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​ = 15 kNP2​ = 15 kNABRA​RB​ = ?1 m1.5 m3.5 m

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

L = 6 mP1 = 15 kN at 1 mP2 = 15 kN at 2.5 m
RB = kN
2.Beam support reactionsHard
w0​ = 15 kN/mP = 60 kNABMA​ = ?5 m5 m

The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is zero at the wall and rises to 15 kN/m at the free end B, plus a point load P at 5 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 10 mw0 = 15 kN/mP = 60 kN at 5 m
MA = kN·m
3.Shear force & bending momentMedium
25 kN15 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.

25 kN @ 1 m15 kN @ 2 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)
4.Shear force & bending momentHard
5 kN/m30 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.

30 kN @ 0 m55 kN @ 12 m5 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
ABBCACACB25 kN4 m1 m4 m

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

span = 4 mheight = 4 mB is 1 m from Avertical load = 25 kN
FBC = kN
6.Truss: method of jointsHard
ABCDEFGH10 kN10 kN15 kN35 kN4 × 3 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.

D: 10 kN ↓F: 10 kN ↓F: 15 kN →G: 35 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
w = 20 kN/mABθ8 m

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θA of the beam at support A, in radians.

w = 20 kN/mL = 8 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴
θA = rad
8.Beam deflectionHard
w = 12 kN/mAB4 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 size of the bending moment MA at the fixed end.

L = 4 mw = 12 kN/m
MA = kN·m
9.Column bucklingMedium
PL = 4 m40 × 80 mmcross-section

A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 4 mE = 200 GPa (steel)b × h = 40 × 80 mm
Pcr = kN
10.Column bucklingHard
P = 135 kNL = 2.5 md = ?cross-section

A solid round column of length L made of aluminium is pinned at both ends. It must carry 135 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 135 kNn = 2L = 2.5 mE = 70 GPa (aluminium)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 8.75 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​ = 15 kNP2​ = 15 kNABRA​RB​1 m1.5 m3.5 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·6 = 15·1 + 15·2.5
RB = 52.5/6 = 8.75 kN
ΣFy = 0: RA = 30 − 8.75 = 21.25 kN

Answer

RB = 8.75 kN
2.Beam support reactions

Answer: MA = 800 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 triangular load

W = ½·15·10 = 75 kN
x̄ = 2L/3 = 6.667 m

Moment about the wall

MA = W·x̄ + P·a = 75·6.667 + 60·5 = 800 kN·m

Answer

MA = 800 kN·m
3.Shear force & bending moment

Answer: C (x = 2 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 = 30.83 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 = 30.83, MC = 36.67 kN·m, MD = 27.5, ME = 18.33, MF = 9.167, MG = 0

Answer

Mark C: MC = 36.67 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 = 18.33 kN, R2 = 96.67 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 = −60, MC = −93.33, MD = −146.7, ME = −220 kN·m, MF = −110, MG = 0

Answer

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

Answer: FBC = −7.813 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.

ABBCACACB25 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·4 = 25·1 → Cy = 6.25 kN
ΣFy = 0: Ay = 25 − 6.25 = 18.75 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 = 4/5 = 0.8, cos γ = 3/5 = 0.6
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −6.25/0.8 = −7.813 kN

Answer

FBC = −7.813 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: 31.88 kN (T), BC: 41.25 kN (T), CD: 41.25 kN (T), DE: 24.37 kN (T), FG: 31.88 kN (C), GH: 24.37 kN (C), AF: 28.13 kN (C), HE: 40.63 kN (C), BF: 12.5 kN (T), CG: 0, DH: 32.5 kN (T), BG: 15.63 kN (C), DG: 28.13 kN (C)

Answer

Zero-force member: CG
7.Beam deflection

Answer: θA = 0.0254 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 20 N/mm, L = 8000 mm
EI = 1.68 × 10¹³ N·mm²

End slope

θA = wL³/(24EI) = 20·8000³/(24·1.68 × 10¹³) = 0.0254 rad

Answer

θA = 0.0254 rad

That is 1.455°.

8.Beam deflection

Answer: MA = 24 kN·m

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 = 3·12·4/8 = 18 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

MA = wL²/2 − RB·L = 96 − 72 = 24 kN·m

Answer

MA = 24 kN·m
9.Column buckling

Answer: Pcr = 107.4 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 = 80·40³/12 = 426667 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·426667/2800² = 107424 N = 107.4 kN

Answer

Pcr = 107.4 kN
10.Column buckling

Answer: d = 83.99 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 = 2·135000 = 270000 N
KL = 1·2500 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·270000·2500²/(π³·70000))^(1/4) = 83.99 mm

Answer

d = 83.99 mm

Check: the slenderness KL/r is 119.1, so the column is slender and Euler applies.

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