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
P1​ = 45 kNP2​ = 45 kNABRA​ = ?RB​1.5 m1.5 m1 m

The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RA.

L = 4 mP1 = 45 kN at 1.5 mP2 = 45 kN at 3 m
RA = kN
2.Beam support reactionsHard
w = 2 kN/mP = 65 kNQ = 75 kNADCMA​ = ?2 m2 m3 m1 m

A compound beam is fixed to the wall at A, has an internal hinge at C (4 m from A) and rests on a roller at D (8 m from A). It carries a point load P at 7 m from A, a uniformly distributed load w on the whole part CD and a point load Q at 2 m from A. Find the magnitude of the reaction moment MA at the wall.

AC = 4 mAD = 8 mP = 65 kN at 7 mw = 2 kN/m from 4 to 8 mQ = 75 kN at 2 m
MA = kN·m
3.Shear force & bending momentMedium
60 kN55 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.

60 kN @ 6 m55 kN @ 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)
4.Shear force & bending momentHard
10 kN/m20 kN20 kN45 kNABCDEFG1 m1 m1 m1 m1 m1 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.

20 kN @ 0 m20 kN @ 6 m45 kN @ 3 m10 kN/m, 1–5 m
  1. 1A (x = 0 m)
  2. 2B (x = 1 m)
  3. 3C (x = 2 m)
  4. 4D (x = 3 m)
  5. 5E (x = 4 m)
  6. 6F (x = 5 m)
  7. 7G (x = 6 m)
5.Truss: method of jointsMedium
ABBCACACB50 kN5 m3 m3 m

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

span = 5 mheight = 3 mB is 3 m from Avertical load = 50 kN
FBC = kN
6.Truss: method of jointsHard
ABCDEFGH20 kN10 kN20 kN10 kN4 × 4 m4 m

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

B: 20 kN ↓D: 10 kN ↓G: 20 kN ↓H: 10 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 = 28 kNABδ1.5 m3.5 m

A simply supported beam carries a point load P at a = 1.5 m from support A. Find the deflection δP under the load.

P = 28 kNa = 1.5 mL = 5 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
P = 26 kNABδ3 m1 m

A simply supported beam carries a point load P at b = 1 m from support B. Find the maximum deflection δmax of the beam.

P = 26 kNb = 1 mL = 4 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 6 md = 60 mmcross-section

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

L = 6 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
P = 80 kNL = 4 md = ?cross-section

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 80 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 80 kNn = 2.5L = 4 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: RA = 39.38 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.

P1​ = 45 kNP2​ = 45 kNABRA​RB​1.5 m1.5 m1 m

Free-body diagram

Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.

Equilibrium

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

ΣMA = 0: RB·4 = 45·1.5 + 45·3
RB = 202.5/4 = 50.63 kN
ΣFy = 0: RA = 90 − 50.63 = 39.38 kN

Answer

RA = 39.38 kN
2.Beam support reactions

Answer: MA = 231 kN·m

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. An internal hinge cannot carry a bending moment, which gives one extra equation: the moment about the hinge of the forces on either side is zero.

Part CD

The hinge has no moment, so moments about C on CD alone must balance.

W = 2·4 = 8 kN at 2 m from C
ΣMC = 0: RD·4 = 65·3 + 8·2
RD = 52.75 kN
VC = 73 − 52.75 = 20.25 kN

Part AC

The hinge force VC acts downward on AC (equal and opposite to its action on CD).

RA = 75 + 20.25 = 95.25 kN
MA = 75·2 + 20.25·4 = 231 kN·m

Answer

MA = 231 kN·m
3.Shear force & bending moment

Answer: D (x = 6 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 = 48.33 kN, R2 = 66.67 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 = 96.67, MC = 193.3, MD = 290 kN·m, ME = 266.7, MF = 133.3, MG = 0

Answer

Mark D: MD = 290 kN·m
4.Shear force & bending moment

Answer: D (x = 3 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 = 62.5 kN, R2 = 62.5 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 = −20, MC = 17.5, MD = 45 kN·m, ME = 17.5, MF = −20, MG = 0

Answer

Mark D: MD = 45 kN·m
5.Truss: method of joints

Answer: FBC = −36.06 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.

ABBCACACB50 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·5 = 50·3 → Cy = 30 kN
ΣFy = 0: Ay = 50 − 30 = 20 kN

Joint C

The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.

sin γ = 3/3.606 = 0.8321, cos γ = 2/3.606 = 0.5547
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −30/0.8321 = −36.06 kN

Answer

FBC = −36.06 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: 20 kN (T), BC: 30 kN (T), CD: 30 kN (T), DE: 20 kN (T), FG: 30 kN (C), GH: 30 kN (C), AF: 42.43 kN (C), HE: 28.28 kN (C), BF: 30 kN (T), CG: 0, DH: 20 kN (T), BG: 14.14 kN (C), DG: 14.14 kN (C)

Answer

Zero-force member: CG
7.Beam deflection

Answer: δP = 6.125 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 = 28000 N, a = 1500 mm, b = 3500 mm, L = 5000 mm
EI = 8.4 × 10¹² N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 28000·1500²·3500²/(3·8.4 × 10¹²·5000) = 6.125 mm

Answer

δP = 6.125 mm
8.Beam deflection

Answer: δmax = 9.613 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 = 26000 N, b = 1000 mm, L = 4000 mm
EI = 2.52 × 10¹² N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((4000² − 1000²)/3) = 2236 mm

Maximum deflection

δmax = Pb(L² − b²)3/2/(9√3·EIL) = 9.613 mm

Answer

δmax = 9.613 mm
9.Column buckling

Answer: Pcr = 12.21 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 round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Effective length

KL = 1·6000 = 6000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·636173/6000² = 12209 N = 12.21 kN

Answer

Pcr = 12.21 kN
10.Column buckling

Answer: d = 82.46 mm

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.

Required critical load

Pcr = n·P = 2.5·80000 = 200000 N
KL = 0.7·4000 = 2800 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·200000·2800²/(π³·70000))^(1/4) = 82.46 mm

Answer

d = 82.46 mm

Check: the slenderness KL/r is 135.8, so the column is slender and Euler applies.

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