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​ = 45 kNP2​ = 45 kNABRA​RB​ = ?2.5 m1.5 m1 m

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

L = 5 mP1 = 45 kN at 2.5 mP2 = 45 kN at 4 m
RB = kN
2.Beam support reactionsHard
P = 80 kNQ = 65 kNADCRD​ = ?1 m2 m4 m1 m

A compound beam is fixed to the wall at A, has an internal hinge at C (3 m from A) and rests on a roller at D (8 m from A). It carries a point load P at 7 m from A and a point load Q at 1 m from A. Find the reaction RD at the roller.

AC = 3 mAD = 8 mP = 80 kN at 7 mQ = 65 kN at 1 m
RD = kN
3.Shear force & bending momentMedium
w = 5 kN/mABsection1 m7 m

Find the bending moment Mx at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mw = 5 kN/m from 0 to 8 mx = 1 m
Mx = kN·m
4.Shear force & bending momentHard
45 kN45 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.

45 kN @ 6 m45 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)
5.Truss: method of jointsMedium
ABBCACACB40 kN6 m4 m4 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 = 4 mB is 4 m from Avertical load = 40 kN
FAB = kN
6.Truss: method of jointsHard
ABBCACACB35 kN25 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 = 35 kNhorizontal load = 25 kN →
FAB = kN
7.Beam deflectionMedium
P = 18 kNABδ3 m5 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 = 18 kNa = 3 mL = 8 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴
δP = mm
8.Beam deflectionHard
w = 8 kN/mAB6 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 = 6 mw = 8 kN/m
RB = kN
9.Column bucklingMedium
PL = 5 md = 60 mmcross-section

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

L = 5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
P = 25 kNL = 5 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 25 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 25 kNn = 3L = 5 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 58.5 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​ = 45 kNP2​ = 45 kNABRA​RB​2.5 m1.5 m1 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·5 = 45·2.5 + 45·4
RB = 292.5/5 = 58.5 kN
ΣFy = 0: RA = 90 − 58.5 = 31.5 kN

Answer

RB = 58.5 kN
2.Beam support reactions

Answer: RD = 64 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. An internal hinge cannot carry a bending moment, which gives one extra equation: the moment about the hinge of the forces on either side is zero.

Part CD

The hinge has no moment, so moments about C on CD alone must balance.

ΣMC = 0: RD·5 = 80·4
RD = 64 kN
VC = 80 − 64 = 16 kN

Answer

RD = 64 kN
3.Shear force & bending moment

Answer: Mx = 17.5 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 = 5·8 = 40 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 5·8·4
RB = 160/8 = 20 kN
ΣFy = 0: RA = 40 − 20 = 20 kN
w = 5 kN/mABRA​ = 20RB​ = 20section1 m7 m

Cut at x = 1 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 = 20·1 − 5·1·0.5 = 17.5 kN·m

Answer

Mx = 17.5 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 = 11.25 kN, R2 = 78.75 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 = 11.25, MC = −22.5, MD = −56.25, ME = −90 kN·m, MF = −45, MG = 0

Answer

Mark E: ME = −90 kN·m
5.Truss: method of joints

Answer: FAB = −18.86 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 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·6 = 40·4 → Cy = 26.67 kN
ΣFy = 0: Ay = 40 − 26.67 = 13.33 kN

Joint A

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

sin α = 4/5.657 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −13.33/0.7071 = −18.86 kN

Answer

FAB = −18.86 kN (compression)
6.Truss: method of joints

Answer: FAB = 1.414 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.

ABBCACACB35 kN25 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·5 = 35·3 + 25·3 → Cy = 36 kN
ΣFy = 0: Ay = 35 − 36 = −1 kN
ΣFx = 0: Ax = −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 = 3/4.243 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −−1/0.7071 = 1.414 kN

Answer

FAB = 1.414 kN (tension)
7.Beam deflection

Answer: δP = 20.09 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 = 18000 N, a = 3000 mm, b = 5000 mm, L = 8000 mm
EI = 8.4 × 10¹² N·mm²

Deflection under the load

δP = Pa²b²/(3EIL) = 18000·3000²·5000²/(3·8.4 × 10¹²·8000) = 20.09 mm

Answer

δP = 20.09 mm
8.Beam deflection

Answer: RB = 18 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 = 3·8·6/8 = 18 kN

Answer

RB = 18 kN
9.Column buckling

Answer: Pcr = 35.88 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·636173/3500² = 35879 N = 35.88 kN

Answer

Pcr = 35.88 kN
10.Column buckling

Answer: d = 55.49 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 = 3·25000 = 75000 N
KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·75000·3500²/(π³·200000))^(1/4) = 55.49 mm

Answer

d = 55.49 mm

Check: the slenderness KL/r is 252.3, so the column is slender and Euler applies.

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