Formula sheets, the key method steps for every topic and 30 practice problems, with full solutions at the back.
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.
Find the bending moment Mx at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment Mmax in the beam.
A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δmax, at midspan.
A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction RB at the prop.
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
A channel section lies with its open side up: a base plate 160 mm wide with two upright legs, 170 mm high overall. Base and legs are all 14 mm thick. Find its second moment of area Ix about the horizontal axis x through its centroid.
A 100 mm × 150 mm plate has a rectangular cut-out 40 mm wide and 40 mm high, centred on the vertical centre line with its centre 97.5 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
For the soil profile shown, find the pore water pressure u at point X.
For the soil profile shown, find the pore water pressure u at point X.
A 7 m thick overconsolidated clay layer has Cc = 0.5, Cr = 0.02 and e₀ = 0.7. At its mid-depth the effective vertical stress is 120 kPa and the overconsolidation ratio is OCR = 1.5. A new building adds 20 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
In a laboratory oedometer test, a clay specimen 25 mm thick, drained at the top and bottom, reaches 50% consolidation in 15 min. In the field the same clay forms a layer 4.5 m thick. It lies between two sand layers, so water drains out at the top and the bottom. How many days t does the field layer take to reach 50% consolidation?
A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 2 m wide sits in the wall of a tank of oil, with its top edge 2 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water flows at 25 L/s through a pipe 800 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 10) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 40 L/s?
A rectangular gravel channel is 3.5 m wide. Water flows uniformly 1.5 m deep. Manning n = 0.025 and the bed slope is S = 0.003. Find the flow rate Q in m³/s.
A trapezoidal gravel channel with a bottom width of 6 m and side slopes of 1:1 (horizontal:vertical) must carry 21 m³/s in uniform flow at a depth of 1.6 m. Manning n = 0.025. Find the required bed slope S (in m/m).
A pump with an efficiency of 85% delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 78 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
A pump moves 60 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 7 m higher. The pipe is 300 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 4 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 75%. Find the shaft power Pshaft in kW.
You invest €8 000 at a nominal rate of 2% per year, compounded monthly. Find the value FV of the investment after 10 years.
You borrow €15 000 at 4.2% per year, compounded monthly, and repay €350 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
A machine costs €50 000 today. It generates €10 000 at the end of each year for 5 years and can then be sold for €5 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €40 000 and returns €12 500, €17 000, €22 000, €12 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 6 of them is below 148.2 kg. Round z to two decimals.
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 25% of them lie (the 25th percentile).
Answer: RB = 28 kN
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Moments about A, counter-clockwise positive. Then vertical force balance.
Answer: RB = 20.83 kN
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Moments about A, counter-clockwise positive. Then vertical force balance.
Answer: FBC = −23.29 kN
Method of joints: every pin of a truss is in equilibrium, so ΣFx = 0 and ΣFy = 0 at each joint. Start at a joint with at most two unknown member forces. A negative result means the member is in compression.
Equilibrium of the whole truss. Moments about A, counter-clockwise positive.
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
Answer: member CG
Zero-force members: at a joint, the forces must balance in every direction. If a joint has no load and no support, and two of its three members are in line, then the balance across that line involves only the third member, so its force is zero. If only two members meet at an unloaded joint and they are not in line, both are zero.
BC and CD lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.
Solving all joints gives these member forces (T = tension, C = compression).
Answer: Mx = 29.25 kN·m
Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.
Equilibrium of the whole beam.
Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.
Answer: Mmax = 102 kN·m
Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.
Equilibrium of the whole beam.
It occurs at x = 6 m from A.
Answer: δmax = 7.99 mm
Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).
Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.
Answer: RB = 11.25 kN
Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition). For a statically indeterminate beam, remove a redundant support and require the deflection there to be zero (compatibility).
Without the prop the beam is a cantilever. Its free end would deflect under the load.
The prop force RB alone lifts the free end by RBL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.
Answer: Pcr = 33.91 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
A round section has the same I about every axis.
Answer: σcr = 17.77 MPa
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
A round section has the same I about every axis.
Answer: Ix = 19.59 × 10⁶ mm⁴
Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.
Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.
Answer: Ix = 27 × 10⁶ mm⁴
Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has Ic = bh³/12 and a circle Ic = πd⁴/64.
Answer: u = 9.81 kPa
Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).
Water pressure grows with depth below the water table.
Answer: u = 49.05 kPa
Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).
Water pressure grows with depth below the water table.
Answer: S = 5.513 mm
Primary consolidation of a clay layer of thickness H, using the effective stress at the middle of the layer: σ′f = σ′0 + Δσ. Normally consolidated clay: S = Cc·H/(1 + e0)·log₁₀(σ′f/σ′0). Overconsolidated clay recompresses along Cr up to the preconsolidation pressure σ′p and follows Cc only beyond it. Time: t = Tv·Hdr²/cv, where the drainage path Hdr is H/2 when water can leave at the top and bottom, and H when only one side drains.
The clay stays on its recompression line.
Answer: t = 337.5 days
Primary consolidation of a clay layer of thickness H, using the effective stress at the middle of the layer: σ′f = σ′0 + Δσ. Normally consolidated clay: S = Cc·H/(1 + e0)·log₁₀(σ′f/σ′0). Overconsolidated clay recompresses along Cr up to the preconsolidation pressure σ′p and follows Cc only beyond it. Time: t = Tv·Hdr²/cv, where the drainage path Hdr is H/2 when water can leave at the top and bottom, and H when only one side drains.
t = Tv·Hdr²/cv with the same T_v and cv in the lab and the field.
Answer: F = 275.9 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Answer: ycp = 3.714 m
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Pressure grows with depth, so the resultant acts below the centroid.
Answer: hf = 11.97 m
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Answer: Δz = 3.479 m
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Answer: Q = 9.976 m³/s
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
Answer: S = 0.001538 m/m
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
That is a fall of 1.538 m per kilometre.
Answer: Pshaft = 18 kW
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Answer: Pshaft = 10.89 kW
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Answer: FV = 9770 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Answer: N = 47 payments
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
46 full payments leave a small balance, which the last, smaller payment clears.
Answer: NPV = −8988 €
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
Answer: MIRR = 15.76 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Answer: P = 0.07078
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Answer: x = 173.3 cm
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
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