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Pancyprian Mechanics & Construction practice questions

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Question 1Engineering Mechanics (statics and dynamics)

A simply supported beam of span 6 m carries a single point load of 12 kN at its midspan. Ignoring the self-weight of the beam, what is the magnitude of the vertical reaction at each support?

  1. 24 kN
  2. 6 kN Correct answer.
  3. 12 kN
  4. 3 kN

Explanation

For a symmetrical simply supported beam with a central point load, the two reactions share the load equally. Static equilibrium requires the sum of the reactions to equal the applied load, so each support carries half of 12 kN, which is 6 kN.

Question 2Engineering Mechanics (statics and dynamics)

Which quantity is a vector in rigid-body statics?

  1. Temperature
  2. Moment of a force about a point Correct answer.
  3. Density
  4. Mass

Explanation

A moment has both magnitude and a rotational sense defined about a specified axis, so it is treated as a vector quantity. Mass, temperature and density are scalars described fully by magnitude alone.

Question 3Engineering Mechanics (statics and dynamics)

A force of 50 N acts perpendicular to a spanner at a perpendicular distance of 0.30 m from the centre of a bolt. What moment does it produce about the bolt centre?

  1. 1.5 N·m
  2. 150 N·m
  3. 0.15 N·m
  4. 15 N·m Correct answer.

Explanation

The moment of a force about a point equals force multiplied by the perpendicular distance from the point to the line of action. Here 50 N × 0.30 m gives 15 N·m.

Question 4Engineering Mechanics (statics and dynamics)

In a concurrent force system in equilibrium, which condition must be satisfied?

  1. The algebraic sum of the forces in each coordinate direction equals zero Correct answer.
  2. The sum of the forces equals a constant nonzero value
  3. The sum of the moments about any point equals the total applied load
  4. The resultant force equals the largest single applied force

Explanation

For a concurrent system in equilibrium, the vector sum of all forces is zero, which resolves into a zero algebraic sum in each coordinate direction. Because all lines of action pass through one point, moment equilibrium about that point is automatically satisfied.

Question 5Engineering Mechanics (statics and dynamics)

A body of mass 20 kg is raised vertically at a constant velocity. Taking g as 9.81 m/s², what is the tension in the lifting cable?

  1. 392.4 N
  2. 20 N
  3. 0 N
  4. 196.2 N Correct answer.

Explanation

At constant velocity the acceleration is zero, so the net vertical force is zero and the cable tension equals the weight. Multiplying 20 kg by 9.81 m/s² gives 196.2 N.

Question 6Engineering Mechanics (statics and dynamics)

A block slides down a plane inclined at 30° to the horizontal with no friction. Taking g as 9.81 m/s², what is its acceleration along the plane?

  1. 9.81 m/s²
  2. 4.91 m/s² Correct answer.
  3. 0 m/s²
  4. 8.50 m/s²

Explanation

On a frictionless incline the only unbalanced force along the plane is the component of weight, mg sin θ, so the acceleration is g sin θ. With θ = 30°, sin 30° = 0.5, giving 9.81 × 0.5 = 4.91 m/s².

Question 7Engineering Mechanics (statics and dynamics)

What is the SI unit of the coefficient of static friction between two dry surfaces?

  1. Newton metre
  2. Newton per metre
  3. Newton
  4. It is dimensionless Correct answer.

Explanation

The coefficient of friction is the ratio of the limiting friction force to the normal reaction, both measured in newtons. The units cancel, so the coefficient is dimensionless.

Question 8Engineering Mechanics (statics and dynamics)

A particle moves in a straight line with a constant acceleration of 2 m/s². If its initial velocity is 5 m/s, what is its velocity after 4 s?

  1. 13 m/s Correct answer.
  2. 21 m/s
  3. 3 m/s
  4. 8 m/s

Explanation

For constant acceleration, final velocity equals initial velocity plus acceleration multiplied by time. Thus 5 + (2 × 4) = 13 m/s.

Question 9Engineering Mechanics (statics and dynamics)

A simply supported beam of span 8 m carries a uniformly distributed load of 5 kN/m over its full length. What is the vertical reaction at each support?

  1. 5 kN
  2. 40 kN
  3. 20 kN Correct answer.
  4. 10 kN

Explanation

The total load is 5 kN/m × 8 m = 40 kN, acting at midspan. By symmetry each of the two supports carries half, giving 20 kN.

Question 10Engineering Mechanics (statics and dynamics)

A force of 100 N acts at 60° to the horizontal. What is its horizontal component?

  1. 0 N
  2. 50 N Correct answer.
  3. 100 N
  4. 86.6 N

Explanation

The horizontal component equals F cos θ, so 100 × cos 60° = 100 × 0.5 = 50 N. The vertical component would be 100 × sin 60° = 86.6 N.

Question 11Engineering Mechanics (statics and dynamics)

A uniform ladder of weight 300 N rests against a smooth vertical wall with its foot on rough horizontal ground. In the free-body diagram, which statement about the wall reaction is correct?

  1. It has both frictional and normal components at the wall
  2. It acts vertically upward to balance the ladder weight
  3. It acts along the ladder axis toward the ground
  4. It acts perpendicular to the wall because the wall is smooth Correct answer.

Explanation

A smooth surface can exert only a normal reaction, so the wall pushes horizontally on the ladder perpendicular to its face. The vertical support and friction are provided at the rough ground contact.

Question 12Engineering Mechanics (statics and dynamics)

A block of weight 200 N rests on a horizontal floor. The coefficient of static friction is 0.30. What horizontal force is just sufficient to start the block moving?

  1. 30 N
  2. 200 N
  3. 60 N Correct answer.
  4. 600 N

Explanation

The limiting static friction equals the coefficient multiplied by the normal reaction, which here equals the weight. Thus 0.30 × 200 = 60 N.

Question 13Engineering Mechanics (statics and dynamics)

A force of 80 N is applied at the end of a lever arm 0.5 m long, acting at 30° to the lever. What moment does it produce about the pivot?

  1. 69.3 N·m
  2. 34.6 N·m
  3. 40 N·m
  4. 20 N·m Correct answer.

Explanation

Only the component perpendicular to the lever creates moment, so M = F L sin θ = 80 × 0.5 × sin 30° = 20 N·m.

Question 14Engineering Mechanics (statics and dynamics)

A car travelling at 20 m/s brakes uniformly to rest over a distance of 50 m. What is the magnitude of its deceleration?

  1. 0.4 m/s²
  2. 8 m/s²
  3. 2 m/s²
  4. 4 m/s² Correct answer.

Explanation

Using v² = u² + 2as with v = 0, we get a = −u²/(2s) = −400/100 = −4 m/s², so the deceleration is 4 m/s².

Question 15Engineering Mechanics (statics and dynamics)

Two forces of 6 N and 8 N act at a point at right angles to each other. What is the magnitude of their resultant?

  1. 2 N
  2. 14 N
  3. 10 N Correct answer.
  4. 48 N

Explanation

Perpendicular forces combine by the Pythagorean relation, so the resultant is the square root of 6² + 8², which is the square root of 100, or 10 N.

Question 16Engineering Mechanics (statics and dynamics)

A body of mass 10 kg hangs from a spring balance inside a lift accelerating upward at 2 m/s². Taking g as 9.81 m/s², what reading does the balance show?

  1. 20 N
  2. 98.1 N
  3. 118.1 N Correct answer.
  4. 78.1 N

Explanation

Newton's second law gives T − mg = ma, so the tension is m(g + a) = 10 × (9.81 + 2) = 118.1 N. The balance reads the tension, which exceeds the weight during upward acceleration.

Question 17Engineering Mechanics (statics and dynamics)

A simply supported beam of span 10 m carries a point load of 30 kN at 4 m from the left support. What is the reaction at the right support?

  1. 7.5 kN
  2. 18 kN
  3. 12 kN Correct answer.
  4. 30 kN

Explanation

Taking moments about the left support gives the right reaction as (30 × 4)/10 = 12 kN. The left reaction then takes the remaining 18 kN.

Question 18Engineering Mechanics (statics and dynamics)

A particle is projected horizontally from a height with a horizontal velocity. Neglecting air resistance, which statement about its motion is correct?

  1. The horizontal velocity decreases while the vertical velocity stays constant
  2. Both horizontal and vertical velocities increase uniformly
  3. Both horizontal and vertical velocities remain constant
  4. The horizontal velocity stays constant while the vertical velocity increases Correct answer.

Explanation

With no air resistance, gravity acts only vertically, so the horizontal velocity is constant while the vertical velocity grows at g. The two components can be treated independently.

Question 19Engineering Mechanics (statics and dynamics)

A uniform beam of weight 400 N is supported at both ends and carries a point load of 200 N at midspan. What is the reaction at each support?

  1. 100 N
  2. 200 N
  3. 300 N Correct answer.
  4. 400 N

Explanation

The total vertical load is the beam weight plus the point load, 400 + 200 = 600 N, acting symmetrically. Each of the two supports therefore carries 300 N.

Question 20Engineering Mechanics (statics and dynamics)

A block of weight 100 N rests on a plane inclined at 20° to the horizontal. What is the component of its weight acting down the plane?

  1. 0 N
  2. 34.2 N Correct answer.
  3. 100 N
  4. 94.0 N

Explanation

The component along the plane is W sin θ = 100 × sin 20° = 34.2 N. The component normal to the plane is W cos 20° = 94.0 N.

Question 21Engineering Mechanics (statics and dynamics)

A rigid body is in complete static equilibrium under a general coplanar force system. Which set of conditions is both necessary and sufficient?

  1. Sum of forces in x and y each equal zero, and sum of moments about any one point equals zero Correct answer.
  2. Sum of forces in x equals zero, and sum of moments about one point equals zero
  3. Sum of forces in x and y each equal zero only
  4. Sum of moments about two points equals zero only

Explanation

In two dimensions there are three independent equilibrium equations: two force sums and one moment sum. Together they are necessary and sufficient for a general coplanar system.

Question 22Engineering Mechanics (statics and dynamics)

A uniform ladder of weight W and length L leans against a smooth vertical wall at angle θ to the horizontal, with its foot on rough ground. Taking moments about the foot, which expression gives the horizontal reaction at the wall?

  1. W L cos θ / (2 L sin θ)
  2. W cos θ / 2
  3. W / (2 tan θ) Correct answer.
  4. W L sin θ / (2 L cos θ)

Explanation

Taking moments about the foot, the weight acts at L/2 with lever arm (L/2)cos θ, and the wall reaction acts at height L sin θ. Equating W(L/2)cos θ = R(L sin θ) gives R = W cos θ/(2 sin θ) = W/(2 tan θ).

Question 23Engineering Mechanics (statics and dynamics)

A block slides down a rough plane inclined at 30° with a coefficient of kinetic friction of 0.25. Taking g as 9.81 m/s², what is its acceleration along the plane?

  1. 4.91 m/s²
  2. 0 m/s²
  3. 7.03 m/s²
  4. 2.78 m/s² Correct answer.

Explanation

The net acceleration is g(sin θ − μ cos θ) = 9.81 × (0.5 − 0.25 × 0.866) = 9.81 × (0.5 − 0.2165) ≈ 2.78 m/s². Friction opposes the downward motion along the plane.

Question 24Engineering Mechanics (statics and dynamics)

A force of 50 N acts vertically downward at a point 2 m to the right of a pivot, and a second force of 30 N acts vertically upward at a point 3 m to the left of the pivot. What is the net moment about the pivot?

  1. 100 N·m clockwise
  2. 10 N·m clockwise
  3. 10 N·m anticlockwise Correct answer.
  4. 190 N·m clockwise

Explanation

The 50 N force gives 50 × 2 = 100 N·m clockwise, and the 30 N force gives 30 × 3 = 90 N·m anticlockwise. The net moment is 100 − 90 = 10 N·m in the clockwise sense, so the anticlockwise option is not correct; the net is clockwise.

Question 25Engineering Mechanics (statics and dynamics)

A flywheel rotating at 300 rev/min is brought to rest uniformly in 10 s. What is the magnitude of its angular deceleration in rad/s²?

  1. 31.4 rad/s²
  2. 0.314 rad/s²
  3. 3.14 rad/s² Correct answer.
  4. 0.0314 rad/s²

Explanation

The initial angular velocity is 300 × 2π/60 = 31.4 rad/s. Dividing by the 10 s stopping time gives a deceleration of about 3.14 rad/s².

Question 26Engineering Mechanics (statics and dynamics)

A particle moves along a straight line such that its displacement s (in metres) from a fixed point is given by s = 2t³ − 9t² + 12t + 5, where t is in seconds. What is the acceleration of the particle at t = 2 s?

  1. 24 m/s²
  2. 12 m/s²
  3. 18 m/s²
  4. 6 m/s² Correct answer.

Explanation

Velocity is the first derivative of displacement: v = ds/dt = 6t² − 18t + 12. Acceleration is the derivative of velocity: a = dv/dt = 12t − 18. Substituting t = 2 s gives a = 24 − 18 = 6 m/s².

Question 27Engineering Mechanics (statics and dynamics)

A particle is projected with a speed of 20 m/s at an angle of 30° above the horizontal from level ground. Taking g = 9.81 m/s² and neglecting air resistance, what is the horizontal range of the projectile?

  1. 35.3 m Correct answer.
  2. 20.4 m
  3. 40.8 m
  4. 70.6 m

Explanation

The horizontal range for projection from and returning to the same level is R = (u² sin 2θ)/g. Here u = 20 m/s and θ = 30°, so sin 60° = 0.8660. Thus R = (400 × 0.8660)/9.81 ≈ 35.3 m.

Question 28Engineering Mechanics (statics and dynamics)

A flywheel starts from rest and accelerates uniformly at 4 rad/s². Through what angle does it turn in the first 5 s?

  1. 20 rad
  2. 50 rad Correct answer.
  3. 40 rad
  4. 10 rad

Explanation

For constant angular acceleration with zero initial angular velocity, the angle turned is θ = ½ α t². Substituting α = 4 rad/s² and t = 5 s gives θ = ½ × 4 × 25 = 50 rad.

Question 29Engineering Mechanics (statics and dynamics)

A force of 50 N acts at a point on a body. What is the moment of this force about a point O that is 3 m from the line of action of the force?

  1. 450 N·m
  2. 53 N·m
  3. 16.7 N·m
  4. 150 N·m Correct answer.

Explanation

The moment of a force about a point is the product of the force magnitude and the perpendicular distance from the point to the line of action: M = F × d. Here M = 50 N × 3 m = 150 N·m.

Question 30Engineering Mechanics (statics and dynamics)

A simply supported beam of span 6 m carries a uniformly distributed load of 4 kN/m over its entire length. What is the maximum bending moment in the beam?

  1. 12 kN·m
  2. 18 kN·m Correct answer.
  3. 24 kN·m
  4. 36 kN·m

Explanation

For a simply supported beam with a uniformly distributed load w over span L, the maximum bending moment occurs at midspan and equals wL²/8. Here w = 4 kN/m and L = 6 m, so M_max = 4 × 36 / 8 = 18 kN·m.

Question 31Engineering Mechanics (statics and dynamics)

A body of mass 5 kg is moving in a straight line with a velocity of 8 m/s. What is its kinetic energy?

  1. 40 J
  2. 80 J
  3. 20 J
  4. 160 J Correct answer.

Explanation

Kinetic energy for a translating body is KE = ½ m v². Substituting m = 5 kg and v = 8 m/s gives KE = ½ × 5 × 64 = 160 J.

Question 32Engineering Mechanics (statics and dynamics)

A particle of mass 2 kg is attached to a string of length 1.5 m and whirled in a horizontal circle at a constant speed of 6 m/s. What is the tension in the string?

  1. 24 N
  2. 48 N Correct answer.
  3. 72 N
  4. 36 N

Explanation

For circular motion, the centripetal force is provided by the tension: T = m v² / r. Here T = 2 × 36 / 1.5 = 48 N.

Question 33Engineering Mechanics (statics and dynamics)

A body of weight 500 N is suspended by two ropes inclined at 30° and 60° to the horizontal. Which statement about the tensions in the ropes is correct?

  1. The tension in the rope at 60° is greater than the tension in the rope at 30°. Correct answer.
  2. The tensions in the two ropes are equal.
  3. The tension in each rope is 500 N.
  4. The tension in the rope at 30° is greater than the tension in the rope at 60°.

Explanation

Applying Lami's theorem or resolving forces horizontally and vertically shows that the tension in a rope is proportional to the sine of the angle between the other rope and the vertical. The rope making the larger angle with the horizontal (60°) has the smaller horizontal component, so it must carry the greater tension to balance the vertical load. Thus the tension in the 60° rope is greater.

Question 34Engineering Mechanics (statics and dynamics)

A uniform rod of length 2 m and weight 100 N is supported at one end by a hinge and held horizontal by a vertical force at the other end. What is the magnitude of the vertical force required?

  1. 50 N Correct answer.
  2. 100 N
  3. 200 N
  4. 25 N

Explanation

Taking moments about the hinge, the weight acts at the centre of the rod (1 m from the hinge) and the vertical force acts at the far end (2 m from the hinge). For equilibrium, F × 2 = 100 × 1, so F = 50 N.

Question 35Engineering Mechanics (statics and dynamics)

A body of mass 10 kg is raised vertically through a height of 5 m at a constant speed. Taking g = 9.81 m/s², what is the work done by the lifting force?

  1. 490.5 J Correct answer.
  2. 98.1 J
  3. 49.05 J
  4. 981 J

Explanation

The lifting force equals the weight when the body moves at constant speed: F = m g = 10 × 9.81 = 98.1 N. Work done is force times distance in the direction of the force: W = 98.1 N × 5 m = 490.5 J.

Question 36Engineering Mechanics (statics and dynamics)

A particle moves in a straight line under a constant resultant force. Which quantity remains constant during the motion?

  1. Velocity
  2. Displacement
  3. Acceleration Correct answer.
  4. Kinetic energy

Explanation

According to Newton's second law, a constant resultant force produces a constant acceleration (a = F/m). Velocity, displacement, and kinetic energy all change as the particle moves under constant acceleration.

Question 37Engineering Mechanics (statics and dynamics)

A block of mass 2 kg is pulled along a rough horizontal surface by a horizontal force of 10 N. The coefficient of kinetic friction between the block and the surface is 0.3. Taking g = 9.81 m/s², what is the acceleration of the block?

  1. 2.06 m/s² Correct answer.
  2. 3.56 m/s²
  3. 5.00 m/s²
  4. 6.94 m/s²

Explanation

The kinetic friction force is f = μ N = μ m g = 0.3 × 2 × 9.81 = 5.886 N. The net horizontal force is 10 − 5.886 = 4.114 N. Acceleration a = F_net / m = 4.114 / 2 ≈ 2.06 m/s².

Question 38Engineering Mechanics (statics and dynamics)

A particle is in equilibrium under the action of three coplanar forces. Which condition must be satisfied by the forces?

  1. The forces must be concurrent.
  2. The forces must be parallel.
  3. The forces must be equal in magnitude.
  4. The vector sum of the forces must be zero. Correct answer.

Explanation

For a particle to be in equilibrium, the resultant force must be zero, which means the vector sum of all forces acting on it is zero. The forces need not be equal, concurrent (they already act at a point for a particle), or parallel.

Question 39Engineering Mechanics (statics and dynamics)

A wheel rotating at 20 rad/s is brought to rest uniformly in 5 s. Through what angle does it turn during this time?

  1. 100 rad
  2. 20 rad
  3. 50 rad Correct answer.
  4. 40 rad

Explanation

For uniform angular deceleration, the average angular velocity is (ω₀ + ω)/2 = (20 + 0)/2 = 10 rad/s. The angle turned is θ = average angular velocity × time = 10 × 5 = 50 rad.

Question 40Engineering Mechanics (statics and dynamics)

A body of mass 4 kg moving at 6 m/s collides head-on with a stationary body of mass 2 kg. If the collision is perfectly elastic, what is the velocity of the 4 kg body immediately after the collision?

  1. 4 m/s in the original direction
  2. 2 m/s in the original direction Correct answer.
  3. 6 m/s opposite to the original direction
  4. 2 m/s opposite to the original direction

Explanation

For a perfectly elastic head-on collision, the velocities after collision are given by v₁' = ((m₁ − m₂)/(m₁ + m₂)) u₁ and v₂' = (2m₁/(m₁ + m₂)) u₁ when the second body is initially at rest. Here m₁ = 4 kg, m₂ = 2 kg, u₁ = 6 m/s, so v₁' = ((4−2)/6) × 6 = 2 m/s in the original direction.

Question 41Strength of Materials and Structural Analysis

Ένας αμφιέρειστος δοκός μήκους L φέρει κεντρικό συγκεντρωμένο φορτίο P. Ποια είναι η μέγιστη ροπή κάμψης;

  1. PL/2
  2. PL/8
  3. PL
  4. PL/4 Correct answer.

Explanation

Σε αμφιέρειστο δοκό με κεντρικό φορτίο, οι αντιδράσεις είναι P/2 και η ροπή αυξάνεται γραμμικά από τα άκρα, με μέγιστη στο μέσο του ανοίγματος. Εκεί M = (P/2)(L/2) = PL/4. Αυτή είναι η κλασική τιμή αναφοράς για τον έλεγχο κάμψης.

Question 42Strength of Materials and Structural Analysis

Ποια ιδιότητα της διατομής εκφράζει τη δυσκαμψία σε κάμψη γύρω από άξονα;

  1. Η επιφάνεια της διατομής
  2. Η ροπή αδράνειας ως προς τον άξονα κάμψης Correct answer.
  3. Η στρεπτική σταθερά
  4. Η ακτίνα σύζευξης

Explanation

Η καμπτική δυσκαμψία δίνεται από το γινόμενο EI, όπου I είναι η ροπή αδράνειας ως προς τον ουδέτερο άξονα κάμψης. Η κατανομή του υλικού μακριά από τον άξονα αυξάνει το I και μειώνει τις τάσεις και τα βέλη.

Question 43Strength of Materials and Structural Analysis

Δοκός από χάλυβα με μέτρο ελαστικότητας E = 200 GPa και ροπή αδράνειας I = 5×10⁻⁵ m⁴ έχει καμπτική δυσκαμψία EI ίση με:

  1. 4×10⁶ kN·m²
  2. 1×10⁴ kN·m² Correct answer.
  3. 1×10⁷ kN·m²
  4. 1×10³ kN·m²

Explanation

Υπολογίζεται EI = 200×10⁹ × 5×10⁻⁵ = 1×10⁷ N·m². Μετατρέποντας σε kN·m² διαιρούμε με 1000 και προκύπτει 1×10⁴ kN·m². Το μέγεθος αυτό χρησιμοποιείται άμεσα στους τύπους βελών.

Question 44Strength of Materials and Structural Analysis

Ποια είναι η μέγιστη διατμητική δύναμη σε αμφιέρειστη δοκό μήκους L με ομοιόμορφο φορτίο w;

  1. wL²/8
  2. wL/4
  3. wL
  4. wL/2 Correct answer.

Explanation

Το συνολικό φορτίο είναι wL και μοιράζεται εξίσου στις δύο στηρίξεις, άρα κάθε αντίδραση ισούται με wL/2. Η διατμητική δύναμη είναι μέγιστη στις στηρίξεις και μηδενίζεται στο μέσο.

Question 45Strength of Materials and Structural Analysis

Σε αξονικό εφελκυσμό ράβδου με διατομή A και δύναμη N, η ορθή τάση είναι:

  1. N/(2A)
  2. A/N
  3. N/A Correct answer.
  4. N·A

Explanation

Η ορθή τάση ορίζεται ως η αξονική δύναμη δια της διατομής, δηλαδή σ = N/A. Η παραμόρφωση στη συνέχεια βρίσκεται από τον νόμο του Hooke με το μέτρο ελαστικότητας.

Question 46Strength of Materials and Structural Analysis

Ποια σχέση εκφράζει τον νόμο του Hooke σε μονοαξονική εντατική κατάσταση;

  1. σ = G·ε
  2. ε = E/σ
  3. σ = E·ε Correct answer.
  4. σ = ε/E

Explanation

Στην ελαστική περιοχή η ορθή τάση είναι ανάλογη της αξονικής παραμόρφωσης, με σταθερά αναλογίας το μέτρο ελαστικότητας E. Άρα σ = E·ε. Το G χρησιμοποιείται για διατμητική σχέση τ = G·γ.

Question 47Strength of Materials and Structural Analysis

Ποιος συντελεστής ασφαλείας ορίζεται ως λόγος τάσης διαρροής προς επιτρεπόμενη τάση;

  1. Ο συντελεστής θερμικής διαστολής
  2. Ο συντελεστής Poisson
  3. Ο λόγος slenderness
  4. Ο συντελεστής ασφαλείας έναντι διαρροής Correct answer.

Explanation

Ο συντελεστής ασφαλείας έναντι διαρροής συγκρίνει την αντοχή διαρροής του υλικού με τη μέγιστη επιτρεπόμενη τάση λειτουργίας. Τιμή μεγαλύτερη της μονάδας σημαίνει περιθώριο ασφαλείας. Ο λόγος slenderness αφορά ευστάθεια λυγισμού.

Question 48Strength of Materials and Structural Analysis

Η μέγιστη ροπή αδράνειας ορθογωνικής διατομής πλάτους b και ύψους h ως προς κεντροβαρικό άξονα παράλληλο στη βάση είναι:

  1. hb³/12
  2. bh³/3
  3. bh²/6
  4. bh³/12 Correct answer.

Explanation

Για ορθογώνιο, η ροπή αδράνειας ως προς κεντροβαρικό άξονα παράλληλο στη βάση είναι bh³/12. Ο εκθέτης 3 στο ύψος δείχνει πόσο ευαίσθητη είναι η δυσκαμψία στο ύψος της διατομής.

Question 49Strength of Materials and Structural Analysis

Σε πρόβολο μήκους L με συγκεντρωμένο φορτίο P στο ελεύθερο άκρο, ποιο είναι το βέλος στο άκρο;

  1. PL/(EI)
  2. PL³/(8EI)
  3. PL²/(2EI)
  4. PL³/(3EI) Correct answer.

Explanation

Ο πρόβολος με φορτίο στο άκρο έχει βέλος PL³/(3EI) στο ελεύθερο άκρο. Η εξάρτηση από το L³ φανερώνει πόσο έντονα μεγαλώνει η παραμόρφωση με το μήκος.

Question 50Strength of Materials and Structural Analysis

Δοκός με άνοιγμα 6 m φέρει ομοιόμορφο φορτίο 10 kN/m. Ποια είναι η μέγιστη ροπή κάμψης;

  1. 45 kN·m Correct answer.
  2. 60 kN·m
  3. 90 kN·m
  4. 30 kN·m

Explanation

Για ομοιόμορφο φορτίο σε αμφιέρειστη δοκό, η μέγιστη ροπή είναι wL²/8. Αντικαθιστώντας, 10×36/8 = 45 kN·m. Η τιμή αυτή ελέγχεται στη διατομή μέσου ανοίγματος.