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
P = 75 kNM = 30 kN·mABRB​ = ?1 m7 m2 m

The simply supported beam has a pin at A and a roller at B. It carries a point load P at 8 m from A and a counter-clockwise couple M at 1 m from A. Find the vertical reaction RB, taking upward as positive.

L = 10 mP = 75 kN at 8 mM = 30 kN·m (counter-clockwise) at 1 m
RB = kN
2.Beam support reactionsHard
P = 15 kN, 40°Q = 40 kNABRA​ = ?2 m2 m4 m

The simply supported beam has a pin at A and a roller at B. A load P acts at 4 m from A, inclined at 40° to the horizontal, and a vertical load Q acts at 2 m from A. Find the magnitude of the total reaction force RA at the pin A.

L = 8 mP = 15 kN at 4 m, θ = 40°Q = 40 kN at 2 m
RA = kN
3.Shear force & bending momentMedium
P1​ = 45 kNP2​ = 35 kNAB3 m4 m3 m

The simply supported beam carries two point loads. Find the maximum bending moment Mmax in the beam.

L = 10 mP1 = 45 kN at 3 mP2 = 35 kN at 7 m
Mmax = kN·m
4.Shear force & bending momentHard
5 kN/m30 kN45 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 m45 kN @ 12 m5 kN/m, 2–10 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
ABBCACACB50 kN8 m7 m4 m

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

span = 8 mheight = 4 mB is 7 m from Avertical load = 50 kN
FAB = kN
6.Truss: method of jointsHard
ABBCACACB40 kN15 kN5 m3 m3 m

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

span = 5 mheight = 3 mB is 3 m from Avertical load = 40 kNhorizontal load = 15 kN ←
FAB = kN
7.Beam deflectionMedium
w = 10 kN/mδL = 2 m

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 10 kN/mL = 2 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴
δ = mm
8.Beam deflectionHard
w = 4 kN/mδL = 1.5 m

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area Ireq that meets this limit.

w = 4 kN/mL = 1.5 mE = 11 GPa (timber)
Ireq = × 10⁶ mm⁴
9.Column bucklingMedium
PL = 1.5 m20 × 40 mmcross-section

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

L = 1.5 mE = 70 GPa (aluminium)b × h = 20 × 40 mm
Pcr = kN
10.Column bucklingHard
P = 145 kNL = 2 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 145 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 145 kNn = 2.5L = 2 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 57 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.

Moments about A

Counter-clockwise positive.

ΣMA = 0: RB·10 − 75·8 + 30 = 0
RB = 570/10 = 57 kN

Vertical equilibrium

RA = 75 − 57 = 18 kN

Answer

RB = 57 kN

A negative reaction acts downward.

2.Beam support reactions

Answer: RA = 36.67 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.

Components of P

Pv = 15·sin 40° = 9.642 kN
Ph = 15·cos 40° = 11.49 kN

Reactions

ΣMA = 0: RB·8 = 9.642·4 + 40·2 → RB = 14.82 kN
Ay = 9.642 + 40 − 14.82 = 34.82 kN
Ax = Ph = 11.49 kN

Resultant at A

RA = √(11.49² + 34.82²) = 36.67 kN

Answer

RA = 36.67 kN
3.Shear force & bending moment

Answer: Mmax = 126 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

ΣMA = 0: RB·10 = 45·3 + 35·7
RB = 380/10 = 38 kN
ΣFy = 0: RA = 80 − 38 = 42 kN

Moment under each load

Keep the part left of each load.

under P1: M = RA·3 = 42·3 = 126 kN·m
under P2: M = RB·3 = 38·3 = 114 kN·m

Largest value

Mmax = 126 kN·m (under P1)

Answer

Mmax = 126 kN·m
4.Shear force & bending moment

Answer: F (x = 10 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 = 46.25 kN, R2 = 68.75 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 = −37.5, MD = −35, ME = −52.5, MF = −90 kN·m, MG = 0

Answer

Mark F: MF = −90 kN·m
5.Truss: method of joints

Answer: FAB = −12.6 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 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·8 = 50·7 → Cy = 43.75 kN
ΣFy = 0: Ay = 50 − 43.75 = 6.25 kN

Joint A

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

sin α = 4/8.062 = 0.4961
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −6.25/0.4961 = −12.6 kN

Answer

FAB = −12.6 kN (compression)
6.Truss: method of joints

Answer: FAB = −35.36 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.

ABBCACACB40 kN15 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·5 = 40·3 − 15·3 → Cy = 15 kN
ΣFy = 0: Ay = 40 − 15 = 25 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 = 3/4.243 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −25/0.7071 = −35.36 kN

Answer

FAB = −35.36 kN (compression)
7.Beam deflection

Answer: δ = 7.937 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 = 10 N/mm, L = 2000 mm
EI = 2.52 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 10·2000⁴/(8·2.52 × 10¹²) = 7.937 mm

Answer

δ = 7.937 mm
8.Beam deflection

Answer: Ireq = 5.753 × 10⁶ mm⁴

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).

Deflection limit

δallow = L/360 = 1500/360 = 4.167 mm

Required second moment of area

I = 5wL⁴/(384·E·δallow) = 5·4·1500⁴/(384·11000·4.167) = 5.753 × 10⁶ mm⁴
Ireq = 5.753 × 10⁶ mm⁴

Answer

Ireq = 5.753 × 10⁶ mm⁴
9.Column buckling

Answer: Pcr = 32.75 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 = 40·20³/12 = 26667 mm⁴

Effective length

KL = 0.5·1500 = 750 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·26667/750² = 32752 N = 32.75 kN

Answer

Pcr = 32.75 kN
10.Column buckling

Answer: d = 87.96 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 = 2.5·145000 = 362500 N
KL = 2·2000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·362500·4000²/(π³·200000))^(1/4) = 87.96 mm

Answer

d = 87.96 mm

Check: the slenderness KL/r is 181.9, so the column is slender and Euler applies.

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