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​ = 35 kNP2​ = 10 kNABCRB​ = ?2 m1 m2 m

The beam has a pin at A and a roller at B, 3 m from A, and overhangs B to the free end C. Find the vertical reaction RB, taking upward as positive.

AB = 3 mAC = 5 mP1 = 35 kN at 2 mP2 = 10 kN at 5 m
RB = kN
2.Beam support reactionsHard
P = 10 kNADCMA​ = ?6 m1 m3 m

A compound beam is fixed to the wall at A, has an internal hinge at C (6 m from A) and rests on a roller at D (10 m from A). It carries a point load P at 7 m from A. Find the magnitude of the reaction moment MA at the wall.

AC = 6 mAD = 10 mP = 10 kN at 7 m
MA = kN·m
3.Shear force & bending momentMedium
M0​ = 10 kN·mAB4 m4 m

The simply supported beam carries a clockwise couple M0 = 10 kN·m at x = 4 m from A, and no other load. Find the bending moment Mx just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 8 mM0 = 10 kN·m clockwise, at 4 m
Mx = kN·m
4.Shear force & bending momentHard
20 kN45 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.

20 kN @ 0 m45 kN @ 6 m35 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
ABBCACACB45 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 = 45 kN
FAB = kN
6.Truss: method of jointsHard
ABBCACACB60 kN15 kN4 m3 m4 m

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

span = 4 mheight = 4 mB is 3 m from Avertical load = 60 kNhorizontal load = 15 kN ←
FAB = kN
7.Beam deflectionMedium
w = 15 kN/mABθ2.5 m

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θA of the beam at support A, in radians.

w = 15 kN/mL = 2.5 mE = 70 GPa (aluminium)I = 8.5 × 10⁶ mm⁴
θA = rad
8.Beam deflectionHard
P = 24 kNABδ2.5 m1.5 m

A simply supported beam carries a point load P at b = 1.5 m from support B. Find the deflection δmid at midspan.

P = 24 kNb = 1.5 mL = 4 mE = 200 GPa (steel)I = 120 × 10⁶ mm⁴
δmid = mm
9.Column bucklingMedium
PL = 4 m20 × 80 mmcross-section

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

L = 4 mE = 200 GPa (steel)b × h = 20 × 80 mm
Pcr = kN
10.Column bucklingHard
P = 35 kNL = 1 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 35 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 35 kNn = 3L = 1 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 40 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·3 = 35·2 + 10·5
RB = 120/3 = 40 kN

Vertical equilibrium

RA = 35 + 10 − 40 = 5 kN

Answer

RB = 40 kN
2.Beam support reactions

Answer: MA = 45 kN·m

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·4 = 10·1
RD = 2.5 kN
VC = 10 − 2.5 = 7.5 kN

Part AC

The hinge force VC acts downward on AC (equal and opposite to its action on CD).

RA = 7.5 = 7.5 kN
MA = 7.5·6 = 45 kN·m

Answer

MA = 45 kN·m
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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.

RB = M0/L = 10/8 = 1.25 kN (upward)
RA = −RB = −1.25 kN (downward)

Cut just right of the couple

The left part now contains RA and the couple.

Mx = RA·a + M0 = −1.25·4 + 10 = 5 kN·m

Answer

Mx = 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 = 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 = −20, MC = −31.67, MD = −43.33, ME = −90 kN·m, MF = −45, MG = 0

Answer

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

Answer: FAB = −21.21 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 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·6 = 45·4 → Cy = 30 kN
ΣFy = 0: Ay = 45 − 30 = 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 = 4/5.657 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −15/0.7071 = −21.21 kN

Answer

FAB = −21.21 kN (compression)
6.Truss: method of joints

Answer: FAB = −37.5 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.

ABBCACACB60 kN15 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·4 = 60·3 − 15·4 → Cy = 30 kN
ΣFy = 0: Ay = 60 − 30 = 30 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 = 4/5 = 0.8
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −30/0.8 = −37.5 kN

Answer

FAB = −37.5 kN (compression)
7.Beam deflection

Answer: θA = 0.01641 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 15 N/mm, L = 2500 mm
EI = 5.95 × 10¹¹ N·mm²

End slope

θA = wL³/(24EI) = 15·2500³/(24·5.95 × 10¹¹) = 0.01641 rad

Answer

θA = 0.01641 rad

That is 0.9404°.

8.Beam deflection

Answer: δmid = 1.219 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 = 24000 N, b = 1500 mm, L = 4000 mm
EI = 2.4 × 10¹³ N·mm²

Midspan deflection

δmid = Pb(3L² − 4b²)/(48EI) = 24000·1500·(3·4000² − 4·1500²)/(48·2.4 × 10¹³) = 1.219 mm

Answer

δmid = 1.219 mm
9.Column buckling

Answer: Pcr = 6.58 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 = 80·20³/12 = 53333 mm⁴

Effective length

KL = 1·4000 = 4000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·53333/4000² = 6580 N = 6.58 kN

Answer

Pcr = 6.58 kN
10.Column buckling

Answer: d = 26.99 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 = 3·35000 = 105000 N
KL = 0.7·1000 = 700 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·105000·700²/(π³·200000))^(1/4) = 26.99 mm

Answer

d = 26.99 mm

Check: the slenderness KL/r is 103.7, so the column is slender and Euler applies.

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