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. A load P acts at 2 m from A, inclined at 30° to the horizontal, and a vertical load Q acts at 6 m from A. Find the angle α between the total reaction force at the pin A and the horizontal.
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
Use the method of joints to find the force FAC in member AC. Give tension as positive and compression as negative.
Find the bending moment Mx at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
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.
A simply supported beam carries a point load P at a = 2 m from support A. Find the deflection δP under the load.
A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area Ireq that meets this limit.
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.
The T-section is used as a beam with the flange on top and carries a sagging bending moment of 40 kN·m. Find the size of the bending stress σtop at the top edge.
For the soil profile shown, find the vertical effective stress σ′ at point X.
For the soil profile shown, find the total vertical stress σ at point X.
A 3.5 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.05 and e₀ = 0.65. At its mid-depth the effective vertical stress is 75 kPa and the overconsolidation ratio is OCR = 2.5. A new building adds 150 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
In a laboratory oedometer test, a clay specimen 19 mm thick, drained at the top and bottom, reaches 50% consolidation in 8 min. In the field the same clay forms a layer 4 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 2 m wide sits in the wall of a tank of water, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of glycerin, with its top edge 1 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water flows at a mean velocity of 1.4 m/s through a pipe 400 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the friction head loss hf.
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.022. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 5) 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 100 L/s?
A rectangular grass-lined earth channel is 3 m wide. Water flows uniformly 1.5 m deep. Manning n = 0.03 and the bed slope is S = 0.001. Find the flow rate Q in m³/s.
A trapezoidal unfinished concrete channel with a bottom width of 4.5 m and side slopes of 1:1 (horizontal:vertical) must carry 43 m³/s in uniform flow at a depth of 2.4 m. Manning n = 0.014. Find the required bed slope S (in m/m).
A pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 16 m. The pump efficiency is 82% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 21 m higher. The pipe is 300 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
You need €50 000 in 6 years. An account pays 2% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
You want €100 000 in 16 years. You save the same amount at the end of every month in an account paying 5.4% per year, compounded monthly. Find the monthly deposit A.
A machine costs €20 000 today, lasts 9 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.
A project costs €50 000 and returns €31 000, €20 500, €31 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 49.5 and 77 points? Give P as a decimal.
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.
Answer: RB = 43.5 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: α = 34.82 °
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Replace each distributed load by its resultant (its area) acting at the centroid of the load diagram, and take moments about a support to remove that support’s reaction from the equation.
Answer: FBC = −32.05 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: FAC = 21.88 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: Mx = 9.375 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: Mx = 51.56 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: δP = 1.014 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: Ireq = 1.875 × 10⁶ mm⁴
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).
Answer: Pcr = 62.01 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 = 1.919 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 = 13.68 × 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.
Measure every y from the bottom edge.
Answer: σtop = 64.76 MPa
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.
Measure every y from the bottom edge.
The top edge is h − ȳ from the axis.
Answer: σ′ = 66.88 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).
Add the weight of each soil band above X. Above the water table use the dry unit weight, below it the saturated one.
Water pressure grows with depth below the water table.
Answer: σ = 102.5 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).
Add the weight of each soil band above X. Above the water table use the dry unit weight, below it the saturated one.
Answer: S = 84.2 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.
Recompression up to σ′p, virgin compression beyond it.
Answer: t = 246.2 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 = 206 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 = 2.481 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 = 7.992 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 = 1.534 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 = 3.916 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 = 0.0007928 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 0.7928 m per kilometre.
Answer: Pel = 2.127 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: H = 28.99 m
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.
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
Answer: PV = 44359 €
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: A = 328.9 €
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: EAC = 5473 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Answer: IRR = 30.12 %
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.
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
Answer: P = 0.8244
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.
Answer: x = 166.5 cm
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
ExamProblems · examproblems.com/courses/fe-exam-civil