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: ΣF_x = 0, ΣF_y = 0, ΣM = 0
Uniform load: resultant w·L at the middle of the load
Moment: M = F·d

Method

Draw a free-body diagram and replace each support by its reactions.
Replace every distributed load by a single force at its centroid.
Take moments about one support to find the reaction at the other.
Use ΣF_y = 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/y_max
Simply supported, uniform load: M_max = wL²/8

Method

Find the support reactions from equilibrium of the whole beam.
Cut the beam at the section you need and keep one side.
Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
For the maximum moment, find where the shear force is zero.

Truss: method of joints

Method of joints: ΣF_x = 0, ΣF_y = 0 at each joint
Method of sections: ΣM = 0 about a cut point
Member force: positive = tension, negative = compression

Method

Find the support reactions for the whole truss.
Method of joints: start at a joint with at most two unknown member forces and use ΣF_x = 0 and ΣF_y = 0.
Method of sections: cut through the member you need and take moments about a point where the other unknown members meet.
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: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

2Practice problems

1.Beam support reactionsMedium

The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w from the wall to 2.5 m and a point load P at 1.5 m from the wall. Find the vertical reaction R_A at the wall.

L = 4 mP = 10 kN at 1.5 mw = 8 kN/m from 0 to 2.5 m

R_A = kN

2.Beam support reactionsHard

The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is zero at the wall and rises to 10 kN/m at the free end B, plus a point load P at 2 m from the wall. Find the magnitude of the reaction moment M_A at the wall.

L = 4 mw₀ = 10 kN/mP = 35 kN at 2 m

M_A = kN·m

3.Shear force & bending momentMedium

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.

25 kN @ 2 m55 kN @ 8 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)

4.Shear force & bending momentHard

Find the bending moment M_x at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP = 40 kN at 2 mw = 10 kN/m from 0 to 10 mx = 2 m

M_x = kN·m

5.Truss: method of jointsMedium

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

span = 4 mheight = 2 mB is 3 m from Avertical load = 60 kN

F_AB = kN

6.Truss: method of jointsHard

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

span = 5 mheight = 3 mB is 1 m from Avertical load = 35 kNhorizontal load = 20 kN ←

F_BC = kN

7.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δ_max, at midspan.

w = 2 kN/mL = 8 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴

δ_max = mm

8.Beam deflectionHard

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

w = 8 kN/mL = 8 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

9.Column bucklingMedium

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 P_cr.

L = 5 mE = 70 GPa (aluminium)b × h = 30 × 50 mm

P_cr = kN

10.Column bucklingHard

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 σ_cr from Euler’s formula.

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

σ_cr = MPa

3Solutions

1.Beam support reactions

Answer: R_A = 30 kN

Cantilever equilibrium: a fixed support can push up and also resist rotation, so it gives a vertical reaction R_A and a reaction moment M_A. ΣF_y = 0 gives R_A, and moments about the wall give M_A = Σ(load × distance from the wall). A distributed load acts as its resultant at the centroid of the load diagram.

Resultant of the distributed load

W = 8·2.5 = 20 kN, x̄ = 1.25 m

Vertical equilibrium

R_A = P + W = 10 + 20 = 30 kN

Answer

R_A = 30 kN

2.Beam support reactions

Answer: M_A = 123.3 kN·m

Resultant of the triangular load

W = ½·10·4 = 20 kN

x̄ = 2L/3 = 2.667 m

Moment about the wall

M_A = W·x̄ + P·a = 20·2.667 + 35·2 = 123.3 kN·m

Answer

M_A = 123.3 kN·m

3.Shear force & bending moment

Answer: E (x = 8 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 = 39.17 kN, R₂ = 40.83 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

M_A = 0, M_B = 78.33, M_C = 106.7, M_D = 135, M_E = 163.3 kN·m, M_F = 81.67, M_G = 0

Answer

Mark E: M_E = 163.3 kN·m

4.Shear force & bending moment

Answer: M_x = 144 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 = 10·10 = 100 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 40·2 + 10·10·5

R_B = 580/10 = 58 kN

ΣF_y = 0: R_A = 140 − 58 = 82 kN

Cut at x = 2 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 = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 82·2 − 10·2·1 = 144 kN·m

Answer

M_x = 144 kN·m

5.Truss: method of joints

Answer: F_AB = −27.04 kN

Method of joints: every pin of a truss is in equilibrium, so ΣF_x = 0 and ΣF_y = 0 at each joint. Start at a joint with at most two unknown member forces. A negative result means the member is in compression.

Support reactions

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

ΣM_A = 0: C_y·4 = 60·3 → C_y = 45 kN

ΣF_y = 0: A_y = 60 − 45 = 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 = 2/3.606 = 0.5547

ΣF_y = 0: A_y + F_AB·sin α = 0

F_AB = −15/0.5547 = −27.04 kN

Answer

F_AB = −27.04 kN (compression)

6.Truss: method of joints

Answer: F_BC = 8.333 kN

Support reactions

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

ΣM_A = 0: C_y·5 = 35·1 − 20·3 → C_y = −5 kN

ΣF_y = 0: A_y = 35 − −5 = 40 kN

ΣF_x = 0: A_x = 20 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 = 3/5 = 0.6, cos γ = 4/5 = 0.8

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −−5/0.6 = 8.333 kN

Answer

F_BC = 8.333 kN (tension)

7.Beam deflection

Answer: δ_max = 12.7 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 = 2 N/mm, L = 8000 mm

EI = 8.4 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·2·8000⁴/(384·8.4 × 10¹²) = 12.7 mm

Answer

δ_max = 12.7 mm

8.Beam deflection

Answer: I_req = 96 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 8000/360 = 22.22 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·8·8000⁴/(384·200000·22.22) = 9.6 × 10⁷ mm⁴

I_req = 96 × 10⁶ mm⁴

Answer

I_req = 96 × 10⁶ mm⁴

9.Column buckling

Answer: P_cr = 6.345 kN

Euler buckling: a slender column becomes unstable at P_cr = 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.

I_min = h·b³/12 = 50·30³/12 = 112500 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·112500/3500² = 6345 N = 6.345 kN

Answer

P_cr = 6.345 kN

10.Column buckling

Answer: σ_cr = 6.294 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²

r = √(I/A) = √(19175/490.9) = 6.25 mm

KL = 0.7·5000 = 3500 mm

KL/r = 3500/6.25 = 560

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/560² = 6.294 MPa

Answer

σ_cr = 6.294 MPa

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