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
w = 6 kN/mP = 55 kNABRA​ = ?2.5 m1 m1.5 m

The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w from the wall to 2.5 m and a point load P at 3.5 m from the wall. Find the vertical reaction RA at the wall.

L = 5 mP = 55 kN at 3.5 mw = 6 kN/m from 0 to 2.5 m
RA = kN
2.Beam support reactionsHard
w = 3 kN/mP = 15 kNABRA​ = ?RB​2 m3 m

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

L = 5 mP = 15 kN at 2 mw = 3 kN/m from 2 to 5 m
RA = kN
3.Shear force & bending momentMedium
15 kN35 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.

15 kN @ 2 m35 kN @ 1 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
10 kN35 kN55 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.

10 kN @ 0 m35 kN @ 6 m55 kN @ 3 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
ABCDEFGH20 kN35 kN20 kN4 × 3 m3 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 ↓F: 35 kN ↓G: 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
6.Truss: method of jointsHard
ABBCACACB45 kN10 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 = 45 kNhorizontal load = 10 kN →
FAC = kN
7.Beam deflectionMedium
P = 28 kNABδ5 m1 m

A simply supported beam carries a point load P at a = 5 m from support A. Find the deflection δP under the load.

P = 28 kNa = 5 mL = 6 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
w = 10 kN/mAB8 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 = 8 mw = 10 kN/m
RB = kN
9.Column bucklingMedium
PL = 6 md = 30 mmcross-section

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

L = 6 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
PL = 1.5 md = 40 mmcross-section

A column of length L made of steel is fixed at the base and pinned at the top. 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 = 70 kN

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 distributed load

W = 6·2.5 = 15 kN, x̄ = 1.25 m

Vertical equilibrium

RA = P + W = 55 + 15 = 70 kN

Answer

RA = 70 kN
2.Beam support reactions

Answer: RA = 11.7 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 = 3 kN/mP = 15 kNABRA​RB​2 m3 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 = 3·3 = 9 kN, acting at x̄ = 3.5 m
ΣMA = 0: RB·5 = 15·2 + 3·3·3.5
RB = 61.5/5 = 12.3 kN
ΣFy = 0: RA = 24 − 12.3 = 11.7 kN

Answer

RA = 11.7 kN
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 = 39.17 kN, R2 = 10.83 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 = 39.17, MC = 43.33 kN·m, MD = 32.5, ME = 21.67, MF = 10.83, MG = 0

Answer

Mark C: MC = 43.33 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 = 8.333 kN, R2 = 91.67 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 = −10, MC = −11.67, MD = −13.33, ME = −70 kN·m, MF = −35, MG = 0

Answer

Mark E: ME = −70 kN·m
5.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: 51.25 kN (T), BC: 47.5 kN (T), CD: 47.5 kN (T), DE: 23.75 kN (T), FG: 51.25 kN (C), GH: 23.75 kN (C), AF: 72.48 kN (C), HE: 33.59 kN (C), BF: 16.25 kN (T), CG: 0, DH: 23.75 kN (T), BG: 5.303 kN (T), DG: 33.59 kN (C)

Answer

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

Answer: FAC = 19.38 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.

ABBCACACB45 kN10 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·4 = 45·3 + 10·2 → Cy = 38.75 kN
ΣFy = 0: Ay = 45 − 38.75 = 6.25 kN
ΣFx = 0: Ax = −10 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 = −38.75/0.8944 = −43.32 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−43.32)·0.4472 = 19.38 kN

Answer

FAC = 19.38 kN (tension)
7.Beam deflection

Answer: δP = 1.215 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 = 5000 mm, b = 1000 mm, L = 6000 mm
EI = 3.2 × 10¹³ N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 28000·5000²·1000²/(3·3.2 × 10¹³·6000) = 1.215 mm

Answer

δP = 1.215 mm
8.Beam deflection

Answer: RB = 30 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 = 3·10·8/8 = 30 kN

Answer

RB = 30 kN
9.Column buckling

Answer: Pcr = 8.721 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/3000² = 8721 N = 8.721 kN

Answer

Pcr = 8.721 kN
10.Column buckling

Answer: σcr = 179 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 = 0.7·1500 = 1050 mm
KL/r = 1050/10 = 105

Critical stress

σcr = π²E/(KL/r)² = π²·200000/105² = 179 MPa

Answer

σcr = 179 MPa

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