Statics · Comprehensive exercise
Exam training: These exercises stand above the seven chapters of the series and therefore have no theory page of their own. What you need is in the subject modules — here it is put together.
To the subject modules →Six exercises in three levels — easy, medium, hard. Each has graded hints: the first names the procedure, the second the balance to set up, the third the decisive intermediate value. No hint gives the answer.
Work through every exercise first on paper. If you get stuck anywhere, the corresponding subject module helps — the link is below each exercise.
Exercise 1 · Deflection pulley on the hall crane
A rope carries the load and runs over a frictionless deflection pulley. It leaves the pulley at the angle from the upward vertical. Find the force with which the pulley loads its axle.
The moment balance about the pulley axle contains only the two rope forces, both with the lever arm : . The radius cancels out, so .
This holds for each deflection angle — a frictionless pulley changes only the direction of a rope force, never its magnitude.
Magnitude of the axle force, symbolically in and .
Allowed symbols: G, gamma (in degrees)
Notation help: sin(gamma/2), brackets allowed.
Both rope forces act at the centre of the pulley — a concurrent force system. One strand pulls vertically downwards, the other at from the vertical:
Check: at the rope runs straight through, both forces cancel — and the formula gives zero.
Its line of action bisects the angle between the two rope strands — it points below the horizontal. For two forces of equal magnitude the resultant always lies on the angle bisector.
Isolate the pulley. The two rope forces and the axle force act on it — a concurrent force system.
For the magnitudes of the rope forces, the moment balance about the axle helps: both have the same lever arm.
Resolve both rope forces into horizontal and vertical components and add componentwise before taking the magnitude.
Conclusion: the pulley changes only the direction
Because the frictionless pulley does not change the magnitude of the rope force, the axle force is purely a matter of geometry. That is why pulley catalogues state the permissible deflection angle and not the permissible axle force.
Deepening: Forces with a common point of application
Exercise 2 · Centroid of a T-section
A T-section consists of a flange and a web . The flange sits on top of the web. The coordinate is counted from the bottom edge of the web counted.
Both part areas happen to be equal — that keeps the later arithmetic tidy but changes nothing about the method.
The flange starts at the top edge of the web at 100 mm; its own centroid lies half a flange thickness above:
Overall centroid, symbolically in the tabular method.
Allowed symbols: A1, A2, y1, y2
Notation help: fraction with /, brackets allowed.
The numerators are the static moments of the part areas (first moments of area); the denominator is the total area.
And needs no calculation at all: both part areas are symmetric about the vertical centre line, so the centroid lies there — 50 mm from the left flange edge.
Split the section into two rectangles and set up a table with and on.
First check whether an axis of symmetry gives one of the two coordinates without calculation.
The sum of the first moments is 320 000 mm³ — divide it by the total area.
Conclusion: symmetry first
Looking for axes of symmetry before calculating halves the work on such sections. Only the coordinate transverse to the axis of symmetry has to be determined at all.
Deepening: centroid
Exercise 3 · Beam under a distributed load and a single moment
A girder of length rests on a pinned support on the left and a roller support on the right . The left half (0 to 2.0 m) carries a distributed load ; at additionally a single moment acts clockwise.
, its line of action lies at the middle of the loaded length, that is at .
For the support reactions this combination is allowed. For internal forces it would only be permitted after the section.
support force , symbolically from the moment balance about .
Allowed symbols: Rq, a (distance from to ), M0, l
Notation help: fraction with /, brackets allowed.
The single moment enters without a lever arm — it is not tied to a line of action and acts the same at every position on the beam.
Without the moment 3.0 kN would come out — the difference of 2.5 kN is exactly .
Check with the moment balance about :
Replace the distributed load by its resultant and isolate the beam.
Take the moment balance about — the support force out.
The single moment gets no lever arm; it appears as its own term in the moment balance. Its sense of rotation decides the sign.
Conclusion: the moment has no line of action
A couple is not tied to a location. In the moment balance it therefore appears simply as a number — without a lever arm but with a sense of rotation. Treating it like a force and multiplying by a distance gives a wrong result.
Deepening: support reactions · General force systems
Exercise 4 · Truss with a Ritter section
A parallel-chord girder of span consists of three panels of and has the height . Lower-chord joints lie at 0, 3, 6 and 9 m, upper-chord joints at 3 and 6 m. Besides the posts and the two end diagonals there is an interior diagonal is between and . The load acts at joint .
How many zero-force members does this truss have?
At the node no external force acts, three members are connected, and two of them (the two lower-chord segments) lie in the same direction. By the third rule this makes the post a zero-force member.
At no other joint does one of the three rules apply: carries four members and the load, four members, three without a collinear pair.
, correspondingly .
The section in the second panel cuts upper chord, diagonal and lower chord. Assign to each sought force its reference point.
The reference point belongs at the intersection of the two other cut members. Lower chord and diagonal meet at , upper chord and diagonal at .
The load stands to the right of the section and does not appear in this balance. Negative means compression.
Check via the horizontal balance: with the diagonal force and its direction cosine date
First look for the zero-force members — each one found is one unknown less.
Calculate on the left part: only the support force stands there, the load lies to the right of the section.
The lever arm of the chord forces is the girder height , not their distance from the section.
Conclusion: strike out first, then cut selectively
Zero-force-member rules and the Ritter section mesh together: the rules remove unknowns, the section fetches exactly the member forces actually needed. For this truss not a single joint had to be calculated.
Deepening: Trusses
Exercise 5 · Gerber beam
A Gerber beam rests on three supports: at 0 (pinned support), at 6.0 m and at 11.0 m (both roller supports). At sits a moment hinge . Over 0 to 6.0 m the beam carries , at additionally the single load .
The secondary beam is statically determinate on its own: 3.0 m long, with the load 2.0 m from .
That is the whole trick of the Gerber beam: the hinge supplies the additional equation and makes the system solvable.
On the main beam the hinge force acts as a single load of 3.0 kN at the cantilever end at 8.0 m:
Check on the solidified whole system: and
Draw the shear force diagram over the whole length. Place points by tapping, drag them into position; a double tap deletes a point.
, then falling linearly to ; at support a jump of on .
Between 6.0 and 10.0 m no load acts, so stays constant — even across the hinge, because a moment hinge does transmit shear force very much so. At 10.0 m, by on and stays there until .
: the shear force becomes zero at zero.
As a check: at the hinge be —
Separate at the hinge and start with the part beam that is statically determinate on its own.
The hinge force acts on the other part beam as a single load — with the opposite sense of direction.
For the shear force diagram: it starts with and ends with . At each support it jumps upwards by that support's force, downwards at each load.
Conclusion: the hinge supplies the missing equation
Three supports mean three unknown reactions — one more than the three equilibrium conditions provide. The moment hinge makes up for that: is the fourth equation.
That is why Gerber beams are built at all: they span several fields and still remain statically determinate — settlement of a support produces no constraint stresses.
Deepening: support reactions · beam
Exercise 6 · Block on an inclined rope
A block of rests on a plane inclined by inclined plane with . A rope holds it and pulls at the angle from the plane upwards. Within what limits must the rope force lie?
First build the free-body diagram of the block.
Four forces: vertically downwards, normal to the plane, along the plane and the rope force under from the plane.
The sense of direction of is assumed up the slope here — that is the case in which the block threatens to slip down.
normal force, symbolically.
Allowed symbols: G, S, alpha, delta
Notation help: sin(delta), cos(alpha), brackets allowed.
Because the rope pulls away from the plane , it reduces the normal force — and with it the limiting static friction force. That is the difference from the horizontally pressing force, where grows.
The block threatens to slip downwards, so and :
Now it threatens to slip upwards, and reverses:
Only in the denominator does the sign change — that is where the influence of the rope angle on the normal force sits.
Put the axes along and normal to the plane, not horizontal and vertical — then only two components.
The rope force appears in both balances: with along and with normally.
After inserting the condition of sticking, appears on both sides. Collect the terms before dividing.
Conclusion: the rope angle has a say
A rope pulling away from the plane takes load off the contact — and thereby weakens exactly the static friction one is relying on. Pulling flat along the plane, by contrast, keeps the full normal force.
That is why lashing straps in freight transport carry a permissible lashing angle: it decides how much of the pre-tension is converted into holding force at all.
Deepening: Static and sliding friction
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| Exercise | Level | Status |
|---|---|---|
| 1 · Deflection pulley | EASY | Open |
| 2 · T-section | EASY | Open |
| 3 · Beam with a moment | MEDIUM | Open |
| 4 · Truss | MEDIUM | Open |
| 5 · Gerber beam | HARD | Open |
| 6 · Friction | HARD | Open |