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 = 8 mP₁ = 25 kN at 2 mP₂ = 60 kN at 4 m

R_A = kN

2.Beam support reactionsHard

The beam has a pin at A and a roller at B, 5 m from A, and overhangs B to the free end C. It carries a point load P at 3 m from A and a uniformly distributed load w from 5 m to the free end at 8 m. Find the vertical reaction R_A, taking upward as positive.

AB = 5 mAC = 8 mP = 55 kN at 3 mw = 4 kN/m from 5 to 8 m

R_A = kN

3.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 35 kN·m at x = 1 m from A, and no other load. Find the bending moment M_x just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 6 mM₀ = 35 kN·m clockwise, at 1 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 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the bending moment M_B at support B. Use sagging moments positive and upward shear on the left face positive.

L = 6 mw = 2 kN/m from 0 to 6 msupports at x = 0 and x = 4 m

M_B = kN·m

5.Truss: method of jointsMedium

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

span = 6 mheight = 3 mB is 5 m from Avertical load = 10 kN

F_AC = kN

6.Truss: method of jointsHard

Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.

B: 25 kN ↓D: 40 kN ↓F: 20 kN →

1member AB
2member BC
3member CD
4member DE
5member FG
6member GH
7member AF
8member HE
9member BF
10member CG
11member DH
12member FC
13member HC

7.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 0.5 m from A. Find the deflection δ_B at the free end B.

P = 9 kNa = 0.5 mL = 2.5 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ_B = mm

8.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 5 kNw = 8 kN/mL = 3 mE = 11 GPa (timber)I = 36 × 10⁶ mm⁴

δ_max = mm

9.Column bucklingMedium

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

L = 1 mE = 70 GPa (aluminium)b × h = 20 × 60 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 1 mE = 70 GPa (aluminium)d = 20 mm

σ_cr = MPa

3Solutions

1.Beam support reactions

Answer: R_A = 48.75 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·8 = 25·2 + 60·4

R_B = 290/8 = 36.25 kN

ΣF_y = 0: R_A = 85 − 36.25 = 48.75 kN

Answer

R_A = 48.75 kN

2.Beam support reactions

Answer: R_A = 18.4 kN

Static equilibrium: a body at rest has ΣF_x = 0, ΣF_y = 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.

Resultant of the distributed load

W = 4·3 = 12 kN, x̄ = (5 + 8)/2 = 6.5 m

Equilibrium

Moments about A, counter-clockwise positive. Then vertical force balance.

ΣM_A = 0: R_B·5 = 55·3 + 12·6.5

R_B = 243/5 = 48.6 kN

R_A = 67 − 48.6 = 18.4 kN

Answer

R_A = 18.4 kN

3.Shear force & bending moment

Answer: M_x = 29.17 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: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 35/6 = 5.833 kN (upward)

R_A = −R_B = −5.833 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −5.833·1 + 35 = 29.17 kN·m

Answer

M_x = 29.17 kN·m

4.Shear force & bending moment

Answer: M_B = −4 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 = 2·6 = 12 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·4 = 2·6·3

R_B = 36/4 = 9 kN

ΣF_y = 0: R_A = 12 − 9 = 3 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.

M_B = −w·a²/2 = −2·2²/2 = −4 kN·m

Answer

M_B = −4 kN·m

5.Truss: method of joints

Answer: F_AC = 2.778 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 = 10·5 → C_y = 8.333 kN

ΣF_y = 0: A_y = 10 − 8.333 = 1.667 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 = 3/3.162 = 0.9487, cos γ = 1/3.162 = 0.3162

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −8.333/0.9487 = −8.784 kN

ΣF_x = 0: −F_AC − F_BC·cos γ = 0 → F_AC = −(−8.784)·0.3162 = 2.778 kN

Answer

F_AC = 2.778 kN (tension)

6.Truss: method of joints

Answer: member CG

Zero-force members: at a joint, the forces must balance in every direction. If a joint has no load and no support, and two of its three members are in line, then the balance across that line involves only the third member, so its force is zero. If only two members meet at an unloaded joint and they are not in line, both are zero.

Joint G

FG and GH lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.

ΣF⊥ = 0 → F_CG = 0

Check with the method of joints

Solving all joints gives these member forces (T = tension, C = compression).

AB: 36.56 kN (T), BC: 36.56 kN (T), CD: 32.19 kN (T), DE: 32.19 kN (T), FG: 34.37 kN (C), GH: 34.37 kN (C), AF: 27.6 kN (C), HE: 53.65 kN (C), BF: 25 kN (T), CG: 0, DH: 40 kN (T), FC: 3.646 kN (C), HC: 3.646 kN (T)

Answer

Zero-force member: CG

7.Beam deflection

Answer: δ_B = 1.989 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 = 9000 N, a = 500 mm, L = 2500 mm

EI = 1.32 × 10¹² N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δ_P = Pa³/(3EI) = 0.2841 mm

θ_P = Pa²/(2EI) = 0.0008523 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 1.989 mm

Answer

δ_B = 1.989 mm

8.Beam deflection

Answer: δ_max = 28.41 mm

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 = 5000 N, w = 8 N/mm, L = 3000 mm, EI = 3.96 × 10¹¹ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 5000·3000³/(48·3.96 × 10¹¹) = 7.102 mm

δ_w = 5wL⁴/(384EI) = 5·8·3000⁴/(384·3.96 × 10¹¹) = 21.31 mm

Add them

δ_max = 7.102 + 21.31 = 28.41 mm

Answer

δ_max = 28.41 mm

9.Column buckling

Answer: P_cr = 6.909 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 = 60·20³/12 = 40000 mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·40000/2000² = 6909 N = 6.909 kN

Answer

P_cr = 6.909 kN

10.Column buckling

Answer: σ_cr = 17.27 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 1·1000 = 1000 mm

KL/r = 1000/5 = 200

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa

Answer

σ_cr = 17.27 MPa

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