Structural Analysis: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Beam support reactions

Equilibrium
ΣFx = 0, ΣFy = 0, ΣM = 0
Uniform load
resultant w·L at the middle of the load
Moment
M = F·d

Method

  1. Draw a free-body diagram and replace each support by its reactions.
  2. Replace every distributed load by a single force at its centroid.
  3. Take moments about one support to find the reaction at the other.
  4. Use ΣFy = 0 to find the remaining reaction, and check with a second moment equation.

Shear force & bending moment

Shear and moment
V = ΣF on one side, M = ΣF·d on one side
Bending stress
σ = M·y/I
Rectangle
I = bh³/12
Section modulus
W = I/ymax
Simply supported, uniform load
Mmax = wL²/8

Method

  1. Find the support reactions from equilibrium of the whole beam.
  2. Cut the beam at the section you need and keep one side.
  3. Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
  4. For the maximum moment, find where the shear force is zero.

Truss: method of joints

Method of joints
ΣFx = 0, ΣFy = 0 at each joint
Method of sections
ΣM = 0 about a cut point
Member force
positive = tension, negative = compression

Method

  1. Find the support reactions for the whole truss.
  2. Method of joints: start at a joint with at most two unknown member forces and use ΣFx = 0 and ΣFy = 0.
  3. Method of sections: cut through the member you need and take moments about a point where the other unknown members meet.
  4. Assume tension; a negative answer means compression.

Beam deflection

Point load at midspan
δ = PL³/(48EI)
Uniform load, simply supported
δ = 5wL⁴/(384EI)
Cantilever, end load
δ = PL³/(3EI)
Cantilever, uniform load
δ = wL⁴/(8EI)
Superposition
δ = δ1 + δ2

Column buckling

Euler load
Pcr = π²EI/(KL)²
Effective length factor
K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress
σcr = Pcr/A
Slenderness
λ = KL/r, r = √(I/A)

2Practice problems

1.Beam support reactionsMedium
w = 8 kN/mP = 45 kNABMA​ = ?3 m2 m

The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w from the wall to 3 m and a point load P at 3 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 5 mP = 45 kN at 3 mw = 8 kN/m from 0 to 3 m
MA = kN·m
2.Beam support reactionsHard
w = 2 kN/mP = 20 kNABCRA​ = ?2.5 m0.5 m1 m2 m

The beam has a pin at A and a roller at B, 4 m from A, and overhangs B to the free end C. It carries a point load P at 2.5 m from A and a uniformly distributed load w from 3 m to the free end at 6 m. Find the vertical reaction RA, taking upward as positive.

AB = 4 mAC = 6 mP = 20 kN at 2.5 mw = 2 kN/m from 3 to 6 m
RA = kN
3.Shear force & bending momentMedium
15 kN20 kNABCDEFG2 m2 m2 m2 m2 m2 m

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

15 kN @ 2 m20 kN @ 4 m
  1. 1A (x = 0 m)
  2. 2B (x = 2 m)
  3. 3C (x = 4 m)
  4. 4D (x = 6 m)
  5. 5E (x = 8 m)
  6. 6F (x = 10 m)
  7. 7G (x = 12 m)
4.Shear force & bending momentHard
6 kN/m50 kN55 kN35 kNABCDEFG2 m2 m2 m2 m2 m2 m

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

50 kN @ 0 m55 kN @ 12 m35 kN @ 4 m6 kN/m, 2–8 m
  1. 1A (x = 0 m)
  2. 2B (x = 2 m)
  3. 3C (x = 4 m)
  4. 4D (x = 6 m)
  5. 5E (x = 8 m)
  6. 6F (x = 10 m)
  7. 7G (x = 12 m)
5.Truss: method of jointsMedium
ABBCACACB45 kN6 m1 m3 m

Use the method of joints to find the force FAB in member AB. Give tension as positive and compression as negative.

span = 6 mheight = 3 mB is 1 m from Avertical load = 45 kN
FAB = kN
6.Truss: method of jointsHard
ABCDEFGH25 kN15 kN4 × 4 m3 m

Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.

G: 25 kN ↓H: 15 kN ←
  1. 1member AB
  2. 2member BC
  3. 3member CD
  4. 4member DE
  5. 5member FG
  6. 6member GH
  7. 7member AF
  8. 8member HE
  9. 9member BF
  10. 10member CG
  11. 11member DH
  12. 12member BG
  13. 13member DG
7.Beam deflectionMedium
P = 4 kNABδ1.5 m0.5 m

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1.5 m from A. Find the deflection δB at the free end B.

P = 4 kNa = 1.5 mL = 2 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴
δB = mm
8.Beam deflectionHard
w = 10 kN/mP = 8 kNδL = 6 m

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δmax.

P = 8 kNw = 10 kN/mL = 6 mE = 200 GPa (steel)I = 120 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 5 m30 × 50 mmcross-section

A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 5 mE = 70 GPa (aluminium)b × h = 30 × 50 mm
Pcr = kN
10.Column bucklingHard
PL = 6 md = 80 mmcross-section

A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 6 mE = 200 GPa (steel)d = 80 mm
σcr = MPa

3Solutions

1.Beam support reactions

Answer: MA = 171 kN·m

Cantilever equilibrium: a fixed support can push up and also resist rotation, so it gives a vertical reaction RA and a reaction moment MA. ΣFy = 0 gives RA, and moments about the wall give MA = Σ(load × distance from the wall). A distributed load acts as its resultant at the centroid of the load diagram.

Resultant of the distributed load

W = 8·3 = 24 kN, x̄ = 1.5 m

Moment about the wall

MA = P·a + W·x̄ = 45·3 + 24·1.5 = 171 kN·m

Answer

MA = 171 kN·m
2.Beam support reactions

Answer: RA = 6.75 kN

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.

Resultant of the distributed load

W = 2·3 = 6 kN, x̄ = (3 + 6)/2 = 4.5 m

Equilibrium

Moments about A, counter-clockwise positive. Then vertical force balance.

ΣMA = 0: RB·4 = 20·2.5 + 6·4.5
RB = 77/4 = 19.25 kN
RA = 26 − 19.25 = 6.75 kN

Answer

RA = 6.75 kN
3.Shear force & bending moment

Answer: C (x = 4 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R1 = 25.83 kN, R2 = 9.167 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

MA = 0, MB = 51.67, MC = 73.33 kN·m, MD = 55, ME = 36.67, MF = 18.33, MG = 0

Answer

Mark C: MC = 73.33 kN·m
4.Shear force & bending moment

Answer: E (x = 8 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R1 = 71.33 kN, R2 = 104.7 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

MA = 0, MB = −100, MC = −69.33, MD = −132.7, ME = −220 kN·m, MF = −110, MG = 0

Answer

Mark E: ME = −220 kN·m
5.Truss: method of joints

Answer: FAB = −39.53 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.

ABBCACACB45 kNAy​Cy​

Support reactions

Equilibrium of the whole truss. Moments about A, counter-clockwise positive.

ΣMA = 0: Cy·6 = 45·1 → Cy = 7.5 kN
ΣFy = 0: Ay = 45 − 7.5 = 37.5 kN

Joint A

Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.

sin α = 3/3.162 = 0.9487
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −37.5/0.9487 = −39.53 kN

Answer

FAB = −39.53 kN (compression)
6.Truss: method of joints

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.

Joint C

BC and CD lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.

ΣF⊥ = 0 → FCG = 0

Check with the method of joints

Solving all joints gives these member forces (T = tension, C = compression).

AB: 5.417 kN (T), BC: 25.83 kN (T), CD: 25.83 kN (T), DE: 12.92 kN (T), FG: 20.42 kN (C), GH: 27.92 kN (C), AF: 25.52 kN (C), HE: 16.15 kN (C), BF: 15.31 kN (T), CG: 0, DH: 9.688 kN (T), BG: 25.52 kN (C), DG: 16.15 kN (C)

Answer

Zero-force member: CG
7.Beam deflection

Answer: δB = 3.835 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).

Consistent units

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.

P = 4000 N, a = 1500 mm, L = 2000 mm
EI = 1.76 × 10¹² N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δP = Pa³/(3EI) = 2.557 mm
θP = Pa²/(2EI) = 0.002557 rad
δB = δP + θP·(L − a) = Pa²(3L − a)/(6EI) = 3.835 mm

Answer

δB = 3.835 mm
8.Beam deflection

Answer: δmax = 8.531 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).

Consistent units

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.

P = 8000 N, w = 10 N/mm, L = 6000 mm, EI = 2.4 × 10¹³ N·mm²

Each load on its own

δP = PL³/(48EI) = 8000·6000³/(48·2.4 × 10¹³) = 1.5 mm
δw = 5wL⁴/(384EI) = 5·10·6000⁴/(384·2.4 × 10¹³) = 7.031 mm

Add them

δmax = 1.5 + 7.031 = 8.531 mm

Answer

δmax = 8.531 mm
9.Column buckling

Answer: Pcr = 12.44 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.

Smallest second moment of area

A rectangle buckles about its weak axis, using the smaller dimension cubed.

Imin = h·b³/12 = 50·30³/12 = 112500 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·112500/2500² = 12436 N = 12.44 kN

Answer

Pcr = 12.44 kN
10.Column buckling

Answer: σcr = 5.483 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.

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴

Radius of gyration and slenderness

A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 2·6000 = 12000 mm
KL/r = 12000/20 = 600

Critical stress

σcr = π²E/(KL/r)² = π²·200000/600² = 5.483 MPa

Answer

σcr = 5.483 MPa

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