Statics
8–12 of 110 itemsNCEES draws between 8 and 12 of the 110 from this area on any legal paper. Its outline lists 7 sub-topics; the bank holds 37 verified templates against them, which expand to 1,142 questions.
Sub-topics, from the NCEES outline
- AResultants of force systems
- BEquivalent force systems
- CEquilibrium of rigid bodies
- DFrames and trusses
- ECentroid of area
- FArea moments of inertia
- GStatic friction
5 worked problems
Real questions from the bank, at the depth the exam asks them. Pick an answer first: the reasoning opens when you do, along with why each wrong answer is tempting.
Question 1 of 5
A 270-lb force acts at the end of a 2-ft lever, inclined 65 degrees from the axis of the lever. The moment about the pivot is most nearly:
A 270 lb force applied at the end of a 2 ft lever, inclined 65° from the axis of the lever; only the component perpendicular to the lever produces a moment about the pin. Show the answer and the reasoning
Why B
Given: F = 270 lb at 65 degrees to a 2-ft lever. Relation: Only the component perpendicular to the lever turns it: M = F sin(theta) d. Substitute: M = 270 sin(65)(2) Result: M = 489.4 ft-lb.
Why A is tempting
Used the component along the lever arm. Only the component perpendicular to the arm produces a moment.
Why C is tempting
Used the full force magnitude, ignoring its inclination.
Why D is tempting
Used a tangent in place of a sine.
Question 2 of 5
A simply supported beam spans 26 ft between supports A and B. It carries a 16-kip concentrated load 8 ft from A, plus a uniformly distributed load of 2 kip/ft over the full span. The vertical reaction at A is most nearly:
Simply supported beam of span 26 ft with a 16 kip load 8 ft from A and a uniform load of 2 kip per foot. Show the answer and the reasoning
Why D
Take moments about B: R_A(26) = 16(18) + 2(26)(26/2), so R_A = 37.1 kips. Note the concentrated load is weighted by its distance from the far support, not the near one.
Why A is tempting
Left the distributed load out of the reaction entirely.
Why B is tempting
Included the distributed load but forgot the concentrated load altogether.
Why C is tempting
Split the concentrated load evenly between the supports. That only holds when the load sits at midspan.
Question 3 of 5
A section is cut through a parallel-chord truss where the internal moment is 425 ft-kips. The truss depth is 12 ft. The force in the chord at that section is most nearly:
Parallel-chord truss 12 ft deep with a section cut through the panel and the chord member highlighted; the internal moment at the cut is 425 ft-kips. Show the answer and the reasoning
Why B
Given: Internal moment 425 ft-kips at the cut; truss depth 12 ft. Relation: Moments about the joint on the opposite chord: F h = M. Substitute: F = 425/12 Result: F = 35.42 kips.
Why A is tempting
Halved the depth, as if the chord force were shared.
Why C is tempting
Used half the depth as the lever arm.
Why D is tempting
Left the truss depth out entirely.
Question 4 of 5
An angle is formed from a vertical leg 12 in long and a horizontal leg 4 in long, both 1 in thick, joined at the bottom-left corner. Measured from the bottom, the centroid lies at most nearly:
Angle formed from a vertical leg 12 in long and a horizontal leg 4 in long, both 1 in thick, joined at their outer corner. Show the answer and the reasoning
Why A
Given: Vertical leg 12 in, horizontal leg 4 in, both 1 in thick. Relation: Split without overlap — the full vertical leg plus the rest of the horizontal leg — then y = sum(A y)/sum(A). Substitute: vertical 12.000 in^2 at 6.00 in, horizontal 3.000 in^2 at 0.500 in Result: y = 4.900 in.
Why B is tempting
Placed the horizontal leg centroid at half the leg length rather than half the leg thickness.
Why C is tempting
Used the mid-height of the vertical leg alone, ignoring the horizontal leg that pulls the centroid down.
Why D is tempting
Left the horizontal leg area out of the denominator.
Question 5 of 5
A 200-lb block sits on a 20-degree incline with a coefficient of static friction of 0.25. The force parallel to the incline needed to push it up the slope is most nearly:
A 200 lb block on a 20° incline with a coefficient of static friction of 0.25, pushed by a force acting parallel to the slope. Show the answer and the reasoning
Why D
Given: W = 200 lb on a 20-degree incline, mu_s = 0.25. Relation: Pushing up the slope, gravity and friction both resist: P = W sin(theta) + mu W cos(theta). Substitute: P = 68.4 + 47.0 Result: P = 115.4 lb.
Why A is tempting
Subtracted the friction. Pushing a block up an incline means friction acts down the slope, opposing the motion.
Why B is tempting
Counted only the friction and ignored the weight component along the slope.
Why C is tempting
Left friction out entirely.
Four mistakes that cost the question
Each of these lands on an answer that is offered, so it costs the question outright. The minutes are our estimate of the time each one burns on top, against an average of 2.9 minutes a question (320 minutes for 110).
Costs the question and about 3 min
Multiplying by the lever length instead of the perpendicular distance
A moment is force times the perpendicular distance from the point to the line of action, or the lever length times the force component perpendicular to the lever. Force times full lever length ignores the angle; using the cosine when the angle is measured from the lever picks the wrong component. Draw the line of action before multiplying.
Costs the question and about 3 min
Placing a triangular load's resultant at mid-length
A uniform load's resultant acts at the middle; a triangular load's acts one third of the length from the heavy end. Putting it at the middle moves the reactions and every moment that follows, and the midpoint answer is always available among the choices.
Costs the question and about 5 min
Missing the zero-force members before solving a truss
At an unloaded joint where two non-collinear members meet, both carry zero; where three meet and two are collinear, the third carries zero. Missing them sends you into joints with more unknowns than equations, which costs the clock as much as the question. For a single member deep in a truss, the method of sections is usually faster.
Costs the question and about 3 min
Measuring a centroid from the wrong reference
A composite centroid is the sum of each area times its centroid distance, divided by the total area, with every distance measured from one datum. Measuring one part from its own edge, or reporting the distance from the top when the question asks from the bottom, gives the other natural answer, which is offered. Mark the datum on the sketch first.
Questions about Statics
Does every statics question have a figure?
On the official NCEES practice exam, which we counted question by question, every statics item did. Reading a free-body diagram is the work in this area, which is why all five problems on this page are drawn, each from the same numbers as its stem.
What is the fastest way to find one member force in a truss?
Usually the method of sections: cut through the member you want and no more than three members in all, then take moments about the point where the other two cut members meet. The method of joints is faster near a support or at a joint with only two unknowns. Find the zero-force members first; they often make the cut obvious.
How is static friction tested?
As impending motion: a block about to slide on a level floor or an incline, a body that may tip before it slides, or a rope wrapped around a fixed drum, where the tension ratio is e raised to the friction coefficient times the wrap angle in radians. Static friction is one of the seven statics sub-topics.
Is the moment of inertia here the same one used in structural questions?
Yes. Statics asks for area moments of inertia of composite sections, using the parallel axis theorem to shift each part to the common centroid. Mechanics of materials and structural engineering then use that same I in bending stress and deflection, so a slip here costs questions in three areas.
How many study hours this area is worth · Using the handbook under the clock