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​ = 20 kNP2​ = 30 kNABRA​RB​ = ?1 m1 m3 m

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

L = 5 mP1 = 20 kN at 1 mP2 = 30 kN at 2 m
RB = kN
2.Beam support reactionsHard
w0​ = 6 kN/mP = 20 kNABMA​ = ?3 m3 m

The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is 6 kN/m at the wall and falls to zero at the free end B, plus a point load P at 3 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 6 mw0 = 6 kN/mP = 20 kN at 3 m
MA = kN·m
3.Shear force & bending momentMedium
w = 6 kN/mABsection1.5 m2.5 m

Find the bending moment Mx at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mw = 6 kN/m from 0 to 4 mx = 1.5 m
Mx = kN·m
4.Shear force & bending momentHard
w = 6 kN/mABC5 m2 m

The beam rests on a pin at A and a roller at B and overhangs B by 2 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the bending moment MB at support B. Use sagging moments positive and upward shear on the left face positive.

L = 7 mw = 6 kN/m from 0 to 7 msupports at x = 0 and x = 5 m
MB = kN·m
5.Truss: method of jointsMedium
ABBCACACB60 kN5 m1 m2 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 = 2 mB is 1 m from Avertical load = 60 kN
FAB = kN
6.Truss: method of jointsHard
ABBCACACB30 kN25 kN5 m3 m3 m

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

span = 5 mheight = 3 mB is 3 m from Avertical load = 30 kNhorizontal load = 25 kN ←
FAC = kN
7.Beam deflectionMedium
w = 15 kN/mABθ4 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 = 4 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴
θA = rad
8.Beam deflectionHard
w = 3 kN/mABδ3 m

A simply supported timber beam of rectangular section, 75 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth hreq of the section.

w = 3 kN/mL = 3 mE = 11 GPa (timber)b = 75 mm
hreq = mm
9.Column bucklingMedium
PL = 3 m30 × 40 mmcross-section

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

L = 3 mE = 200 GPa (steel)b × h = 30 × 40 mm
Pcr = kN
10.Column bucklingHard
P = 115 kNL = 3 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 115 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 115 kNn = 2.5L = 3 mE = 200 GPa (steel)
d = mm

3Solutions

1.Beam support reactions

Answer: RB = 16 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​ = 20 kNP2​ = 30 kNABRA​RB​1 m1 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.

ΣMA = 0: RB·5 = 20·1 + 30·2
RB = 80/5 = 16 kN
ΣFy = 0: RA = 50 − 16 = 34 kN

Answer

RB = 16 kN
2.Beam support reactions

Answer: MA = 96 kN·m

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 triangular load

W = ½·6·6 = 18 kN
x̄ = L/3 = 2 m

Moment about the wall

MA = W·x̄ + P·a = 18·2 + 20·3 = 96 kN·m

Answer

MA = 96 kN·m
3.Shear force & bending moment

Answer: Mx = 11.25 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 = 6·4 = 24 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 6·4·2
RB = 48/4 = 12 kN
ΣFy = 0: RA = 24 − 12 = 12 kN
w = 6 kN/mABRA​ = 12RB​ = 12section1.5 m2.5 m

Cut at x = 1.5 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 = 12·1.5 − 6·1.5·0.75 = 11.25 kN·m

Answer

Mx = 11.25 kN·m
4.Shear force & bending moment

Answer: MB = −12 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 6·7 = 42 kN, acting at x̄ = 3.5 m
ΣMA = 0: RB·5 = 6·7·3.5
RB = 147/5 = 29.4 kN
ΣFy = 0: RA = 42 − 29.4 = 12.6 kN

Cut at B and keep the overhang

The load on the overhang, w·a, acts at a/2 from B and makes the beam hog over the support.

MB = −w·a²/2 = −6·2²/2 = −12 kN·m

Answer

MB = −12 kN·m
5.Truss: method of joints

Answer: FAB = −53.67 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 kNAy​Cy​

Support reactions

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

ΣMA = 0: Cy·5 = 60·1 → Cy = 12 kN
ΣFy = 0: Ay = 60 − 12 = 48 kN

Joint A

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

sin α = 2/2.236 = 0.8944
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −48/0.8944 = −53.67 kN

Answer

FAB = −53.67 kN (compression)
6.Truss: method of joints

Answer: FAC = 2 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.

ABBCACACB30 kN25 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·5 = 30·3 − 25·3 → Cy = 3 kN
ΣFy = 0: Ay = 30 − 3 = 27 kN
ΣFx = 0: Ax = 25 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 = 3/3.606 = 0.8321, cos γ = 2/3.606 = 0.5547
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −3/0.8321 = −3.606 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−3.606)·0.5547 = 2 kN

Answer

FAC = 2 kN (tension)
7.Beam deflection

Answer: θA = 0.04329 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 = 4000 mm
EI = 9.24 × 10¹¹ N·mm²

End slope

θA = wL³/(24EI) = 15·4000³/(24·9.24 × 10¹¹) = 0.04329 rad

Answer

θA = 0.04329 rad

That is 2.48°.

8.Beam deflection

Answer: hreq = 156.5 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 = 3 N/mm, L = 3000 mm, E = 11000 N/mm²

Deflection limit

δallow = L/250 = 3000/250 = 12 mm

Required second moment of area

Ireq = 5wL⁴/(384·E·δallow) = 5·3·3000⁴/(384·11000·12) = 2.397 × 10⁷ mm⁴

Required depth

bh³/12 = Ireq → h = (12·2.397 × 10⁷/75)1/3 = 156.5 mm

Answer

hreq = 156.5 mm

In practice, choose the next standard depth above this value.

9.Column buckling

Answer: Pcr = 40.28 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 = 40·30³/12 = 90000 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·90000/2100² = 40284 N = 40.28 kN

Answer

Pcr = 40.28 kN
10.Column buckling

Answer: d = 60.14 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 = 2.5·115000 = 287500 N
KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·287500·2100²/(π³·200000))^(1/4) = 60.14 mm

Answer

d = 60.14 mm

Check: the slenderness KL/r is 139.7, so the column is slender and Euler applies.

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