Structural Analysis: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 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 R_A, taking upward as positive.

AB = 3 mAC = 5 mP₁ = 65 kN at 1 mP₂ = 45 kN at 5 m

R_A = kN

2.Beam support reactionsHard

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

L = 5 mP = 30 kN at 2.5 mw = 8 kN/m from 3 to 4 m

R_B = kN

3.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 60 kN·m at x = 5 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 6 mM₀ = 60 kN·m clockwise, at 5 m

M_x = kN·m

4.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment M_max in the beam.

L = 6 mw = 4 kN/m from 0 to 2 m

M_max = kN·m

5.Truss: method of jointsMedium

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

F: 25 kN ↓G: 15 kN ↓H: 10 kN ↓

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

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 = 8 mheight = 4 mB is 1 m from Avertical load = 30 kNhorizontal load = 25 kN →

F_AB = kN

7.Beam deflectionMedium

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

w = 20 kN/mL = 3 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴

θ_A = rad

8.Beam deflectionHard

A simply supported timber beam of rectangular section, 125 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth h_req of the section.

w = 3 kN/mL = 6 mE = 11 GPa (timber)b = 125 mm

h_req = mm

9.Column bucklingMedium

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 P_cr.

L = 1 mE = 200 GPa (steel)b × h = 20 × 100 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel 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 = 3 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

3Solutions

1.Beam support reactions

Answer: R_A = 13.33 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.

Moments about A

Counter-clockwise positive.

ΣM_A = 0: R_B·3 = 65·1 + 45·5

R_B = 290/3 = 96.67 kN

Vertical equilibrium

R_A = 65 + 45 − 96.67 = 13.33 kN

Answer

R_A = 13.33 kN

2.Beam support reactions

Answer: R_B = 20.6 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.

UDL resultant: W = w·ℓ = 8·1 = 8 kN, acting at x̄ = 3.5 m

ΣM_A = 0: R_B·5 = 30·2.5 + 8·1·3.5

R_B = 103/5 = 20.6 kN

ΣF_y = 0: R_A = 38 − 20.6 = 17.4 kN

Answer

R_B = 20.6 kN

3.Shear force & bending moment

Answer: M_x = −50 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 = 60/6 = 10 kN (upward)

R_A = −R_B = −10 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −10·5 = −50 kN·m

Answer

M_x = −50 kN·m

4.Shear force & bending moment

Answer: M_max = 5.556 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.

UDL resultant: W = w·ℓ = 4·2 = 8 kN, acting at x̄ = 1 m

ΣM_A = 0: R_B·6 = 4·2·1

R_B = 8/6 = 1.333 kN

ΣF_y = 0: R_A = 8 − 1.333 = 6.667 kN

Zero shear

The shear force starts at R_A, stays constant over any unloaded part and falls with slope w inside the load.

R_A − w·(x₀ − 0) = 0

x₀ = 0 + 6.667/4 = 1.667 m

Maximum moment

M_max = R_A·x₀ − w·(x₀ − 0)²/2 = 6.667·1.667 − 4·1.667²/2 = 5.556 kN·m

Answer

M_max = 5.556 kN·m

5.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 C

BC and CD 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: 38.33 kN (T), BC: 43.33 kN (T), CD: 43.33 kN (T), DE: 28.33 kN (T), FG: 38.33 kN (C), GH: 28.33 kN (C), AF: 47.92 kN (C), HE: 35.42 kN (C), BF: 3.75 kN (T), CG: 0, DH: 11.25 kN (T), BG: 6.25 kN (C), DG: 18.75 kN (C)

Answer

Zero-force member: CG

6.Truss: method of joints

Answer: F_AB = −14.17 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·8 = 30·1 + 25·4 → C_y = 16.25 kN

ΣF_y = 0: A_y = 30 − 16.25 = 13.75 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 = 4/4.123 = 0.9701

ΣF_y = 0: A_y + F_AB·sin α = 0

F_AB = −13.75/0.9701 = −14.17 kN

Answer

F_AB = −14.17 kN (compression)

7.Beam deflection

Answer: θ_A = 0.004891 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 = 20 N/mm, L = 3000 mm

EI = 4.6 × 10¹² N·mm²

End slope

θ_A = wL³/(24EI) = 20·3000³/(24·4.6 × 10¹²) = 0.004891 rad

Answer

θ_A = 0.004891 rad

That is 0.2803°.

8.Beam deflection

Answer: h_req = 264 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.

w = 3 N/mm, L = 6000 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/250 = 6000/250 = 24 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·3·6000⁴/(384·11000·24) = 1.918 × 10⁸ mm⁴

Required depth

bh³/12 = I_req → h = (12·1.918 × 10⁸/125)^1/3 = 264 mm

Answer

h_req = 264 mm

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

9.Column buckling

Answer: P_cr = 268.6 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 = 100·20³/12 = 66667 mm⁴

Effective length

KL = 0.7·1000 = 700 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·66667/700² = 268561 N = 268.6 kN

Answer

P_cr = 268.6 kN

10.Column buckling

Answer: σ_cr = 5.483 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 2·3000 = 6000 mm

KL/r = 6000/10 = 600

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/600² = 5.483 MPa

Answer

σ_cr = 5.483 MPa

ExamProblems · examproblems.com/courses/structural-analysis