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​ = 12 kN/mP = 75 kNABRB​ = ?4.5 m1.5 m

The simply supported beam (pin at A, roller at B) carries a linearly varying load that falls from 12 kN/m at A to zero at B. A point load P also acts at 4.5 m from A. Find the vertical support reaction RB.

L = 6 mw0 = 12 kN/mP = 75 kN at 4.5 m
RB = kN
2.Beam support reactionsHard
P = 75 kN, 40°Q = 35 kNABα = ?0.5 m2 m1.5 m

The simply supported beam has a pin at A and a roller at B. A load P acts at 0.5 m from A, inclined at 40° to the horizontal, and a vertical load Q acts at 2.5 m from A. Find the angle α between the total reaction force at the pin A and the horizontal.

L = 4 mP = 75 kN at 0.5 m, θ = 40°Q = 35 kN at 2.5 m
α = °
3.Shear force & bending momentMedium
w = 2 kN/mP = 10 kNABsection1.5 m1 m

The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 2.5 m from A. Find the bending moment Mx at the section x = 1.5 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 2.5 mP = 10 kN at 2.5 mw = 2 kN/m from 0 to 2.5 mx = 1.5 m
Mx = kN·m
4.Shear force & bending momentHard
10 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.

10 kN @ 0 m15 kN @ 6 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
ABCDEFGH30 kN15 kN25 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.

D: 30 kN ↓G: 15 kN ↓H: 25 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
ABBCACACB50 kN15 kN6 m2 m2 m

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

span = 6 mheight = 2 mB is 2 m from Avertical load = 50 kNhorizontal load = 15 kN →
FAB = kN
7.Beam deflectionMedium
P = 12 kNABδ3 m1 m

A simply supported beam carries a point load P at a = 3 m from support A. Find the deflection δP under the load.

P = 12 kNa = 3 mL = 4 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
P = 14 kNABδ4 m1 m

A simply supported beam carries a point load P at b = 1 m from support B. Find the maximum deflection δmax of the beam.

P = 14 kNb = 1 mL = 5 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴
δmax = mm
9.Column bucklingMedium
PL = 3 m30 × 80 mmcross-section

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

L = 3 mE = 70 GPa (aluminium)b × h = 30 × 80 mm
Pcr = kN
10.Column bucklingHard
PL = 2.5 md = 50 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 = 2.5 mE = 200 GPa (steel)d = 50 mm
σcr = MPa

3Solutions

1.Beam support reactions

Answer: RB = 68.25 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 = ½·12·6 = 36 kN
x̄ = L/3 = 2 m

Equilibrium

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

ΣMA = 0: RB·6 = 36·2 + 75·4.5
RB = 68.25 kN
RA = 111 − 68.25 = 42.75 kN

Answer

RB = 68.25 kN
2.Beam support reactions

Answer: α = 43.91 °

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 = 75·sin 40° = 48.21 kN
Ph = 75·cos 40° = 57.45 kN

Reactions

ΣMA = 0: RB·4 = 48.21·0.5 + 35·2.5 → RB = 27.9 kN
Ay = 48.21 + 35 − 27.9 = 55.31 kN
Ax = Ph = 57.45 kN

Direction at A

α = atan(Ay/Ax) = atan(55.31/57.45) = 43.91°

Answer

α = 43.91°
3.Shear force & bending moment

Answer: Mx = −11 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.

Keep the free-end part

P: −10·1 = −10 kN·m
w: −2·1²/2 = −1 kN·m
Mx = −10 − 1 = −11 kN·m

Answer

Mx = −11 kN·m

Negative means hogging: tension on the top face.

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 = 3.333 kN, R2 = 21.67 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 = −16.67, MD = −23.33, ME = −30 kN·m, MF = −15, MG = 0

Answer

Mark E: ME = −30 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: 21.25 kN (T), BC: 21.25 kN (T), CD: 48.75 kN (T), DE: 48.75 kN (T), FG: 42.5 kN (C), GH: 42.5 kN (C), AF: 30.05 kN (C), HE: 68.94 kN (C), BF: 0, CG: 15 kN (C), DH: 30 kN (T), FC: 30.05 kN (T), HC: 8.839 kN (C)

Answer

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

Answer: FAB = −40.07 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.

ABBCACACB50 kN15 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·6 = 50·2 + 15·2 → Cy = 21.67 kN
ΣFy = 0: Ay = 50 − 21.67 = 28.33 kN
ΣFx = 0: Ax = −15 kN

Joint A

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

sin α = 2/2.828 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −28.33/0.7071 = −40.07 kN

Answer

FAB = −40.07 kN (compression)
7.Beam deflection

Answer: δP = 9.74 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 = 12000 N, a = 3000 mm, b = 1000 mm, L = 4000 mm
EI = 9.24 × 10¹¹ N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 12000·3000²·1000²/(3·9.24 × 10¹¹·4000) = 9.74 mm

Answer

δP = 9.74 mm
8.Beam deflection

Answer: δmax = 12 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 = 14000 N, b = 1000 mm, L = 5000 mm
EI = 1.76 × 10¹² N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((5000² − 1000²)/3) = 2828 mm

Maximum deflection

δmax = Pb(L² − b²)3/2/(9√3·EIL) = 12 mm

Answer

δmax = 12 mm
9.Column buckling

Answer: Pcr = 3.454 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·180000/6000² = 3454 N = 3.454 kN

Answer

Pcr = 3.454 kN
10.Column buckling

Answer: σcr = 100.7 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·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 0.7·2500 = 1750 mm
KL/r = 1750/12.5 = 140

Critical stress

σcr = π²E/(KL/r)² = π²·200000/140² = 100.7 MPa

Answer

σcr = 100.7 MPa

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