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​ = 55 kNP2​ = 55 kNABRA​ = ?RB​3 m1 m2 m

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

L = 6 mP1 = 55 kN at 3 mP2 = 55 kN at 4 m
RA = kN
2.Beam support reactionsHard
P = 10 kN, 40°Q = 15 kNABα = ?2.5 m1.5 m1 m

The simply supported beam has a pin at A and a roller at B. A load P acts at 2.5 m from A, inclined at 40° to the horizontal, and a vertical load Q acts at 4 m from A. Find the angle α between the total reaction force at the pin A and the horizontal.

L = 5 mP = 10 kN at 2.5 m, θ = 40°Q = 15 kN at 4 m
α = °
3.Shear force & bending momentMedium
P1​ = 30 kNP2​ = 50 kNABC2 m2 m1 m

The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. It carries a point load P1 between the supports and a point load P2 at C. Find the bending moment Mx under the load P1. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP1 = 30 kN at 2 mP2 = 50 kN at 5 msupports at x = 0 and x = 4 m
Mx = kN·m
4.Shear force & bending momentHard
w = 2 kN/mP = 30 kNABsection3 m3 m2 m

Find the bending moment Mx at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 30 kN at 6 mw = 2 kN/m from 0 to 8 mx = 3 m
Mx = kN·m
5.Truss: method of jointsMedium
ABBCACACB15 kN4 m3 m2 m

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

span = 4 mheight = 2 mB is 3 m from Avertical load = 15 kN
FAC = kN
6.Truss: method of jointsHard
ABCDEFGH40 kN35 kN10 kN20 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: 40 kN ↓G: 35 kN ↓H: 10 kN ↓H: 20 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 = 13 kNABδ3.5 m4.5 m

A simply supported beam carries a point load P at a = 3.5 m from support A. Find the deflection δP under the load.

P = 13 kNa = 3.5 mL = 8 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
P = 10 kNAB2.5 m2.5 m

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the size of the bending moment MA at the fixed end.

L = 5 mP = 10 kN at 2.5 m
MA = kN·m
9.Column bucklingMedium
PL = 2 md = 50 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 = 2 mE = 200 GPa (steel)d = 50 mm
Pcr = kN
10.Column bucklingHard
P = 45 kNL = 1 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 45 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 45 kNn = 2L = 1 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RA = 45.83 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​ = 55 kNP2​ = 55 kNABRA​RB​3 m1 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·6 = 55·3 + 55·4
RB = 385/6 = 64.17 kN
ΣFy = 0: RA = 110 − 64.17 = 45.83 kN

Answer

RA = 45.83 kN
2.Beam support reactions

Answer: α = 39.05 °

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.

Components of P

Pv = 10·sin 40° = 6.428 kN
Ph = 10·cos 40° = 7.66 kN

Reactions

ΣMA = 0: RB·5 = 6.428·2.5 + 15·4 → RB = 15.21 kN
Ay = 6.428 + 15 − 15.21 = 6.214 kN
Ax = Ph = 7.66 kN

Direction at A

α = atan(Ay/Ax) = atan(6.214/7.66) = 39.05°

Answer

α = 39.05°
3.Shear force & bending moment

Answer: Mx = 5 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.

Support reactions

Equilibrium of the whole beam. The overhang load also acts about A.

ΣMA = 0: RB·4 = 30·2 + 50·5
RB = 310/4 = 77.5 kN
ΣFy = 0: RA = 80 − 77.5 = 2.5 kN

Cut under P1 and keep the left part

Mx = RA·2 = 2.5·2 = 5 kN·m

Answer

Mx = 5 kN·m
4.Shear force & bending moment

Answer: Mx = 37.5 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 2·8 = 16 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 30·6 + 2·8·4
RB = 244/8 = 30.5 kN
ΣFy = 0: RA = 46 − 30.5 = 15.5 kN
w = 2 kN/mP = 30 kNABRA​ = 15.5RB​ = 30.5section3 m3 m2 m

Cut at x = 3 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 15.5·3 − 2·3·1.5 = 37.5 kN·m

Answer

Mx = 37.5 kN·m
5.Truss: method of joints

Answer: FAC = 5.625 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.

ABBCACACB15 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·4 = 15·3 → Cy = 11.25 kN
ΣFy = 0: Ay = 15 − 11.25 = 3.75 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 = 2/2.236 = 0.8944, cos γ = 1/2.236 = 0.4472
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −11.25/0.8944 = −12.58 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−12.58)·0.4472 = 5.625 kN

Answer

FAC = 5.625 kN (tension)
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: 35 kN (T), BC: 50 kN (T), CD: 50 kN (T), DE: 30 kN (T), FG: 55 kN (C), GH: 50 kN (C), AF: 77.78 kN (C), HE: 42.43 kN (C), BF: 55 kN (T), CG: 0, DH: 20 kN (T), BG: 21.21 kN (C), DG: 28.28 kN (C)

Answer

Zero-force member: CG
7.Beam deflection

Answer: δP = 55.99 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 = 13000 N, a = 3500 mm, b = 4500 mm, L = 8000 mm
EI = 2.4 × 10¹² N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 13000·3500²·4500²/(3·2.4 × 10¹²·8000) = 55.99 mm

Answer

δP = 55.99 mm
8.Beam deflection

Answer: MA = 9.375 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 = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

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) = 5PL³/(48EI) → RB = 5P/16 = 5·10/16 = 3.125 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

MA = P·L/2 − RB·L = 25 − 15.63 = 9.375 kN·m

Answer

MA = 9.375 kN·m
9.Column buckling

Answer: Pcr = 37.85 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 2·2000 = 4000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·306796/4000² = 37849 N = 37.85 kN

Answer

Pcr = 37.85 kN
10.Column buckling

Answer: d = 31.04 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 = 2·45000 = 90000 N
KL = 1·1000 = 1000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·90000·1000²/(π³·200000))^(1/4) = 31.04 mm

Answer

d = 31.04 mm

Check: the slenderness KL/r is 128.8, so the column is slender and Euler applies.

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