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
w0​ = 10 kN/mABRA​ = ?10 m

The simply supported beam (pin at A, roller at B) carries a linearly varying load that rises from zero at A to 10 kN/m at B. Find the vertical support reaction RA.

L = 10 mw0 = 10 kN/m
RA = kN
2.Beam support reactionsHard
w0​ = 15 kN/mP = 15 kNABMA​ = ?3.5 m0.5 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 3.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 4 mw0 = 15 kN/mP = 15 kN at 3.5 m
MA = kN·m
3.Shear force & bending momentMedium
M0​ = 30 kN·mAB2 m8 m

The simply supported beam carries a clockwise couple M0 = 30 kN·m at x = 2 m from A, and no other load. Find the bending moment Mx just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 10 mM0 = 30 kN·m clockwise, at 2 m
Mx = kN·m
4.Shear force & bending momentHard
35 kN50 kN20 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.

35 kN @ 0 m50 kN @ 6 m20 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)
5.Truss: method of jointsMedium
ABCDEFGH40 kN15 kN35 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.

D: 40 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 FC
  13. 13member HC
6.Truss: method of jointsHard
ABBCACACB55 kN25 kN4 m3 m2 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 = 2 mB is 3 m from Avertical load = 55 kNhorizontal load = 25 kN ←
FBC = kN
7.Beam deflectionMedium
w = 15 kN/mδL = 8 m

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

w = 15 kN/mL = 8 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴
δmax = mm
8.Beam deflectionHard
w = 15 kN/mP = 5 kNδL = 3 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 = 5 kNw = 15 kN/mL = 3 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 5 m20 × 50 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 = 5 mE = 200 GPa (steel)b × h = 20 × 50 mm
Pcr = kN
10.Column bucklingHard
PL = 1.5 md = 40 mmcross-section

A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 1.5 mE = 200 GPa (steel)d = 40 mm
σcr = MPa

3Solutions

1.Beam support reactions

Answer: RA = 16.67 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 triangular load

W = ½·10·10 = 50 kN
x̄ = 2L/3 = 6.667 m

Equilibrium

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

ΣMA = 0: RB·10 = 50·6.667
RB = 33.33 kN
RA = 50 − 33.33 = 16.67 kN

Answer

RA = 16.67 kN
2.Beam support reactions

Answer: MA = 132.5 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 = ½·15·4 = 30 kN
x̄ = 2L/3 = 2.667 m

Moment about the wall

MA = W·x̄ + P·a = 30·2.667 + 15·3.5 = 132.5 kN·m

Answer

MA = 132.5 kN·m
3.Shear force & bending moment

Answer: Mx = −6 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.

RB = M0/L = 30/10 = 3 kN (upward)
RA = −RB = −3 kN (downward)

Cut just left of the couple

The left part contains only RA.

Mx = RA·a = −3·2 = −6 kN·m

Answer

Mx = −6 kN·m
4.Shear force & bending moment

Answer: E (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 = 26.67 kN, R2 = 78.33 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 = −35, MC = −43.33, MD = −71.67, ME = −100 kN·m, MF = −50, MG = 0

Answer

Mark E: ME = −100 kN·m
5.Truss: method of joints

Answer: member BF

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 B

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

ΣF⊥ = 0 → FBF = 0

Check with the method of joints

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

AB: 38.75 kN (T), BC: 38.75 kN (T), CD: 51.25 kN (T), DE: 51.25 kN (T), FG: 62.5 kN (C), GH: 62.5 kN (C), AF: 54.8 kN (C), HE: 72.48 kN (C), BF: 0, CG: 35 kN (C), DH: 40 kN (T), FC: 33.59 kN (T), HC: 15.91 kN (T)

Answer

Zero-force member: BF
6.Truss: method of joints

Answer: FBC = −32.14 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.

ABBCACACB55 kN25 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·4 = 55·3 − 25·2 → Cy = 28.75 kN
ΣFy = 0: Ay = 55 − 28.75 = 26.25 kN
ΣFx = 0: Ax = 25 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 = 2/2.236 = 0.8944, cos γ = 1/2.236 = 0.4472
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −28.75/0.8944 = −32.14 kN

Answer

FBC = −32.14 kN (compression)
7.Beam deflection

Answer: δmax = 25 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 = 15 N/mm, L = 8000 mm
EI = 3.2 × 10¹³ N·mm²

Deflection

δmax = 5wL⁴/(384EI) = 5·15·8000⁴/(384·3.2 × 10¹³) = 25 mm

Answer

δmax = 25 mm
8.Beam deflection

Answer: δmax = 1.664 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 = 5000 N, w = 15 N/mm, L = 3000 mm, EI = 1.12 × 10¹³ N·mm²

Each load on its own

δP = PL³/(48EI) = 5000·3000³/(48·1.12 × 10¹³) = 0.2511 mm
δw = 5wL⁴/(384EI) = 5·15·3000⁴/(384·1.12 × 10¹³) = 1.413 mm

Add them

δmax = 0.2511 + 1.413 = 1.664 mm

Answer

δmax = 1.664 mm
9.Column buckling

Answer: Pcr = 5.371 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 = 50·20³/12 = 33333 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·33333/3500² = 5371 N = 5.371 kN

Answer

Pcr = 5.371 kN
10.Column buckling

Answer: σcr = 87.73 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·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 1·1500 = 1500 mm
KL/r = 1500/10 = 150

Critical stress

σcr = π²E/(KL/r)² = π²·200000/150² = 87.73 MPa

Answer

σcr = 87.73 MPa

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