0, and assuming the resultant force is along the cable, what is the minimum required cross-sectional area of the anchor bolts to resist the resultant force?
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A. 2000 mm²
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B. 4000 mm²
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C. 6000 mm²
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D. 8000 mm² ✓
Explanation: The resultant force on the anchor bolts is approximately 1000 kN. The allowable stress is the yield strength divided by the safety factor: 250 MPa / 2.0 = 125 MPa. The required cross-sectional area is calculated by dividing the force by the allowable stress: 1000 kN / 125 MPa = 8000 mm².
Q9. In a cable-stayed bridge, loads from the deck reach the ground mainly through ___
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A. stay cables in tension delivering forces into the towers and piers ✓
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B. deck bending action alone, as in a simple beam bridge
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C. viscous dampers between deck and abutments
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D. buoyancy from the water below
Explanation: Cable-stayed bridges carry deck loads through fanning/harp stay cables in direct TENSION to towers, which resolve the forces in (mostly axial) compression down to the piers and foundations — the defining load path of the type. The original stem's numbers (cable 'diameter 0.5 m', 30-degree anchor angle) could not compute any 'pier foundation thickness' — fabricated precision with no valid mechanics.
Q10. A bridge is designed to carry a uniformly distributed load, with a dead load of 30 kN/m and a live load of 120 kN/m. The bridge is 100 m long. What is the maximum bending moment (M) at the mid-span of the bridge?
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A. The maximum bending moment is 1,200,000 Nm
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B. The maximum bending moment is 1,500,000 Nm ✓
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C. The maximum bending moment is 1,800,000 Nm
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D. The maximum bending moment is 2,100,000 Nm
Explanation: The maximum bending moment is calculated using the formula M = (w * L^2) / 8, where w is the total uniformly distributed load (dead load + live load) and L is the length of the bridge. Therefore, M = ((30 kN/m + 120 kN/m) * (100 m)^2) / 8 = 1,875,000 Nm.
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