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

L = 10 mP₁ = 40 kN at 1 mP₂ = 25 kN at 3 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 8 kN/m at the free end B, plus a point load P at 2.5 m from the wall. Find the magnitude of the reaction moment M_A at the wall.

L = 4 mw₀ = 8 kN/mP = 70 kN at 2.5 m

M_A = kN·m

3.Shear force & bending momentMedium

The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 5 m from A. Find the bending moment M_x at the section x = 2 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 10 kN at 5 mw = 6 kN/m from 0 to 5 mx = 2 m

M_x = kN·m

4.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 2.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment M_max between the supports.

L = 7.5 mw = 10 kN/m from 0 to 7.5 msupports at x = 0 and x = 5 m

M_max = 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 = 6 mheight = 3 mB is 2 m from Avertical load = 45 kN

F_AB = kN

6.Truss: method of jointsHard

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

span = 5 mheight = 2 mB is 2 m from Avertical load = 40 kNhorizontal load = 25 kN →

F_AB = kN

7.Beam deflectionMedium

A simply supported beam carries a point load P at a = 2 m from support A. Find the deflection δ_P under the load.

P = 4 kNa = 2 mL = 6 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴

δ_P = 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 pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 4 mE = 70 GPa (aluminium)b × h = 25 × 40 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 4 mE = 70 GPa (aluminium)d = 60 mm

σ_cr = MPa

3Solutions

1.Beam support reactions

Answer: R_A = 53.5 kN

Static equilibrium: a body at rest has ΣF_x = 0, ΣF_y = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.

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.

ΣM_A = 0: R_B·10 = 40·1 + 25·3

R_B = 115/10 = 11.5 kN

ΣF_y = 0: R_A = 65 − 11.5 = 53.5 kN

Answer

R_A = 53.5 kN

2.Beam support reactions

Answer: M_A = 217.7 kN·m

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 triangular load

W = ½·8·4 = 16 kN

x̄ = 2L/3 = 2.667 m

Moment about the wall

M_A = W·x̄ + P·a = 16·2.667 + 70·2.5 = 217.7 kN·m

Answer

M_A = 217.7 kN·m

3.Shear force & bending moment

Answer: M_x = −57 kN·m

Keep the free-end part

P: −10·3 = −30 kN·m

w: −6·3²/2 = −27 kN·m

M_x = −30 − 27 = −57 kN·m

Answer

M_x = −57 kN·m

Negative means hogging: tension on the top face.

4.Shear force & bending moment

Answer: M_max = 17.58 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 = 10·7.5 = 75 kN, acting at x̄ = 3.75 m

ΣM_A = 0: R_B·5 = 10·7.5·3.75

R_B = 281.3/5 = 56.25 kN

ΣF_y = 0: R_A = 75 − 56.25 = 18.75 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 18.75/10 = 1.875 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 18.75²/(2·10) = 17.58 kN·m

Answer

M_max = 17.58 kN·m

5.Truss: method of joints

Answer: F_AB = −36.06 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·6 = 45·2 → C_y = 15 kN

ΣF_y = 0: A_y = 45 − 15 = 30 kN

Joint A

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

sin α = 3/3.606 = 0.8321

ΣF_y = 0: A_y + F_AB·sin α = 0

F_AB = −30/0.8321 = −36.06 kN

Answer

F_AB = −36.06 kN (compression)

6.Truss: method of joints

Answer: F_AB = −19.8 kN

Support reactions

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

ΣM_A = 0: C_y·5 = 40·2 + 25·2 → C_y = 26 kN

ΣF_y = 0: A_y = 40 − 26 = 14 kN

ΣF_x = 0: A_x = −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 = 2/2.828 = 0.7071

ΣF_y = 0: A_y + F_AB·sin α = 0

F_AB = −14/0.7071 = −19.8 kN

Answer

F_AB = −19.8 kN (compression)

7.Beam deflection

Answer: δ_P = 1.693 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 = 4000 N, a = 2000 mm, b = 4000 mm, L = 6000 mm

EI = 8.4 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 4000·2000²·4000²/(3·8.4 × 10¹²·6000) = 1.693 mm

Answer

δ_P = 1.693 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 = 2.249 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 = 40·25³/12 = 52083 mm⁴

Effective length

KL = 1·4000 = 4000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·52083/4000² = 2249 N = 2.249 kN

Answer

P_cr = 2.249 kN

10.Column buckling

Answer: σ_cr = 2.429 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²

r = √(I/A) = √(636173/2827) = 15 mm

KL = 2·4000 = 8000 mm

KL/r = 8000/15 = 533.3

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/533.3² = 2.429 MPa

Answer

σ_cr = 2.429 MPa

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