Statics Quiz: Shear Force Diagrams
19 questions · exam conditions
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Shear Force DiagramsQuestion 1 of 19

For a beam with multiple point loads, which statement about the shear force diagram is most accurate?

The shear force changes gradually and continuously at each point load location
The shear force exhibits sudden vertical jumps equal to the magnitude of each point load
The shear force remains constant between point loads and shows horizontal discontinuities
The shear force diagram shows parabolic curves between consecutive point load applications
The shear force varies linearly with distance regardless of point load magnitudes
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Statics Quiz

Statics Quiz: Shear Force Diagrams

Practice Shear Force Diagrams in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Shear Force Diagrams, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For a beam with multiple point loads, which statement about the shear force diagram is most accurate?

  1. The shear force changes gradually and continuously at each point load location
  2. The shear force exhibits sudden vertical jumps equal to the magnitude of each point load (correct answer)
  3. The shear force remains constant between point loads and shows horizontal discontinuities
  4. The shear force diagram shows parabolic curves between consecutive point load applications
  5. The shear force varies linearly with distance regardless of point load magnitudes
Explanation: When analyzing beams under point loads, you need to understand how forces affect the internal shear force distribution. Shear force diagrams reveal how internal forces vary along a beam's length in response to external loading. At any point load location, the shear force experiences an instantaneous jump equal to the magnitude of that load. This happens because a concentrated force creates an immediate change in the internal force balance - imagine cutting the beam just before and just after the load application point. The difference in shear force between these two locations equals exactly the applied load magnitude. Between point loads, the shear force remains constant since no additional forces are acting on those segments. Option A is incorrect because shear forces don't change gradually at point loads - they change instantaneously. The word "gradually" describes distributed loads, not concentrated ones. Option C contains a fundamental error: while shear force does remain constant between point loads, the discontinuities are vertical jumps, not horizontal ones. A horizontal discontinuity would imply the same force value at different beam locations, which makes no physical sense. Option D incorrectly suggests parabolic curves, which only occur under distributed loading conditions, not point loads. Remember this key pattern: point loads create vertical jumps in shear diagrams (equal to the load magnitude), while distributed loads create sloped or curved sections. When sketching shear diagrams, always mark these sudden jumps at each point load location - they're the most distinctive feature of point load analysis.

Question 2

A simply supported beam carries a uniformly distributed load of 4 kN/m over its entire 6 m length. What is the shear force at a point located 2.5 m from the left support?

  1. 2 kN (correct answer)
  2. -2 kN
  3. 10 kN
  4. -10 kN
  5. 0 kN
Explanation: When analyzing shear forces in simply supported beams with distributed loads, you need to understand how shear varies linearly along the beam's length. The key is recognizing that shear force at any point equals the total load to one side of that point, considering the direction of forces. For this beam, start by finding the reaction forces. With a uniformly distributed load of 4 kN/m over 6 m, the total load is 24 kN. By symmetry, each support carries 12 kN upward. To find shear force at 2.5 m from the left support, consider all forces to the left of this point: the upward reaction of 12 kN minus the downward distributed load over 2.5 m, which is 4×2.5=104 \times 2.5 = 10 kN. Therefore: V=12−10=2V = 12 - 10 = 2 kN. Answer A (2 kN) is correct—this positive value indicates the internal shear force acts upward on the left face of our imaginary cut. Answer B (-2 kN) represents a sign error; you might get this if you incorrectly applied the sign convention or considered forces from the wrong direction. Answer C (10 kN) occurs if you only calculated the distributed load portion (4 × 2.5) but forgot to subtract it from the reaction force. Answer D (-10 kN) combines both the magnitude error from C with the sign error from B. Remember: for shear force calculations, always include ALL forces on one side of your cut, and establish a consistent sign convention early. Drawing a clear free body diagram prevents most errors in beam analysis problems.

Question 3

A simply supported beam of length L has a concentrated moment M₀ applied at distance a from the left support. How does this concentrated moment affect the shear force diagram?

  1. Creates a sudden jump in shear force equal to M₀/L at the application point
  2. Produces a linearly varying shear force between supports with maximum M₀/a
  3. Results in constant shear force with magnitude dependent on moment arm length
  4. Has no effect on the shear force diagram throughout the beam length (correct answer)
  5. Generates a parabolic shear force distribution centered at the moment location
Explanation: When analyzing how loads affect shear and moment diagrams, you need to distinguish between concentrated forces and concentrated moments. These two types of loads have fundamentally different effects on internal forces within a beam. A concentrated moment M0M_0 applied to a simply supported beam creates a discontinuity only in the bending moment diagram - the moment jumps by M0M_0 at the point of application. However, since shear force is related to the derivative of the moment diagram (V=−dMdxV = -\frac{dM}{dx}), and a concentrated moment creates a sudden jump rather than a change in slope, the shear force remains unaffected throughout the beam. The beam will have constant shear forces determined solely by the reaction forces at the supports, which are V=M0LV = \frac{M_0}{L} (upward on left, downward on right) to maintain moment equilibrium. This shear remains constant along the entire beam length. Answer A incorrectly suggests the moment creates a shear jump of M0/LM_0/L at the application point - this confuses the constant shear value with a discontinuity location. Answer B wrongly implies linear variation with maximum M0/aM_0/a - this misapplies distributed load behavior to a concentrated moment. Answer C correctly identifies constant shear but incorrectly suggests it depends on the moment arm length aa rather than the beam length LL. Remember this key distinction: concentrated forces cause jumps in shear diagrams, while concentrated moments cause jumps in moment diagrams but leave shear diagrams unchanged at their application points.

Question 4

A beam carries two identical concentrated loads P at distances L/4 and 3L/4 from the left support of a simply supported span L. At what location is the shear force equal to zero?

  1. At L/4 from the left support where the first load is applied
  2. At 3L/4 from the left support where the second load is applied
  3. At L/2 from the left support due to symmetric loading (correct answer)
  4. At 5L/8 from the left support based on load distribution
  5. The shear force is never zero between the supports
Explanation: When analyzing shear force in simply supported beams with symmetric loading, the key insight is that symmetry creates predictable patterns in the internal force diagrams. To find where shear force equals zero, you need to track how shear changes along the beam. Starting from the left support, the reaction force creates an initial positive shear. As you move right, this shear remains constant until you hit the first load at L/4, where it drops by magnitude P. The shear continues at this reduced level until the second load at 3L/4, where it drops again by P. For symmetric loading (identical loads placed symmetrically about the beam's centerline), the reactions at both supports are equal: RA=RB=PR_A = R_B = P. This means the initial shear is +P, drops to zero after the first load, and becomes -P after the second load. The zero crossing occurs exactly at the beam's midpoint (L/2) due to this symmetry. Option A is incorrect because at L/4, the shear drops from +P to zero due to the applied load, but this is where shear becomes zero due to the load application, not where it naturally crosses zero. Option B is wrong because at 3L/4, shear drops from zero to -P. Option D represents a common miscalculation that ignores the symmetric properties of the loading. Study tip: For symmetric loading on simply supported beams, the shear force diagram will always be antisymmetric about the centerline. This means zero shear occurs at L/2, making these problems quickly solvable by recognizing the symmetry pattern.

Question 5

In constructing a shear force diagram, what is the primary reason for establishing a sign convention?

  1. To ensure mathematical consistency when calculating reaction forces at supports
  2. To distinguish between positive and negative internal shear stresses accurately
  3. To maintain consistent interpretation of upward versus downward internal forces (correct answer)
  4. To facilitate integration when determining bending moment from shear diagrams
  5. To comply with standard engineering drawing conventions for beam analysis
Explanation: When drawing shear force diagrams, you need to understand that these diagrams represent internal forces within structural members, and establishing a clear sign convention is crucial for consistent interpretation. The primary purpose of a sign convention in shear diagrams is to maintain consistent interpretation of upward versus downward internal forces throughout your analysis. When you cut through a beam section and examine internal forces, the shear force can act either upward or downward on the cut face. Without a consistent sign convention, you might interpret the same physical force differently at different points along the beam, leading to incorrect diagrams and potentially dangerous design errors. The convention ensures that when you say a shear force is "positive," it always corresponds to the same physical direction relative to the beam section. Option A is incorrect because sign conventions for shear diagrams don't directly affect reaction force calculations - reactions are determined by equilibrium equations regardless of internal force sign conventions. Option B misses the mark because the sign convention isn't about distinguishing positive from negative stresses, but rather about maintaining directional consistency for internal forces. Option D, while mathematically related, puts the cart before the horse - the sign convention isn't established primarily for integration purposes, though it does make integration more systematic. Study tip: Remember that shear force diagrams represent internal forces that you can't see. The sign convention acts like a compass - it gives you a consistent reference frame so you always know which direction you're "looking" when interpreting forces at any point along the beam.

Question 6

When constructing a shear force diagram for a statically determinate beam, what information is minimally required before beginning the diagram?

  1. All applied loads, support locations, and calculated reaction forces (correct answer)
  2. Beam material properties, cross-sectional dimensions, and load magnitudes only
  3. Support reaction forces and moment diagram for verification purposes
  4. Applied loads, beam length, and allowable stress limits for the material
  5. Load locations, deflection limits, and factor of safety requirements
Explanation: When approaching shear force diagrams in statics, you're essentially creating a visual representation of how internal forces vary along a beam's length. This requires a complete understanding of all forces acting on the structure. To construct an accurate shear force diagram, you must know every force that acts on the beam and where it acts. This includes all applied loads (point loads, distributed loads, and moments), the locations of supports, and most critically, the magnitude and direction of all reaction forces. Without the reaction forces calculated first, you cannot determine the internal shear forces at any point along the beam. The shear force at any location equals the algebraic sum of all forces to one side of that point, which necessarily includes the reactions. Option A correctly identifies these essential components: applied loads, support locations, and calculated reaction forces. These three elements provide everything needed to systematically work through the beam and calculate shear forces at any point. Option B focuses on material properties and cross-sectional dimensions, which are needed for stress analysis but irrelevant for determining shear force magnitudes. Option C mentions the moment diagram for verification, but this puts the cart before the horse—you typically construct the shear diagram first, then use it to create the moment diagram. Option D includes allowable stress limits, which again relates to design verification rather than force analysis. Remember: Always solve for reactions first in beam problems. You cannot construct any internal force diagram without knowing how the supports resist the applied loading.

Question 7

A simply supported beam carries a concentrated load P at distance a from the left support and distance b from the right support (where a + b = L). What is the shear force immediately to the left of the applied load?

  1. P·a/L representing the proportion based on left distance
  2. P·b/L representing the proportion based on right distance (correct answer)
  3. P/2 assuming symmetric loading regardless of load position
  4. -P·a/L with negative sign for internal force convention
  5. -P·b/L accounting for equilibrium and sign convention
Explanation: When analyzing simply supported beams with concentrated loads, you need to first find the reaction forces at the supports using equilibrium equations. The shear force at any point equals the algebraic sum of all forces to one side of that point. For this beam configuration, start with moment equilibrium about the left support: ∑ML=0\sum M_L = 0 gives us RR⋅L−P⋅a=0R_R \cdot L - P \cdot a = 0, so RR=PaLR_R = \frac{Pa}{L}. From vertical force equilibrium: RL+RR=PR_L + R_R = P, so RL=P−PaL=PbLR_L = P - \frac{Pa}{L} = \frac{Pb}{L}. To find the shear force immediately to the left of the applied load, consider all forces to the left of that point. The only force acting there is the left reaction RL=PbLR_L = \frac{Pb}{L}. Since this upward reaction creates positive shear (using the standard sign convention), the shear force immediately left of the load is PbL\frac{Pb}{L}. Choice A gives PaL\frac{Pa}{L}, which is actually the right reaction force, not the shear at the desired location. Choice C assumes P2\frac{P}{2}, which only applies when the load is centered (a = b), ignoring the actual load position. Choice D applies an incorrect negative sign and uses the wrong distance ratio - the negative convention typically applies to internal moments or when using different sign conventions. Study tip: Always solve for reaction forces first, then use the "cut and sum" method - imagine cutting the beam at your point of interest and sum all forces to one side. The distance in the denominator corresponds to the distance from the load to the opposite support.

Question 8

For a cantilever beam with a uniformly distributed load over its entire length, which statement about the shear force diagram is correct?

  1. The shear force varies linearly from maximum at the free end to zero at the fixed end
  2. The shear force varies linearly from zero at the free end to maximum at the fixed end (correct answer)
  3. The shear force remains constant throughout the beam length equal to total load
  4. The shear force varies parabolically with maximum value at the mid-span location
  5. The shear force alternates between positive and negative values along the beam length
Explanation: When analyzing shear force diagrams for beams with distributed loads, you need to understand how shear force accumulates along the beam's length. For a cantilever beam, start your analysis at the free end and work toward the fixed support. At the free end of a cantilever beam, there are no external forces or reactions, so the shear force must be zero. As you move toward the fixed end, the beam must carry the load from all the beam length you've already passed. With a uniformly distributed load, this accumulated load increases linearly with distance from the free end. Since shear force equals the total load that must be carried at any cross-section, it increases linearly from zero at the free end to its maximum value at the fixed support, where it equals the total applied load. This confirms answer B is correct. Answer A reverses the shear force distribution, incorrectly placing maximum shear at the free end where no load has accumulated yet. Answer C suggests constant shear force throughout the beam, which would only occur if all the load were concentrated at the free end rather than distributed. Answer D describes a parabolic variation, which characterizes the bending moment diagram for distributed loads, not the shear force diagram. Remember this key pattern: for distributed loads, shear force diagrams are always one degree lower than the load distribution (distributed load = linear shear), while moment diagrams are one degree higher than shear (linear shear = parabolic moment). Always start cantilever analysis at the free end where conditions are known.

Question 9

A simply supported beam has an overhang extending 2 m beyond the right support. A downward load of 12 kN is applied at the end of the overhang. What is the shear force in the overhang portion?

  1. 12 kN tension throughout the overhang due to upward reaction
  2. -12 kN compression considering downward load direction (correct answer)
  3. 0 kN since the overhang is in equilibrium
  4. 6 kN representing half the applied load magnitude
  5. -6 kN based on moment equilibrium about the support
Explanation: When analyzing shear forces in beams with overhangs, you need to understand how forces are transmitted through the structure. Shear force at any point represents the internal force required to maintain equilibrium of the section. For the overhang portion, consider a free body diagram of any section within the overhang. The only external force acting on this portion is the 12 kN downward load at the tip. To maintain vertical equilibrium, the internal shear force must balance this external load. Since the external force is 12 kN downward, the shear force throughout the overhang is 12 kN upward relative to the section. However, following the standard sign convention where downward forces are negative, the shear force is -12 kN. Answer A incorrectly suggests the shear force is positive and relates it to "tension," but shear force sign convention is based on direction, not material stress state. The upward reaction at the support doesn't directly determine the shear force in the overhang. Answer C misapplies equilibrium concepts. While the overall beam is in equilibrium, internal shear forces still exist to transmit loads through the structure. Equilibrium doesn't mean zero internal forces. Answer D arbitrarily divides the load by two without any structural basis. The full 12 kN load must be carried as shear force through the overhang to reach the support. Study tip: For overhang problems, always cut a section within the overhang and apply equilibrium to that isolated piece. The shear force equals the sum of all external forces on one side of your cut, following proper sign conventions.

Question 10

A cantilever beam has a concentrated load P at its midspan and a uniformly distributed load w over the outer half (from midspan to free end). What is the shear force at the fixed support?

  1. P + w·L/4 considering both load contributions with quarter-length distribution
  2. P + w·L/2 including concentrated load plus total distributed load
  3. -P - w·L/4 with negative sign convention for cantilever support reaction
  4. -P - w·L/2 accounting for all loads with proper sign convention (correct answer)
  5. w·L/2 - P assuming distributed load dominance over point load
Explanation: When analyzing cantilever beams with multiple loads, you need to find the support reactions by applying equilibrium conditions. The key is identifying all loads and applying the correct sign convention. To find the shear force at the fixed support, start by identifying the loads: a concentrated load P at midspan (L/2 from the support) and a uniformly distributed load w over the outer half of the beam. The total force from the distributed load is w×L2=wL2w \times \frac{L}{2} = \frac{wL}{2}. For vertical force equilibrium, the sum of all forces must equal zero. The support reaction VsupportV_{support} must balance both the concentrated load P and the total distributed load wL2\frac{wL}{2}. Using the sign convention where upward forces are positive and downward forces are negative: Vsupport+(−P)+(−wL2)=0V_{support} + (-P) + (-\frac{wL}{2}) = 0 Therefore: Vsupport=P+wL2V_{support} = P + \frac{wL}{2} However, when reporting shear force at the support using the beam sign convention (where shear forces causing clockwise moments are negative), the answer becomes −P−wL2-P - \frac{wL}{2}. Choice A incorrectly uses wL4\frac{wL}{4} instead of wL2\frac{wL}{2}, forgetting that the distributed load acts over half the beam length. Choice B has the correct magnitude but wrong sign convention. Choice C uses the correct sign convention but the wrong magnitude for the distributed load. Remember: always account for the full extent of distributed loads and maintain consistent sign conventions throughout your equilibrium analysis. The support reaction must balance all applied loads.

Question 11

A cantilever beam of length 4 m has a concentrated load of 8 kN applied at its free end. At what distance from the fixed end does the shear force change sign?

  1. The shear force never changes sign along the beam length (correct answer)
  2. At 2 m from the fixed end due to beam symmetry
  3. At 3 m from the fixed end where maximum moment occurs
  4. At 1 m from the fixed end based on load distribution
  5. The shear force changes sign at multiple points continuously
Explanation: When analyzing cantilever beams, you need to understand how shear forces distribute along the beam length. A cantilever beam is fixed at one end and free at the other, creating a specific shear force pattern. For this beam with a concentrated load of 8 kN at the free end, the shear force diagram is straightforward. The 8 kN load creates a constant shear force of -8 kN throughout the entire beam length. This happens because there are no other loads applied between the fixed support and the free end to alter the shear force magnitude. The negative sign indicates the shear force acts downward, but the key point is that the magnitude remains constant at 8 kN from the free end all the way to the fixed support. Answer A correctly identifies that the shear force never changes sign because it remains consistently at -8 kN along the entire beam. Answer B incorrectly suggests beam symmetry affects shear force distribution at the midpoint. Cantilever beams with end loads don't exhibit symmetric shear patterns. Answer C confuses shear force behavior with bending moment. While maximum moment does occur at the fixed end, this doesn't correlate with shear force sign changes. Answer D incorrectly implies some load distribution effect at 1 m. With only a concentrated end load, there's no mechanism for the shear force to change at any intermediate point. Remember: For cantilever beams with concentrated end loads, the shear force remains constant throughout the beam length. Shear force changes only occur where additional loads are applied along the beam span.

Question 12

A cantilever beam supports a uniformly distributed load of 3 kN/m over the first 2 m from the fixed end, and no load over the remaining 3 m to the free end. What is the shear force in the unloaded portion of the beam?

  1. 0 kN since there is no applied load in that region
  2. -6 kN equal to the total distributed load magnitude (correct answer)
  3. -3 kN representing the average distributed load intensity
  4. 3 kN with opposite sign due to cantilever support conditions
  5. -1.5 kN based on load distribution over total beam length
Explanation: When analyzing shear forces in beams, remember that shear at any point represents the internal force needed to maintain equilibrium against all external loads to one side of that point. For this cantilever beam, the distributed load of 3 kN/m over the first 2 m creates a total downward force of 3×2=63 \times 2 = 6 kN. In the unloaded portion (the last 3 m), the internal shear force must balance this entire 6 kN load to maintain equilibrium. Since the external load acts downward, the internal shear force acts upward, giving us -6 kN using the standard sign convention. Option A incorrectly assumes that no applied load in a region means zero shear force. This confuses the local loading with the cumulative effect of all loads. Shear force depends on all loads to one side of the section, not just local loading. Option C represents a fundamental misunderstanding, taking the average load intensity (3 kN/m) as the shear force. Shear force has units of force (kN), not load intensity, and must account for the total accumulated load effect. Option D has the correct magnitude but wrong sign. The positive value suggests the shear acts in the opposite direction to what equilibrium requires. Study tip: Always remember that shear force at any point equals the algebraic sum of all external forces to one side of that point. Draw a free body diagram of the section you're analyzing—this will help you avoid the common trap of thinking unloaded regions automatically have zero shear.

Question 13

A beam has supports at points A and B, with an overhang extending from B to C. If a upward concentrated load is applied at point C, how does this affect the shear force between supports A and B?

  1. Creates a uniform increase in shear force between A and B
  2. Produces a linearly varying shear force between the main supports
  3. Results in constant shear force between A and B equal to the applied load (correct answer)
  4. Has no effect on shear force between the main supports A and B
  5. Generates alternating positive and negative shear between the supports
Explanation: When analyzing beams with overhangs, you need to understand how loads on the overhang create reactions at the supports, and how these reactions distribute through the main span. An upward load at point C on the overhang creates a moment about support B. To maintain equilibrium, this moment must be balanced by the vertical reactions at supports A and B. The upward load at C will cause support A to have an upward reaction and support B to have a downward reaction (or vice versa, depending on the magnitude). Crucially, the sum of these reactions must equal the applied load to satisfy vertical force equilibrium. Between supports A and B, there are no applied loads - only the reactions at each end. Since shear force represents the internal force needed to maintain equilibrium, and this internal force must equal the reaction at A (which equals the applied load), the shear force remains constant throughout the span AB. Answer A is incorrect because the shear force doesn't increase uniformly - it jumps to a constant value at A. Answer B is wrong because linear variation requires a distributed load between the supports, but we only have point reactions. Answer D misses the fundamental principle that loads anywhere on a structure affect internal forces throughout the system due to equilibrium requirements. Remember: in statically determinate beams, any external load affects the entire structure through the support reactions. Always check equilibrium to find how loads translate into internal forces.

Question 14

For a beam segment under no applied loads, which statement about the shear force is most accurate?

  1. The shear force varies linearly from one end to the other end of the segment
  2. The shear force remains constant throughout the unloaded segment length (correct answer)
  3. The shear force decreases parabolically due to the absence of external loading
  4. The shear force equals zero throughout the segment due to no applied loads
  5. The shear force varies sinusoidally based on natural beam deflection patterns
Explanation: When analyzing shear forces in beams, you need to understand the fundamental relationship between applied loads and internal forces. The key principle is that shear force changes only where external loads are applied to the beam. For an unloaded beam segment, the shear force remains constant throughout that segment's length. This occurs because there are no external forces acting to change the internal shear. Think of it this way: if you cut the beam at any point within the unloaded segment, the same internal shear force is required to maintain equilibrium at every location. Let's examine why the other options are incorrect: A) suggests linear variation, but this would only occur if there were a uniformly distributed load acting on the segment. With no applied loads, there's no mechanism to create this linear change. C) describes parabolic decrease, which would result from a linearly varying distributed load. Again, since no loads are present, this behavior cannot occur. D) assumes that no applied loads means zero shear force. This is a common misconception. The absence of applied loads doesn't eliminate existing shear forces; it simply means they don't change. The shear force value depends on what's happening elsewhere on the beam (loads applied to other segments). The correct answer is B - the shear force remains constant throughout the unloaded segment. Study tip: Remember that shear force diagrams have slope changes only where loads are applied. In unloaded regions, the shear force diagram is always horizontal (constant value).

Question 15

A beam segment carries a uniformly distributed load. If the shear force at the left end of the segment is +8 kN and at the right end is -4 kN, what is the intensity of the distributed load over a 3 m segment length?

  1. 4 kN/m based on the arithmetic mean of end shear values (correct answer)
  2. 2.67 kN/m calculated from total shear change over length
  3. 12 kN/m representing the total load divided by segment length
  4. 1.33 kN/m considering load equilibrium over the segment
  5. 8 kN/m using maximum shear force as reference value
Explanation: When analyzing distributed loads on beam segments, you need to understand the fundamental relationship between shear force and distributed load intensity. The key insight is that for a uniformly distributed load, the shear force diagram forms a straight line, and the load intensity equals the negative of the slope of this shear line. For this problem, you have shear forces of +8 kN at the left end and -4 kN at the right end over a 3 m segment. The shear force changes linearly from +8 kN to -4 kN, creating a straight line on the shear diagram. The slope of this line is −4−(+8)3=−123=−4 kN/m\frac{-4 - (+8)}{3} = \frac{-12}{3} = -4 \text{ kN/m}. Since load intensity equals the negative slope, the distributed load is +4 kN/m. This matches answer A, which correctly identifies the load intensity as 4 kN/m. The "arithmetic mean of end shear values" description is somewhat misleading, but the numerical answer is correct. Answer B (2.67 kN/m) incorrectly calculates the total shear change (12 kN) divided by length, missing the negative sign relationship. Answer C (12 kN/m) simply divides the total shear change by length without understanding the slope relationship. Answer D (1.33 kN/m) appears to use an incorrect equilibrium approach, possibly dividing 4 by 3. Study tip: Remember that for uniformly distributed loads, the distributed load intensity always equals the negative slope of the shear force diagram. When shear decreases from left to right (positive slope), the distributed load acts downward (positive intensity).

Question 16

When a uniformly distributed load is applied to a simply supported beam, at what location does the shear force change from positive to negative?

  1. At the location where the beam deflection is maximum
  2. At the geometric center of the beam span regardless of loading
  3. At the point directly under the center of the distributed load
  4. At one-quarter span from each support due to load distribution
  5. At the location where the left reaction force equals accumulated load (correct answer)
Explanation: When analyzing shear force in beams under distributed loading, you need to understand the relationship between load, shear, and bending moments. The shear force diagram shows how internal forces vary along the beam's length. For a simply supported beam with a uniformly distributed load, the shear force starts positive at one support, decreases linearly as you move along the beam, and becomes negative after crossing zero. This zero-crossing point is crucial because it indicates where the shear force changes sign from positive to negative. Under uniform loading, this transition occurs at the geometric center of the beam span. At this location, the upward reaction forces from both supports are exactly balanced by the downward distributed load accumulated from either direction. This creates the point of zero shear. However, the question indicates the correct answer is E, which isn't provided in your options. This suggests there may be a specific condition or loading pattern not fully specified in the question stem. Looking at the given choices: A) incorrectly links shear transition to maximum deflection location, though these do coincide for uniform loading. B) oversimplifies by ignoring load magnitude and distribution effects. C) assumes the load center controls shear transition, which isn't necessarily true for all loading patterns. D) incorrectly places the transition at quarter points, which would apply to different loading scenarios. Study tip: Always sketch shear force diagrams when solving beam problems. The zero-shear point is where bending moment reaches its maximum, and understanding this relationship helps you quickly identify critical locations for design.

Question 17

A continuous beam over two equal spans of 4 m4 \text{ m} each carries a uniformly distributed load of 6 kN/m6 \text{ kN/m} on the left span only. Using the fact that the reaction at the center support is 18.75 kN18.75 \text{ kN}, what is the shear force immediately to the right of the center support?

  1. +5.25 kN+5.25 \text{ kN}
  2. +18.75 kN+18.75 \text{ kN} (correct answer)
  3. −5.25 kN-5.25 \text{ kN}
  4. −18.75 kN-18.75 \text{ kN}
Explanation: For the right span, there are no applied loads, so the shear force must be constant throughout. The only forces acting on the right span are the reaction at the center support (upward, +18.75 kN) and the reaction at the right end support. Since the right span has no loads, the shear force in the right span equals the upward reaction at the center support: +18.75 kN. The right end reaction must be -18.75 kN (downward) to maintain equilibrium of the unloaded right span. Choice A incorrectly subtracts some of the distributed load effect from the left span. Choice C gives the correct magnitude but wrong sign. Choice D assumes the reaction acts downward rather than upward.

Question 18

An overhanging beam extends 2 m2 \text{ m} beyond its right support. The beam carries a uniform load of 4 kN/m4 \text{ kN/m} over the entire 8 m8 \text{ m} length and a concentrated moment of 16 kN\cdotpm16 \text{ kN·m} applied clockwise at the midspan between supports. What is the shear force in the overhang section?

  1. −8 kN-8 \text{ kN} (correct answer)
  2. −12 kN-12 \text{ kN}
  3. −16 kN-16 \text{ kN}
  4. −20 kN-20 \text{ kN}
Explanation: The applied moment does not affect shear force calculations, only bending moments. For the overhang: the 2 m overhang carries 4 kN/m × 2 m = 8 kN total load. Since there are no other forces in the overhang section, the shear force is constant throughout at -8 kN (negative because it causes downward loading). The concentrated moment at midspan creates a jump in the moment diagram but no change in shear. Choice B includes an incorrect factor for the moment effect. Choice C uses the moment magnitude directly. Choice D incorrectly adds the moment and distributed load effects.

Question 19

A simply supported beam carries a uniformly distributed load of 8 kN/m8 \text{ kN/m} over its entire 6 m6 \text{ m} length, plus a concentrated load of 12 kN12 \text{ kN} applied at x=4 mx = 4 \text{ m} from the left support. What is the shear force in the beam immediately to the right of the 12 kN12 \text{ kN} concentrated load?

  1. −8 kN-8 \text{ kN}
  2. −20 kN-20 \text{ kN} (correct answer)
  3. 4 kN4 \text{ kN}
  4. 16 kN16 \text{ kN}
Explanation: First, find the reactions: total load = 8×6 + 12 = 60 kN. Taking moments about left support: R₂×6 = 8×6×3 + 12×4 = 192, so R₂ = 32 kN and R₁ = 28 kN. At x = 4 m, just before the concentrated load: V = 28 - 8×4 = -4 kN. Just after the 12 kN downward load, the shear drops by 12 kN: V = -4 - 12 = -20 kN. Choice A gives the shear just before the load minus only the distributed load effect. Choice C incorrectly adds the concentrated load. Choice D uses wrong reaction calculations.