Simple Stress and Strain

Result: $E = 3K(1 - 2\mu)$ where $E$ is Young's modulus, $K$ the bulk modulus and $\mu$ Poisson's ratio.

Definitions

• Bulk modulus: $K = \dfrac{\text{volumetric (hydrostatic) stress}}{\text{volumetric strain}} = \dfrac{p}{\varepsilon_v}$
• Volumetric strain $\varepsilon_v = \dfrac{\delta V}{V}$.

Derivation

Take a cube of side $L$ subjected to equal stress $\sigma$ on all six faces (uniform hydrostatic compression/tension), so $\sigma_x = \sigma_y = \sigma_z = \sigma$.

Strain in the x-direction due to all three stresses (Hooke's law with Poisson effect):

$$\varepsilon_x = \frac{\sigma}{E} - \mu\frac{\sigma}{E} - \mu\frac{\sigma}{E} = \frac{\sigma}{E}(1 - 2\mu)$$

By symmetry $\varepsilon_y = \varepsilon_z = \varepsilon_x$.

Volume $V = L^3$; for small strains the volumetric strain is the sum of the linear strains:

$$\varepsilon_v = \varepsilon_x + \varepsilon_y + \varepsilon_z = \frac{3\sigma}{E}(1 - 2\mu)$$

(Check: $V + \delta V = L^3(1+\varepsilon)^3 \approx L^3(1 + 3\varepsilon)$.)

From the definition of bulk modulus, with $p = \sigma$:

$$K = \frac{\sigma}{\varepsilon_v} = \frac{\sigma}{\dfrac{3\sigma}{E}(1 - 2\mu)} = \frac{E}{3(1 - 2\mu)}$$ $$\boxed{E = 3K(1 - 2\mu)}$$

Notes: since $K$ must be positive, $1 - 2\mu > 0$, so $\mu < 0.5$. For an incompressible material, $\mu = 0.5$ and $K \to \infty$. For example, for steel with $\mu = 0.3$, $K = E/(3\times0.4) = 0.833E$.

Result: $E = 2G(1 + \mu)$, where $G$ is the modulus of rigidity.

Setting up

Consider a square element ABCD of side $a$ subjected to pure shear: shear stress $\tau$ on its faces. Then the diagonals experience equal tensile and compressive stresses of magnitude $\tau$ at $45^\circ$ (principal stresses $\sigma_1 = +\tau$ and $\sigma_2 = -\tau$).

   A────────────D       Shear τ on all faces
   │            │       shear strain φ = τ/G
   │    ╲       │       diagonal BD stretches
   B────────────C       diagonal AC shortens

Strain in the diagonal in terms of E and μ

The tensile principal stress along diagonal BD is $\tau$ and the compressive one along AC is $-\tau$. The strain of diagonal BD is

$$\varepsilon_{BD} = \frac{\tau}{E} - \mu\left(\frac{-\tau}{E}\right) = \frac{\tau}{E}(1 + \mu) \qquad (1)$$

Strain in the diagonal in terms of the shear strain

Under shear strain $\phi$, the face DC moves to DC' by $\delta = a\phi$ (the face CD shifts parallel to AB). The diagonal BD changes length by the component of this displacement along BD, i.e.

$$\delta_{BD} = CC'\cos45^\circ = a\phi\cos45^\circ$$

and the original diagonal length is $BD = a\sqrt{2} = a/\cos45^\circ$. Therefore

$$\varepsilon_{BD} = \frac{a\phi\cos45^\circ}{a/\cos45^\circ} = \phi\cos^245^\circ = \frac{\phi}{2} = \frac{\tau}{2G} \qquad (2)$$

(using $\phi = \tau/G$).

Equating (1) and (2)

$$\frac{\tau}{2G} = \frac{\tau}{E}(1 + \mu) \;\Rightarrow\; \boxed{E = 2G(1 + \mu)}$$

For steel with $\mu = 0.3$: $G = E/2.6 = 0.385E$.

The three elastic constants $E$, $G$, $K$ are connected with Poisson's ratio $\mu$ by:

$$E = 2G(1 + \mu),\qquad E = 3K(1 - 2\mu),\qquad E = \frac{9KG}{3K + G},\qquad \mu = \frac{3K - 2G}{2(3K + G)}$$

(a) Relation between E, G and μ

Take a square element under pure shear $\tau$. The diagonals carry principal stresses $+\tau$ (tension) and $-\tau$ (compression).

Strain of the tension diagonal from Hooke's law:

$$\varepsilon = \frac{\tau}{E} + \mu\frac{\tau}{E} = \frac{\tau}{E}(1 + \mu)$$

Geometry: the shear strain $\phi = \tau/G$ stretches the diagonal by $\varepsilon = \phi/2 = \dfrac{\tau}{2G}$.

Equating:

$$\frac{\tau}{2G} = \frac{\tau}{E}(1 + \mu) \;\Rightarrow\; E = 2G(1 + \mu)\qquad (1)$$

(b) Relation between E, K and μ

A cube under equal stress $\sigma$ on all faces has

$$\varepsilon_x = \frac{\sigma}{E}(1 - 2\mu),\qquad \varepsilon_v = 3\varepsilon_x = \frac{3\sigma}{E}(1 - 2\mu)$$

and $K = \sigma/\varepsilon_v$, so

$$K = \frac{E}{3(1 - 2\mu)} \;\Rightarrow\; E = 3K(1 - 2\mu)\qquad (2)$$

(c) Eliminating μ to relate E, G and K

From (1): $\mu = \dfrac{E}{2G} - 1$. Substitute in (2):

$$E = 3K\left[1 - 2\left(\frac{E}{2G} - 1\right)\right] = 3K\left(3 - \frac{E}{G}\right)$$ $$E\left(1 + \frac{3K}{G}\right) = 9K \;\Rightarrow\; E = \frac{9KG}{3K + G}$$

Also, equating (1) and (2): $2G(1 + \mu) = 3K(1 - 2\mu)$, so

$$\mu = \frac{3K - 2G}{2(3K + G)}$$

Example: for $E = 200$ GPa and $\mu = 0.3$: $G = \dfrac{200}{2.6} = 76.9$ GPa and $K = \dfrac{200}{3(0.4)} = 166.7$ GPa.

Consider a circular bar of length $L$ whose diameter varies uniformly from $d_1$ at one end to $d_2$ at the other, subjected to an axial pull $P$. Young's modulus is $E$.

   d1                         d2
  ┌──┐                       ┌──┐
 P│  └───────────────────────┘  │P
 →│   ╲_____________________╱    │→
  └──┘            x           └──┘
        L

Derivation

Measure $x$ from the small end (diameter $d_1$, taking $d_2 > d_1$). The diameter at distance $x$ is

$$d_x = d_1 + \frac{d_2 - d_1}{L}\,x = d_1 + kx,\qquad k = \frac{d_2 - d_1}{L}$$

Cross-sectional area: $A_x = \dfrac{\pi}{4}d_x^2$, so the stress is $\sigma_x = \dfrac{4P}{\pi d_x^2}$.

Elongation of a small length $dx$:

$$d\delta = \frac{\sigma_x}{E}dx = \frac{4P}{\pi E d_x^2}dx$$

Total elongation:

$$\delta = \frac{4P}{\pi E}\int_0^L\frac{dx}{(d_1 + kx)^2} = \frac{4P}{\pi E}\left[-\frac{1}{k(d_1 + kx)}\right]_0^L = \frac{4P}{\pi E k}\left[\frac{1}{d_1} - \frac{1}{d_2}\right]$$

Since $d_1 + kL = d_2$, and $\dfrac{1}{d_1} - \dfrac{1}{d_2} = \dfrac{d_2 - d_1}{d_1d_2} = \dfrac{kL}{d_1d_2}$:

$$\boxed{\delta = \frac{4PL}{\pi E\,d_1\,d_2}}$$

Checks

• If $d_1 = d_2 = d$, then $\delta = \dfrac{4PL}{\pi Ed^2} = \dfrac{PL}{AE}$, the formula for a prismatic bar.
• The formula holds for a gradual taper only, away from the ends, where the stress is assumed uniform over each section (Saint-Venant's principle).

Example: $P = 50$ kN, $L = 1000$ mm, $d_1 = 20$ mm, $d_2 = 40$ mm, $E = 200$ GPa gives $\delta = \dfrac{4(50\times10^3)(1000)}{\pi(200\times10^3)(20)(40)} = 0.398$ mm.

Take bar A as the left bar and bar B as the right bar. The bars first expand freely; if the free expansion exceeds the gap, the walls prevent the excess and a compressive force $P$ (equal in both bars) develops.

Free thermal expansion

$$\begin{aligned} \delta_A &= \alpha_A\,\Delta T\,L_A = 12\times10^{-6}\times40\times400 = 0.192\ \text{mm}\\ \delta_B &= \alpha_B\,\Delta T\,L_B = 23\times10^{-6}\times40\times300 = 0.276\ \text{mm}\\ \delta_{total} &= 0.468\ \text{mm} > \text{gap} = 0.26\ \text{mm} \end{aligned}$$

The expansion that is prevented is $\delta_{prevented} = 0.468 - 0.26 = 0.208$ mm, so the bars are in contact and stresses are induced.

Compatibility

The compressive force $P$ shortens the bars by exactly the prevented expansion:

$$\frac{P L_A}{A_A E_A} + \frac{P L_B}{A_B E_B} = 0.208$$ $$P\left(\frac{400}{1000\times2\times10^5} + \frac{300}{800\times1\times10^5}\right) = P\,(5.7500\times10^{-6}) = 0.208$$ $$P = 36173.9\ \text{N} = 36.17\ \text{kN}$$

Stresses

$$\sigma_A = \frac{P}{A_A} = \frac{36173.9}{1000} = 36.17\ \text{N/mm}^2,\qquad \sigma_B = \frac{P}{A_B} = \frac{36173.9}{800} = 45.22\ \text{N/mm}^2$$

Answer: both bars are in compression: $\sigma_A = 36.17$ N/mm² (bar A) and $\sigma_B = 45.22$ N/mm² (bar B); common force $P = 36.17$ kN.

Consider a uniform vertical bar of length $L$, area $A$, Young's modulus $E$ and unit weight (weight density) $\gamma = \rho g$, hanging from a fixed support at the top and carrying only its own weight.

 ═══════  fixed support
   │ ↑ x measured from the free (bottom) end
   │
   │  dx
   │
   ○  free end

Derivation

Take a section at distance $x$ from the free end. The load it carries is the weight of the bar below it:

$$P_x = \gamma A x$$

The stress at the section is $\sigma_x = \dfrac{P_x}{A} = \gamma x$, which varies linearly from zero at the free end to $\gamma L$ at the support.

The elongation of a small element of length $dx$:

$$d\delta = \frac{\sigma_x}{E}dx = \frac{\gamma x}{E}dx$$

Total elongation:

$$\delta = \int_0^L\frac{\gamma x}{E}dx = \frac{\gamma L^2}{2E}$$

With the total weight $W = \gamma A L$:

$$\boxed{\delta = \frac{\gamma L^2}{2E} = \frac{WL}{2AE}}$$

Meaning: the elongation due to self-weight is half the elongation that the same weight $W$ would cause if it were applied as a point load at the free end ($\delta = WL/AE$). Maximum stress $\sigma_{max} = \gamma L$ occurs at the support.

Example: a steel bar of $L = 100$ m with $\gamma = 77$ kN/m³ and $E = 200$ GPa: $\delta = \dfrac{77\times10^{-6}\ \text{N/mm}^3\times(10^5)^2}{2\times200\times10^3} = 1.925$ mm, with $\sigma_{max} = 7.7$ N/mm².

The load is carried by a rigid plate resting on the three rods, so all rods shorten equally. The steel rod stands in a recess 1 m deeper, so it is 5 m long (copper 4 m).

Compatibility

$$\delta_c = \delta_s \Rightarrow \frac{\sigma_c L_c}{E_c} = \frac{\sigma_s L_s}{E_s} \Rightarrow \frac{\sigma_c(400)}{10^6} = \frac{\sigma_s(500)}{2\times10^6} \Rightarrow \sigma_s = 1.60\,\sigma_c$$

Equilibrium

Area of each rod $A = \dfrac{\pi}{4}(4)^2 = 12.566$ cm².

$$P = 2\sigma_cA + \sigma_sA = \sigma_cA\,(2 + 1.60)$$ $$3000 = \sigma_c(12.566)(3.60) \Rightarrow \sigma_c = 66.315\ \text{kg/cm}^2,\quad \sigma_s = 106.103\ \text{kg/cm}^2$$

Check: $2(833.3) + 1333.3 = 3000.0$ kg.

Answer: stress in each copper rod = 66.31 kg/cm²; stress in the steel rod = 106.10 kg/cm² (compressive).

The rigid plate rests on three rods, so all rods shorten equally. The steel rod stands in a recess 1 m deeper: copper is 3 m long, steel 4 m.

Compatibility

$$\frac{\sigma_c L_c}{E_c} = \frac{\sigma_s L_s}{E_s} \Rightarrow \frac{\sigma_c(300)}{10^6} = \frac{\sigma_s(400)}{2\times10^6} \Rightarrow \sigma_s = 1.50\,\sigma_c$$

Equilibrium

$A = \dfrac{\pi}{4}(3)^2 = 7.069$ cm² for each rod.

$$5000 = \sigma_cA(2+1.50) \Rightarrow \sigma_c = \frac{5000}{7.069\times3.50} = 202.102\ \text{kg/cm}^2$$ $$\sigma_s = 1.50\times202.102 = 303.152\ \text{kg/cm}^2$$

Answer: each copper rod 202.10 kg/cm²; steel rod 303.15 kg/cm² (compressive).

Principle: both tubes have the same length and are loaded together, so they shorten equally: $\varepsilon_s = \varepsilon_b$. The total load is shared: $P_s + P_b = 750$ kN.

Areas

Steel: inner $d = 130$ mm, outer $D = 150$ mm. Brass: inner $d = 160$ mm, outer $D = 180$ mm.

$$A_s = \frac{\pi}{4}(150^2 - 130^2) = 4398.2\ \text{mm}^2, \qquad A_b = \frac{\pi}{4}(180^2 - 160^2) = 5340.7\ \text{mm}^2$$

Stresses

Equal strain gives

$$\frac{\sigma_s}{E_s} = \frac{\sigma_b}{E_b} \Rightarrow \sigma_s = \frac{E_s}{E_b}\sigma_b = 2\sigma_b$$ $$P = \sigma_sA_s + \sigma_bA_b = \sigma_b(2A_s + A_b) \Rightarrow 750\,000 = \sigma_b(2\times4398.2 + 5340.7)$$ $$\sigma_b = 53.05\ \text{MPa}, \qquad \sigma_s = 106.10\ \text{MPa}$$

Loads carried

$$P_s = \sigma_sA_s = 466.67\ \text{kN}, \qquad P_b = \sigma_bA_b = 283.33\ \text{kN}$$

Check: $466.67 + 283.33 = 750.00$ kN.

Shortening

$$\delta = \frac{\sigma_s L}{E_s} = \frac{106.10\times250}{2\times10^5} = 0.1326\ \text{mm}$$

(Brass gives the same: $53.05\times250/10^5 = 0.1326$ mm.)

Answer: $\sigma_s = 106.10$ MPa, $\sigma_b = 53.05$ MPa; $P_s = 466.67$ kN, $P_b = 283.33$ kN; shortening $= 0.1326$ mm.

Both tubes carry the load together and have the same length, so their strains are equal and $P_s + P_b = 1000$ kN $= 10^6$ N.

Areas

Steel: inner 15 cm, outer 17 cm. Brass: inner 17 cm, outer 19 cm.

$$A_s = \frac{\pi}{4}(17^2 - 15^2) = 50.27\ \text{cm}^2, \qquad A_b = \frac{\pi}{4}(19^2 - 17^2) = 56.55\ \text{cm}^2$$

Stresses

$$\frac{\sigma_s}{E_s} = \frac{\sigma_b}{E_b} \Rightarrow \sigma_s = \frac{2\times10^7}{10^7}\sigma_b = 2\sigma_b$$ $$10^6 = \sigma_b(2A_s + A_b) = \sigma_b(2\times50.27 + 56.55) \Rightarrow \sigma_b = 6366.2\ \text{N/cm}^2$$ $$\sigma_s = 12732.4\ \text{N/cm}^2$$

Loads

$$P_s = \sigma_sA_s = 640.00\ \text{kN}, \qquad P_b = \sigma_bA_b = 360.00\ \text{kN}$$

Check: $640.00 + 360.00 = 1000.00$ kN.

Answer: steel: $12732.4\ \text{N/cm}^2$ ($127.32$ MPa), load $640.0$ kN; brass: $6366.2\ \text{N/cm}^2$ ($63.66$ MPa), load $360.0$ kN.

Stress concentration and stress concentration factor

Stress concentration is the localised rise of stress near a sudden change in cross-section or geometry (holes, notches, grooves, keyways, fillets, sharp corners, threads), where the flow of internal force is disturbed. The stress at such a place is much higher than the average stress calculated from $P/A$.

The stress concentration factor is

$$K_t = \frac{\sigma_{max}}{\sigma_{nom}}$$

where $\sigma_{max}$ is the peak stress at the discontinuity and $\sigma_{nom}$ is the nominal stress based on the net area. $K_t \ge 1$ and depends on geometry (not on the material, as long as it is elastic). For a small circular hole in a wide plate in tension $K_t = 3$.

 Stress flow lines crowd near a hole:
  ─────┐   ┌─────
  ─────┤ ○ ├─────   σ_max at the edge of the hole
  ─────┘   └─────

Effects are serious for brittle materials and for fatigue loading, whereas ductile materials under static loading yield locally and redistribute stress.

Saint-Venant's principle

It states that the stress and strain produced at points sufficiently far from the place of application of a load (or from a discontinuity) depend only on the resultant of the load, not on the exact way it is distributed; the differences between two statically equivalent loadings become negligible at a distance about equal to the largest cross-sectional dimension of the member.

With regard to stress concentration, it tells us that the disturbance is local: high stresses exist only in the neighbourhood of the hole, notch or point load, and at a distance of about one width from it the stress becomes uniform ($P/A$). So the formula $\sigma = P/A$ may be used away from these places and the concentration need be considered only locally.

Why a factor of safety is used

The factor of safety ($FOS = \dfrac{\text{ultimate or yield stress}}{\text{permissible (working) stress}}$) is used because of:

1. Uncertainty in the actual loads (overloading, impact, dynamic effects).
2. Variation in material properties, defects and non-uniformity.
3. Approximations in analysis and stress concentrations, workmanship and fabrication errors.
4. Deterioration with time (corrosion, fatigue, wear).
5. The need for safety of life and property, and to keep stresses in the elastic range.

Typical values: 1.5 to 2 for steel (yield basis), 3 to 4 for brittle materials (ultimate basis), higher for dynamic loads.

The load $P$ at C rotates the rigid bar clockwise about the pin B: A moves up (stretching the aluminium rod fixed to the ground below) and D moves down (stretching the steel rod hung from the ceiling). Both rods are in tension; let the forces be $F_{al}$ and $F_{st}$.

 F_al                    F_st
  |                        ^
  v                        |
  A---1.5m---B---0.75---C---0.75---D
   (Al, 1.5 m)  pin     P=500 N  (steel, 0.9 m)

Equilibrium (moments about B)

Aluminium pulls A down and steel pulls D up; both give anticlockwise moments about B, balancing the clockwise moment of $P$:

$$F_{al}(1.5) + F_{st}(1.5) = P(0.75) = 500\times0.75$$ $$F_{al} + F_{st} = 250\ \text{N}$$

Compatibility

The bar is rigid, and A and D are both 1.5 m from the pin, so the upward movement of A equals the downward movement of D: $\delta_{al} = \delta_{st}$.

$$\frac{F_{al}(1500)}{300\times100\times10^3} = \frac{F_{st}(900)}{200\times200\times10^3} \;\Rightarrow\; 5\times10^{-5}F_{al} = 2.25\times10^{-5}F_{st} \;\Rightarrow\; F_{al} = 0.450\,F_{st}$$

Solution

$$\begin{aligned} F_{st}(1+0.450) &= 250 \Rightarrow F_{st} = 172.41\ \text{N}\\ F_{al} &= 77.59\ \text{N}\\ \sigma_{st} &= \frac{172.41}{200} = 0.862\ \text{N/mm}^2\\ \sigma_{al} &= \frac{77.59}{300} = 0.259\ \text{N/mm}^2 \end{aligned}$$

Answer: $\sigma_{steel} = 0.862$ N/mm² (tension) and $\sigma_{aluminium} = 0.259$ N/mm² (tension).

Assumed arrangement: both bars are 6 m long (3 m + 3 m) and pinned at P and R. The aluminium rod hangs from the ceiling to Q (upper bar), the steel rod joins Q to S (lower bar), and $W$ acts at the middle of RS.

Equilibrium

• Lower bar RS (moments about R): the steel rod pulls S up with $F_{st}$, so $F_{st}(6) = W(3)$ and $F_{st} = W/2$.
• Upper bar PQ (moments about P): the steel rod pulls Q down with $F_{st}$ and the aluminium rod pulls Q up with $F_{al}$, so $F_{al}(6) = F_{st}(6)$ and $F_{al} = F_{st} = W/2$.

Let $F = W/2$. Both rods are in tension.

Deformations

$$\begin{aligned} \delta_{al} &= \frac{F L}{AE} = \frac{F(2000)}{500\times70\times10^3} = 5.7143e-05\,F\ \text{mm}\\ \delta_{st} &= \frac{F(2000)}{300\times200\times10^3} = 3.3333e-05\,F\ \text{mm} \end{aligned}$$

Geometry of movement

• Q moves down by $\delta_{al}$ (bar PQ rotates about P).
• S moves down by $\delta_Q + \delta_{st} = \delta_{al} + \delta_{st}$ (bar RS rotates about R).
• The load point is at the middle of RS, so it moves $\tfrac12(\delta_{al}+\delta_{st})$.

Limiting the movement of W to 5 mm:

$$\tfrac12(\delta_{al}+\delta_{st}) = 5 \Rightarrow (5.7143e-05+3.3333e-05)F = 10$$ $$F = 110526.3\ \text{N},\qquad W = 2F = 221052.6\ \text{N} = 221.05\ \text{kN}$$

Answer: $W_{max} = 221.05$ kN (rod force 110.53 kN in each rod).

The steel and concrete shorten by the same amount, so the strain is equal and $\sigma_s = m\,\sigma_c$ with the modular ratio $m = E_s/E_c = 18.67$.

Areas

$$\begin{aligned} A_s &= 8\times\frac{\pi}{4}(12)^2 = 904.78\ \text{mm}^2\\ A_g &= 230\times400 = 92000\ \text{mm}^2\\ A_c &= A_g - A_s = 91095.22\ \text{mm}^2 \end{aligned}$$

Equilibrium

$$P = \sigma_c A_c + \sigma_s A_s = \sigma_c\,(A_c + mA_s)$$ $$600\times10^3 = \sigma_c\,(91095.22 + 18.67\times904.78) = \sigma_c\,(107987.44)$$ $$\sigma_c = 5.556\ \text{N/mm}^2,\qquad \sigma_s = m\sigma_c = 18.67\times5.556 = 103.73\ \text{N/mm}^2$$

Check: load carried by concrete 506.14 kN + steel 93.86 kN = 600.00 kN.

Answer: stress in concrete = 5.56 N/mm² and stress in steel = 103.73 N/mm² (both compressive).

Order from the fixed wall: aluminium (300 mm), copper (250 mm), steel (100 mm). Loads: 50 kN (←) at the aluminium/copper junction, 25 kN (←) at the copper/steel junction and 100 kN (→) at the free end. Check: 100 kN to the right balances 50 + 25 = 75 kN to the left plus the wall reaction of 25 kN (←), so the wall pulls on the bar with 25 kN.

Axial force in each part (tension +ve, from the free end)

|=== Al ===|=== Cu ===|== St ==>  100 kN
Wall   <-50 kN     <-25 kN

• Steel: $N_s = +100$ kN (tension)
• Copper: $N_c = 100 - 25 = +75$ kN (tension)
• Aluminium: $N_a = 100 - 25 - 50 = +25$ kN (tension)

Areas

$$A_a = \frac{\pi}{4}(105)^2 = 8659.0,\quad A_c = \frac{\pi}{4}(85)^2 = 5674.5,\quad A_s = \frac{\pi}{4}(50)^2 = 1963.5\ \text{mm}^2$$

Stresses

$$\begin{aligned} \sigma_a &= \frac{25\times10^3}{8659.0} = 2.887\ \text{N/mm}^2\\ \sigma_c &= \frac{75\times10^3}{5674.5} = 13.217\ \text{N/mm}^2\\ \sigma_s &= \frac{100\times10^3}{1963.5} = 50.930\ \text{N/mm}^2 \end{aligned}$$

Change in length ($\delta = NL/AE$)

$$\begin{aligned} \delta_a &= \frac{25\times10^3\times300}{8659.0\times70\times10^3} = 0.0124\ \text{mm}\\ \delta_c &= \frac{75\times10^3\times250}{5674.5\times100\times10^3} = 0.0330\ \text{mm}\\ \delta_s &= \frac{100\times10^3\times100}{1963.5\times200\times10^3} = 0.0255\ \text{mm}\\ \delta_{total} &= 0.0709\ \text{mm} \end{aligned}$$

Answer: $\sigma_{Al} = 2.89$, $\sigma_{Cu} = 13.22$, $\sigma_{St} = 50.93$ N/mm² (all tensile); total elongation = 0.0709 mm.

The load at D turns the bar clockwise about the pin A: B moves down and stretches the brass rod BE (tension, hung from the ceiling); C moves down and compresses the steel rod CF (standing on the ground). Let the forces be $F_b$ (tension) and $F_s$ (compression); the distances from A are B = 2 m, C = 4 m, D = 5 m.

 ceiling
   |E
   |            brass (2 m)
 A-+--2m--B--2m--C--1m--D
 pin      |      |      | 15 kN
          E      |      v
                 | steel (3 m) F on ground

Equilibrium (moments about A)

Both rods resist the load (brass pulls B up, steel pushes C up):

$$F_b(2) + F_s(4) = 15\times5 = 75\ \text{kN·m}$$

Compatibility

The bar is rigid, so the deflection is proportional to the distance from A: $\delta_C = 2\delta_B$.

$$\frac{F_s(3000)}{600\times200} = 2\,\frac{F_b(2000)}{1000\times100} \;\Rightarrow\; 25\,F_s = 40\,F_b \;\Rightarrow\; F_s = 1.60\,F_b$$

(forces in kN with E in kN/mm²).

Solution

$$\begin{aligned} 2F_b + 4(1.6F_b) &= 75 \Rightarrow F_b = 8.929\ \text{kN},\quad F_s = 14.286\ \text{kN}\\ \sigma_b &= \frac{8928.6}{1000} = 8.93\ \text{N/mm}^2\\ \sigma_s &= \frac{14285.7}{600} = 23.81\ \text{N/mm}^2 \end{aligned}$$

Answer: $\sigma_{brass} = 8.93$ N/mm² (tension) and $\sigma_{steel} = 23.81$ N/mm² (compression).

Let the bar have uniform area $A$ and modulus $E$. The bar is fixed at the left end, and the loads are (positions from the fixed end): 1.5 t (→) at 1 m, 1.5 t (→) at 3 m, 1.5 t (←) at the free end (5 m). The 2 t force is applied at distance $x$ from the fixed end; to cancel a net shortening it must act away from the wall (→).

Force in each portion (tension +ve) without the 2 t force

Take the sum of the forces to the right of the section:

$$\delta_{without} = \frac{1}{AE}\left[1.5(1) + 0(2) + (-1.5)(2)\right] = \frac{-1.5}{AE}\ \text{(t·m)}$$

The bar is net shortened, so a tensile 2 t force must lengthen it by $1.5/AE$.

Effect of the 2 t force at distance $x$

It adds $+2$ t of tension in every portion between the wall and $x$, so it adds $2x/AE$ to the length:

$$-1.5 + 2x = 0 \Rightarrow x = 0.75\ \text{m}$$

Since $x = 0.75 < 1$ m, the force lies in the first portion, as assumed.

Check: portions become 0 – 0.75 m: $+3.5$ t; 0.75 – 1 m: $+1.5$ t; 1 – 3 m: 0; 3 – 5 m: $-1.5$ t. Sum $= 3.5(0.75) + 1.5(0.25) + 0 - 1.5(2) = 2.625 + 0.375 - 3 = 0$.

Answer: $x = 0.75$ m from the fixed end, with the 2 t force acting away from the fixed end.

The members are bolted together, so they extend equally: $\varepsilon_t = \varepsilon_s$ and $\sigma_s/E_s = \sigma_t/E_t$.

$$m = \frac{E_s}{E_t} = \frac{210}{8.4} = 25.0,\qquad \sigma_s = m\,\sigma_t$$

Areas

$$A_t = 2\times(150\times75) = 22500\ \text{mm}^2,\qquad A_s = 150\times6 = 900\ \text{mm}^2$$

Which material governs?

• If timber reaches 6 N/mm², steel would be at $25\times6 = 150$ N/mm² $> 130$ N/mm² (not permitted).
• If steel reaches 130 N/mm², timber is at $130/25 = 5.2$ N/mm² $< 6$ N/mm² (safe).

Hence the steel plate governs: $\sigma_s = 130$ N/mm² and $\sigma_t = 5.2$ N/mm².

Permissible load

$$P = \sigma_s A_s + \sigma_t A_t = 130\times900 + 5.2\times22500 = 117000 + 117000 = 234000\ \text{N} = 234.0\ \text{kN}$$

Elongation (L = 4 m = 4000 mm)

$$\delta = \frac{\sigma_s}{E_s}L = \frac{130}{210\times10^3}\times4000 = 2.476\ \text{mm}$$

Answer: permissible tensile load = 234.0 kN; elongation = 2.476 mm.

Longitudinal quantities

$$\begin{aligned} A &= \frac{\pi}{4}(30)^2 = 706.86\ \text{mm}^2\\ \sigma &= \frac{P}{A} = \frac{60\times10^3}{706.86} = 84.883\ \text{N/mm}^2\\ \varepsilon_{long} &= \frac{\delta L}{L} = \frac{0.09}{200} = 0.000450\\ E &= \frac{\sigma}{\varepsilon} = \frac{84.883}{0.000450} = 188628\ \text{N/mm}^2 = 188.6\ \text{kN/mm}^2 \end{aligned}$$

Poisson's ratio

$$\varepsilon_{lat} = \frac{\delta d}{d} = \frac{0.0039}{30} = 0.000130,\qquad \mu = \frac{\varepsilon_{lat}}{\varepsilon_{long}} = \frac{0.000130}{0.000450} = 0.2889$$

Other moduli

$$\begin{aligned} G &= \frac{E}{2(1+\mu)} = \frac{188628}{2(1+0.2889)} = 73175\ \text{N/mm}^2\\ K &= \frac{E}{3(1-2\mu)} = \frac{188628}{3(1-2\times0.2889)} = 148917\ \text{N/mm}^2 \end{aligned}$$

Answer: $\mu = 0.289$; $E = 188.6$ kN/mm²; $G = 73.2$ kN/mm²; $K = 148.9$ kN/mm².

The bars are joined in series between rigid supports. On heating, both expand; the supports resist and the same compressive force $P$ acts in both bars.

Data

$$A_s = \frac{\pi}{4}(40)^2 = 1256.64\ \text{mm}^2,\qquad A_a = \frac{\pi}{4}(60)^2 = 2827.43\ \text{mm}^2,\qquad \Delta T = 50-25 = 25^\circ\text{C}$$

Free thermal expansion

$$\begin{aligned} \delta_{T,s} &= 13\times10^{-6}\times25\times500 = 0.1625\ \text{mm}\\ \delta_{T,a} &= 23.1\times10^{-6}\times25\times700 = 0.4042\ \text{mm}\\ \delta_T &= 0.5667\ \text{mm} \end{aligned}$$

Flexibility of the two bars in series:

$$\frac{L_s}{A_sE_s}+\frac{L_a}{A_aE_a} = \frac{500}{1256.64\times200\times10^3}+\frac{700}{2827.43\times70\times10^3} = 5.52621e-06\ \text{mm/N}$$

(i) Unyielding supports

$$P\times5.52621e-06 = 0.5667 \Rightarrow P = 102556.7\ \text{N} = 102.56\ \text{kN}$$ $$\sigma_s = \frac{102556.7}{1256.64} = 81.61\ \text{N/mm}^2,\qquad \sigma_a = \frac{102556.7}{2827.43} = 36.27\ \text{N/mm}^2\ \text{(both compressive)}$$

(ii) Supports move apart by 0.1 mm

Only the expansion not accommodated by the yielding is prevented: $\delta = 0.5667 - 0.1 = 0.4667$ mm.

$$P = \frac{0.4667}{5.52621e-06} = 84461.1\ \text{N} = 84.46\ \text{kN}$$ $$\sigma_s = 67.21\ \text{N/mm}^2,\qquad \sigma_a = 29.87\ \text{N/mm}^2\ \text{(compressive)}$$

Answer: (i) steel 81.61 N/mm², aluminium 36.27 N/mm²; (ii) steel 67.21 N/mm², aluminium 29.87 N/mm² (all compressive).

Loads: 90 kN (←) at A, 20 kN (→) at B, 30 kN (→) at C, 40 kN (→) at D. Check: 20 + 30 + 40 = 90 kN, so the bar is in equilibrium.

Axial force in each portion (tension +ve)

 90 kN <--A-----750-----B--------1000--------C----750----D--> 40 kN
                        |                    |
                      20 kN                30 kN
                        v (to the right)

Using the section method from the left end:

Elongation

$$\begin{aligned} \delta &= \frac{1}{AE}\sum N_iL_i = \frac{90\times10^3(750)+70\times10^3(1000)+40\times10^3(750)}{600\times200\times10^3}\\ &= \frac{167500000}{120000000}\\ \delta_{AB} &= 0.5625\ \text{mm},\quad \delta_{BC} = 0.5833\ \text{mm},\quad \delta_{CD} = 0.2500\ \text{mm}\\ \delta_{total} &= 1.3958\ \text{mm} \end{aligned}$$

Answer: total elongation = 1.396 mm.