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​ = 60 kNP2​ = 25 kNABRA​ = ?RB​1 m7 m2 m

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

L = 10 mP1 = 60 kN at 1 mP2 = 25 kN at 8 m
RA = kN
2.Beam support reactionsHard
w = 4 kN/mP = 55 kNABRA​RB​ = ?1 m2 m2 m

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

L = 5 mP = 55 kN at 3 mw = 4 kN/m from 1 to 3 m
RB = kN
3.Shear force & bending momentMedium
6 kN/m40 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.

40 kN @ 6 m6 kN/m, 0–12 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/m60 kN25 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 @ 12 m25 kN @ 4 m10 kN/m, 0–8 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
ABBCACACB10 kN4 m3 m3 m

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

span = 4 mheight = 3 mB is 3 m from Avertical load = 10 kN
FAB = kN
6.Truss: method of jointsHard
ABCDEFGH35 kN10 kN30 kN40 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: 35 kN ↓F: 10 kN ↓F: 30 kN →G: 40 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 FC
  13. 13member HC
7.Beam deflectionMedium
w = 5 kN/mδL = 3 m

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δmax, at midspan.

w = 5 kN/mL = 3 mE = 11 GPa (timber)I = 8.5 × 10⁶ mm⁴
δmax = mm
8.Beam deflectionHard
w = 6 kN/mAB6 m

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.

L = 6 mw = 6 kN/m
RB = kN
9.Column bucklingMedium
PL = 3 md = 80 mmcross-section

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.

L = 3 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
P = 70 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 70 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 70 kNn = 2.5L = 2.5 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RA = 59 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​ = 60 kNP2​ = 25 kNABRA​RB​1 m7 m2 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·10 = 60·1 + 25·8
RB = 260/10 = 26 kN
ΣFy = 0: RA = 85 − 26 = 59 kN

Answer

RA = 59 kN
2.Beam support reactions

Answer: RB = 36.2 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.

w = 4 kN/mP = 55 kNABRA​RB​1 m2 m2 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.

UDL resultant: W = w·ℓ = 4·2 = 8 kN, acting at x̄ = 2 m
ΣMA = 0: RB·5 = 55·3 + 4·2·2
RB = 181/5 = 36.2 kN
ΣFy = 0: RA = 63 − 36.2 = 26.8 kN

Answer

RB = 36.2 kN
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 = 56 kN, R2 = 56 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 = 176, MD = 228 kN·m, ME = 176, MF = 100, MG = 0

Answer

Mark D: MD = 228 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 = 22.5 kN, R2 = 142.5 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 = 25, MC = 10, MD = −95, ME = −240 kN·m, MF = −120, MG = 0

Answer

Mark E: ME = −240 kN·m
5.Truss: method of joints

Answer: FAB = −3.536 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.

ABBCACACB10 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·4 = 10·3 → Cy = 7.5 kN
ΣFy = 0: Ay = 10 − 7.5 = 2.5 kN

Joint A

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

sin α = 3/4.243 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −2.5/0.7071 = −3.536 kN

Answer

FAB = −3.536 kN (compression)
6.Truss: method of joints

Answer: member DH

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 D

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

ΣF⊥ = 0 → FDH = 0

Check with the method of joints

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

AB: 76.25 kN (T), BC: 76.25 kN (T), CD: 38.75 kN (T), DE: 38.75 kN (T), FG: 77.5 kN (C), GH: 77.5 kN (C), AF: 65.41 kN (C), HE: 54.8 kN (C), BF: 35 kN (T), CG: 40 kN (C), DH: 0, FC: 1.768 kN (T), HC: 54.8 kN (T)

Answer

Zero-force member: DH
7.Beam deflection

Answer: δmax = 56.4 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.

w = 5 N/mm, L = 3000 mm
EI = 9.35 × 10¹⁰ N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·5·3000⁴/(384·9.35 × 10¹⁰) = 56.4 mm

Answer

δmax = 56.4 mm
8.Beam deflection

Answer: RB = 13.5 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).

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δload = wL⁴/(8EI)

Compatibility at the prop

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.

RB·L³/(3EI) = wL⁴/(8EI) → RB = 3wL/8 = 3·6·6/8 = 13.5 kN

Answer

RB = 13.5 kN
9.Column buckling

Answer: Pcr = 110.2 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/6000² = 110245 N = 110.2 kN

Answer

Pcr = 110.2 kN
10.Column buckling

Answer: d = 81.97 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 = 2.5·70000 = 175000 N
KL = 2·2500 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·175000·5000²/(π³·200000))^(1/4) = 81.97 mm

Answer

d = 81.97 mm

Check: the slenderness KL/r is 244, so the column is slender and Euler applies.

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