Theory of Flexure

Bending equation (flexure formula): $\dfrac{\sigma}{y} = \dfrac{M}{I} = \dfrac{E}{R}$, where $\sigma$ is the bending stress at distance $y$ from the neutral axis, $M$ the bending moment, $I$ the moment of inertia of the section about the neutral axis, $E$ Young's modulus and $R$ the radius of curvature of the beam.

Assumptions

1. The beam is initially straight, of uniform section, symmetrical about the plane of loading, and of homogeneous, isotropic material.
2. Plane sections remain plane after bending (Bernoulli).
3. The material obeys Hooke's law; stress is within the elastic limit.
4. $E$ is the same in tension and compression.
5. The beam bends in a circular arc (pure bending) and the radius of curvature is large compared with the depth.

Derivation

    M ( __________________ ) M
        N______________A        before bending
                  
         bent into an arc:
            R
        O   .  (centre of curvature)
          \ | /
    ----- -N----A-- neutral layer NA (unchanged length)
          layer EF at distance y

Consider a short length $dx$ of the beam bent by moment $M$ into an arc of radius $R$ subtending angle $d\theta$ at the centre O. The neutral layer NA does not change in length:

$$NA = R\,d\theta = dx$$

A layer EF at distance $y$ from the neutral layer has the length $(R+y)\,d\theta$ after bending (original length $dx = R\,d\theta$). Its strain is

$$\varepsilon = \frac{(R+y)\,d\theta - R\,d\theta}{R\,d\theta} = \frac{y}{R}$$

By Hooke's law, $\sigma = E\varepsilon$:

$$\sigma = E\,\frac{y}{R} \quad\Rightarrow\quad \frac{\sigma}{y} = \frac{E}{R} \qquad (1)$$

So the stress varies linearly with $y$.

Neutral axis

For equilibrium of axial forces there is no net force on the section: $\int\sigma\,dA = \frac{E}{R}\int y\,dA = 0$. Hence $\int y\,dA = 0$, so the neutral axis passes through the centroid.

Moment of resistance

The force on a strip of area $dA$ is $\sigma\,dA$ and its moment about the neutral axis is $\sigma\,dA\cdot y$. The total moment of resistance equals the applied bending moment:

$$M = \int\sigma\,y\,dA = \frac{E}{R}\int y^2\,dA = \frac{E}{R}\,I$$ $$\frac{M}{I} = \frac{E}{R} \qquad (2)$$

Combining (1) and (2):

$$\boxed{\frac{\sigma}{y} = \frac{M}{I} = \frac{E}{R}}$$

The maximum stress occurs at the extreme fibre: $\sigma_{max} = \dfrac{M}{I}y_{max} = \dfrac{M}{Z}$, where $Z = \dfrac{I}{y_{max}}$ is the section modulus.

Flexural rigidity

Flexural rigidity is the product $EI$ of Young's modulus and the moment of inertia of the section about the neutral axis. It measures the resistance of a beam to bending: from $\dfrac{M}{EI} = \dfrac{1}{R} = \dfrac{d^2y}{dx^2}$, the curvature and deflection of a beam are inversely proportional to $EI$.

Importance: a larger $EI$ gives smaller deflection and curvature for the same loading. It is used to check deflection limits, to compare sections (deep sections have large $I$), and it appears in the deflection and column formulas ($\delta \propto \dfrac{WL^3}{EI}$, $P_{cr} = \dfrac{\pi^2EI}{L_e^2}$).

Bending stresses in the beam

Reading of the figure: simply supported beam with span $AD = 1 + 1.5 + 1 = 3.5$ m and two 60 kN loads at B (1 m from A) and C (1 m from D). The T-section has a flange $7.5 + 5 + 7.5 = 20$ cm wide and 5 cm thick, and a web 5 cm wide and 15 cm deep (total depth 20 cm). The flange is taken at the top (if the flange is at the bottom, the tensile and compressive values below are interchanged).

Reactions and maximum moment: by symmetry $R_A = R_D = 60$ kN. The maximum moment is constant between B and C:

$$M_{max} = 60\times1 = 60\ \text{kN·m} = 60\times10^6\ \text{N·mm}$$

Centroid (from the bottom of the web):

$$\bar y = \frac{(20\times5)(17.5) + (5\times15)(7.5)}{100 + 75} = 13.214\ \text{cm}$$

Moment of inertia about the neutral axis:

$$I = \left[\frac{20(5)^3}{12} + 100(17.5-13.214)^2\right] + \left[\frac{5(15)^3}{12} + 75(13.214-7.5)^2\right] = 5900.30\ \text{cm}^4$$

Stresses ($\sigma = My/I$, $M = 60\times10^5$ N·cm):

$$\sigma_{tension}\ (\text{bottom},\ y = 13.214\ \text{cm}) = \frac{60\times10^5 \times 13.214}{5900.30} = 13437.6\ \text{N/cm}^2 = 134.38\ \text{N/mm}^2$$ $$\sigma_{compression}\ (\text{top},\ y = 6.786\ \text{cm}) = \frac{60\times10^5 \times 6.786}{5900.30} = 6900.4\ \text{N/cm}^2 = 69.00\ \text{N/mm}^2$$

Answer: maximum tensile stress $= 134.38$ N/mm² (bottom of the web); maximum compressive stress $= 69.00$ N/mm² (top of the flange).

Reading of the figure: simple span $L = 6$ m. The 100 kN and 10 kN loads both act at 3 m from the supports, i.e. at mid-span, so together they form a 110 kN central load. The T-section has a flange 200 mm $\times$ 50 mm and a web 50 mm thick extending 200 mm below the flange (total depth 250 mm). The flange is on top.

Maximum bending moment

$$M_{max} = \frac{WL}{4} = \frac{110\times6}{4} = 165.0\ \text{kN·m} = 1.6500e+08\ \text{N·mm}$$

Section properties (distances from the bottom of the web)

$$A = 200(50) + 50(200) = 20000\ \text{mm}^2$$ $$\bar y = \frac{(200\times50)(225) + (50\times200)(100)}{20\,000} = 162.50\ \text{mm} \quad(\text{from the bottom}), \qquad H - \bar y = 87.50\ \text{mm}\ (\text{from the top})$$ $$I = \left[\frac{200(50)^3}{12} + 10\,000(225-162.50)^2\right] + \left[\frac{50(200)^3}{12} + 10\,000(162.50-100)^2\right] = 1.1354e+08\ \text{mm}^4$$

Bending stresses ($\sigma = My/I$)

$$\sigma_{compression}\ (\text{top}) = \frac{1.6500e+08\times87.50}{1.1354e+08} = 127.16\ \text{N/mm}^2$$ $$\sigma_{tension}\ (\text{bottom}) = \frac{1.6500e+08\times162.50}{1.1354e+08} = 236.15\ \text{N/mm}^2$$

Answer: maximum compressive stress $= 127.2$ N/mm² (top of the flange); maximum tensile stress $= 236.1$ N/mm² (bottom of the web).

Reading of the figure: $AB = 1$ m, $BC = 1$ m, $CD = 3$ m, $DE = 2$ m; hinge at A (x = 0) and roller at D (x = 5 m). Loads: 5 kN at B (x = 1), UDL 1 kN/m over CD (x = 2 to 5), and a 2 kN·m couple at the free end E (taken clockwise). $EI$ is constant. Deflection is taken positive upward.

Reactions

Moments about A (anticlockwise positive):

$$R_D(5) - 5(1) - 3(3.5) - 2 = 0 \Rightarrow R_D = 3.5\ \text{kN}, \qquad R_A = 5 + 3 - 3.5 = 4.5\ \text{kN}$$

Macaulay's method

The UDL ends at D, so it is continued to the end and cancelled by an equal upward load (+1 kN/m beyond D). Bending moment (sagging positive) at a section $x$ from A:

$$EI\frac{d^2y}{dx^2} = 4.5x - 5\langle x-1\rangle - \frac{1}{2}\langle x-2\rangle^2 + 3.5\langle x-5\rangle + \frac{1}{2}\langle x-5\rangle^2 + 2\langle x-7\rangle^0$$

Integrating:

$$EI\,\theta = 2.25x^2 - 2.5\langle x-1\rangle^2 - \frac{1}{6}\langle x-2\rangle^3 + 1.75\langle x-5\rangle^2 + \frac{1}{6}\langle x-5\rangle^3 + 2\langle x-7\rangle + C_1$$ $$EI\,y = 0.75x^3 - \frac{5}{6}\langle x-1\rangle^3 - \frac{1}{24}\langle x-2\rangle^4 + \frac{7}{12}\langle x-5\rangle^3 + \frac{1}{24}\langle x-5\rangle^4 + \langle x-7\rangle^2 + C_1x + C_2$$

Constants

• At A, $y = 0$ at $x = 0$: $C_2 = 0$.
• At D, $y = 0$ at $x = 5$: $0.75(125) - \frac{5}{6}(64) - \frac{1}{24}(81) + 5C_1 = 0 \Rightarrow C_1 = -7.4083$

Slopes at A and D

$$EI\,\theta_A = C_1 = -7.408\ \text{kN·m}^2 \quad\Rightarrow\quad \theta_A = \frac{7.408}{EI}\ \text{rad (clockwise)}$$ $$EI\,\theta_D = 2.25(25) - 2.5(16) - \frac{27}{6} + C_1 = 4.342\ \text{kN·m}^2 \quad\Rightarrow\quad \theta_D = \frac{4.342}{EI}\ \text{rad (anticlockwise)}$$

Deflections at B and C

$$EI\,y_B = 0.75(1) + C_1(1) = -6.658\ \text{kN·m}^3 \Rightarrow y_B = \frac{6.658}{EI}\ \text{downward}$$ $$EI\,y_C = 0.75(8) - \frac{5}{6}(1) + C_1(2) = -9.650\ \text{kN·m}^3 \Rightarrow y_C = \frac{9.650}{EI}\ \text{downward}$$

Answer (with $EI$ in kN·m²): $y_B = 6.66/EI$ and $y_C = 9.65/EI$ (downward); $\theta_A = 7.41/EI$ (clockwise) and $\theta_D = 4.34/EI$ (anticlockwise).

Data: flanges 40 cm $\times$ 6 cm each (width $\times$ thickness), web 60 cm deep $\times$ 6 cm thick, so total depth $= 6 + 60 + 6 = 72$ cm. Span $L = 6$ m, UDL $w = 25$ kN/m.

Maximum bending moment

$$M_{max} = \frac{wL^2}{8} = \frac{25(6)^2}{8} = 112.5\ \text{kN·m} = 1.1250e+08\ \text{N·mm}$$

Moment of inertia about the neutral axis (symmetrical section, in mm)

$$I = 2\left[\frac{400(60)^3}{12} + (400\times60)(330)^2\right] + \frac{60(600)^3}{12}$$

where the flange centroid is $300 + 30 = 330$ mm from the neutral axis.

$$I = 2\left[7.2000e+06 + 2.6136e+09\right] + 1.0800e+09 = 6.3216e+09\ \text{mm}^4$$

Stress at the flange-web junction

The junction is at $y = 300$ mm from the neutral axis:

$$\sigma = \frac{My}{I} = \frac{1.1250e+08\times300}{6.3216e+09} = 5.339\ \text{N/mm}^2$$

The stress is compressive at the top junction and tensile at the bottom junction. For comparison, the extreme fibre stress is $\dfrac{M(360)}{I} = 6.407\ \text{N/mm}^2$.

Answer: bending stress at the flange-web junction $= 5.34\ \text{N/mm}^2$ (about 5.34 MPa).

Pure bending

Pure bending is the bending of a beam under a constant bending moment with zero shear force over the length considered. Only normal (bending) stresses act on the cross-section; there is no shear stress.

Example: a simply supported beam carrying two equal loads $P$ placed symmetrically at distance $a$ from each support. Between the loads the shear force is zero and $M = Pa$ is constant, so that middle portion is in pure bending.

   P                P
   |<--a-->|        |
   v                v
 A ^----------------------^ B
   SFD:  +P  | 0 |  -P
   BMD:  rises to Pa, flat, falls

Maximum deflection of SS beam under UDL

Take A as origin, span $L$, load $w$ per unit length. Reactions $R_A = R_B = wL/2$.

Bending moment at distance $x$:

$$EI\frac{d^2y}{dx^2} = M = \frac{wL}{2}x - \frac{w x^2}{2}$$

Integrate once (slope):

$$EI\frac{dy}{dx} = \frac{wL x^2}{4} - \frac{w x^3}{6} + C_1$$

Integrate again (deflection):

$$EI\,y = \frac{wL x^3}{12} - \frac{w x^4}{24} + C_1 x + C_2$$

Boundary conditions: at $x=0$, $y=0$ gives $C_2 = 0$. At $x=L$, $y=0$:

$$0 = \frac{wL^4}{12} - \frac{wL^4}{24} + C_1 L \;\Rightarrow\; C_1 = -\frac{wL^3}{24}$$

So

$$y = \frac{w}{24EI}\left(2Lx^3 - x^4 - L^3 x\right)$$

By symmetry the slope is zero and deflection is maximum at mid-span $x = L/2$:

$$y_{max} = \frac{w}{24EI}\left(\frac{2L^4}{8} - \frac{L^4}{16} - \frac{L^4}{2}\right) = -\frac{5wL^4}{384EI}$$

Answer: $y_{max} = \dfrac{5wL^4}{384EI}$ downward, at mid-span.

Assumption: flange 30 cm × 10 cm on top, web 10 cm wide, overall depth 25 cm (web depth 15 cm). Self-weight ignored.

 10kN
 4 kN/m                      |
 vvvvvvvvvvvvvvvvvv          v
 ^-------------------^------------
 A        6 m        B    2 m

Reactions

Taking moments about A:

$$R_B \times 6 = 4\times 6\times 3 + 10\times 8 \Rightarrow R_B = 25.33\ \text{kN}$$

$R_A = 24 + 10 - 25.33 = 8.67$ kN (check: $\Sigma F_y = 0$).

Bending moments

• At B (hogging): $M_B = -10\times 2 = -20$ kNm.
• Zero shear at $x = R_A/4 = 2.167$ m from A, so the sagging moment is maximum there:

$$M_{max}^{+} = 8.67\times 2.167 - 4\times\frac{2.167^2}{2} = 9.39\ \text{kNm}$$

Section properties

Measure from the base (cm):

$$\bar y = \frac{30\times10\times20 + 10\times15\times7.5}{300+150} = 15.83\ \text{cm}$$ $$I = \frac{30\times10^3}{12} + 300(20-15.83)^2 + \frac{10\times15^3}{12} + 150(7.5-15.83)^2 = 20937.5\ \text{cm}^4$$

Distance to top fibre $= 25 - 15.83 = 9.17$ cm; to bottom fibre $= 15.83$ cm.

Stresses ($\sigma = My/I$)

$$\sigma = \frac{20\times10^6\times 158.3}{20937.5\times10^4} = 15.12\ \text{N/mm}^2$$

Answer: maximum bending stress = 15.12 N/mm² (compressive) at the bottom fibre over support B.

Given: $L = 6$ m, $w = 5$ kN/m, $P = 15$ kN at centre, $\sigma_b = 12\ \text{N/mm}^2$. Self-weight is ignored.

Maximum bending moment (at mid-span)

$$M_{max} = \frac{wL^2}{8} + \frac{PL}{4} = \frac{5\times 6^2}{8} + \frac{15\times 6}{4} = 22.5 + 22.5 = 45\ \text{kNm}$$

Required section modulus

$$Z = \frac{M}{\sigma_b} = \frac{45\times10^6}{12} = 3.75\times10^6\ \text{mm}^3$$

Dimensions

A joist is deeper than it is wide; assume $d = 2b$ (common practice).

$$Z = \frac{bd^2}{6} = \frac{b(2b)^2}{6} = \frac{2b^3}{3}$$ $$b^3 = \frac{3\times 3.75\times10^6}{2} = 5.625\times10^6 \Rightarrow b = 177.8\ \text{mm}$$

$d = 2b = 355.6$ mm.

Check with $180\times360$ mm: $Z = 180\times360^2/6 = 3.888\times10^6\ \text{mm}^3$, $\sigma = 45\times10^6/3.888\times10^6 = 11.6\ \text{N/mm}^2 < 12$, safe.

Answer: provide a joist 180 mm wide × 360 mm deep (calculated 177.8 mm × 355.6 mm).

Section: flanges 100 mm × 25 mm, web 50 mm thick, web height 50 mm. Overall depth $= 25 + 50 + 25 = 100$ mm. The section is symmetrical, so the neutral axis is at mid-depth, $y = 50$ mm from top and bottom.

 |<--100-->|
 +---------+  25
    | | |
    | | |     50
 +---------+  25

Moment of inertia

$$I = \frac{100\times100^3}{12} - \frac{(100-50)\times 50^3}{12} = 8.333\times10^6 - 0.521\times10^6 = 7.8125\times10^6\ \text{mm}^4$$

Moment of resistance

Both extreme fibres are 50 mm from the neutral axis. The permissible stress in the weaker fibre governs (the smaller value, 150 MPa):

$$M_r = \frac{\sigma I}{y} = \frac{150\times 7.8125\times10^6}{50} = 23.44\times10^6\ \text{Nmm} = 23.44\ \text{kNm}$$

(At 180 MPa the other face would allow 28.1 kNm, which is more, so 150 MPa governs.)

Safe UDL

$$M_{max} = \frac{wL^2}{8} = M_r \Rightarrow w = \frac{8\times 23.44}{10^2} = 1.875\ \text{kN/m}$$

Answer: moment of resistance = 23.44 kNm; safe UDL $w = 1.875$ kN/m.

Given: $L = 10$ m, $w = 20$ kN/m, $P = 50$ kN at centre, $d = 2b$, permissible $\sigma_t = 120$, $\sigma_c = 100\ \text{N/mm}^2$. Self-weight is ignored.

Maximum bending moment

$$M = \frac{wL^2}{8} + \frac{PL}{4} = \frac{20\times100}{8} + \frac{50\times10}{4} = 250 + 125 = 375\ \text{kNm}$$

Governing stress

A rectangle is symmetrical, so the tension and compression fibres have equal stress. The lower permissible value, $100\ \text{N/mm}^2$ (compression), governs.

$$Z = \frac{M}{\sigma} = \frac{375\times10^6}{100} = 3.75\times10^6\ \text{mm}^3$$

Dimensions

$$Z = \frac{bd^2}{6} = \frac{b(2b)^2}{6} = \frac{2b^3}{3} \Rightarrow b^3 = 5.625\times10^6 \Rightarrow b = 177.8\ \text{mm}$$

$d = 2b = 355.6$ mm.

Check with $180\times360$ mm: $Z = 3.888\times10^6\ \text{mm}^3$, $\sigma = 96.5\ \text{N/mm}^2 < 100$. Safe.

Answer: $b \approx 180$ mm, $d \approx 360$ mm (calculated 177.8 mm × 355.6 mm).

Section: flange 18 cm × 6 cm on top, stem 4 cm × 18 cm below; overall depth 24 cm. $\sigma_t = \sigma_c = 150\ \text{N/mm}^2$.

 P
 |      w = 1.5 kN/m
 v  vvvvvvvvvvvvvvvvv
 |##################  free end
 fixed    3 m

Section properties (from base, cm)

$$\bar y = \frac{18\times6\times21 + 4\times18\times9}{108+72} = 16.2\ \text{cm}$$ $$I = \frac{18\times6^3}{12} + 108(21-16.2)^2 + \frac{4\times18^3}{12} + 72(9-16.2)^2 = 8488.8\ \text{cm}^4$$

Top fibre distance $y_t = 24 - 16.2 = 7.8$ cm; bottom $y_b = 16.2$ cm.

$$Z_t = \frac{I}{y_t} = 1088.3\ \text{cm}^3,\qquad Z_b = \frac{I}{y_b} = 524.0\ \text{cm}^3$$

Moment of resistance

A downward load causes hogging: top in tension, bottom in compression. Both have the same permissible stress, so the smaller $Z$ governs:

$$M_r = 150\times 524.0\times10^3 = 78.6\times10^6\ \text{Nmm} = 78.6\ \text{kNm}$$

Value of P

$$M_{max} = P\times 3 + \frac{1.5\times 3^2}{2} = 3P + 6.75$$ $$3P + 6.75 = 78.6 \Rightarrow P = 23.95\ \text{kN}$$

Answer: $P = 23.95$ kN; moment of resistance $= 78.6$ kNm.

Assumption: the figure gives only the flange width (10 cm), web thickness (3 cm) and overall depth (42 cm). The flange thickness is taken as 4 cm, flange on top, so the web is 38 cm deep.

Bending moment

$$M = \frac{wL^2}{8} = \frac{4\times 5^2}{8} = 12.5\ \text{kNm (sagging, constant over whole section)}$$

Neutral axis (from base, cm)

$$\bar y = \frac{3\times38\times19 + 10\times4\times40}{114+40} = \frac{2166+1600}{154} = 24.45\ \text{cm}$$

Moment of inertia

$$I = \frac{3\times38^3}{12} + 114(19-24.45)^2 + \frac{10\times4^3}{12} + 40(40-24.45)^2 = 26829.5\ \text{cm}^4 = 2.683\times10^8\ \text{mm}^4$$

Stress at 15 cm above the base

Distance from the neutral axis: $y = 24.45 - 15 = 9.45$ cm $= 94.5$ mm, below the neutral axis, so the fibre is in tension (sagging).

$$\sigma = \frac{My}{I} = \frac{12.5\times10^6\times 94.5}{2.683\times10^8} = 4.40\ \text{N/mm}^2$$

Answer: bending stress at 15 cm above the base = 4.40 N/mm² (tensile).

Section: flanges 300 mm × 30 mm, web 20 mm × 140 mm, overall depth 200 mm. It is symmetrical, so the neutral axis is at mid-depth, 100 mm above the base.

Bending moment

$$M = \frac{wL^2}{8} = \frac{3\times 4^2}{8} = 6\ \text{kNm}$$

Moment of inertia (about neutral axis)

$$I = 2\left[\frac{300\times30^3}{12} + 300\times30\times(85)^2\right] + \frac{20\times140^3}{12}$$ $$I = 2\,(0.675\times10^6 + 65.025\times10^6) + 4.573\times10^6 = 135.97\times10^6\ \text{mm}^4$$

Stress 80 mm above the base

Distance from neutral axis $y = 100 - 80 = 20$ mm, below the NA, so the fibre is in tension (sagging beam).

$$\sigma = \frac{My}{I} = \frac{6\times10^6\times 20}{135.97\times10^6} = 0.88\ \text{N/mm}^2$$

Answer: bending stress at 80 mm above base = 0.88 N/mm² (tensile).

Given: $AB = 2$ m with 50 kN at B; $BC = 3$ m with 40 kN at free end C; total length 5 m. $E = 200\ \text{kN/mm}^2 = 2\times10^5\ \text{N/mm}^2$.

$$I = \frac{50\times100^3}{12} = 4.167\times10^6\ \text{mm}^4,\qquad EI = 8.333\times10^{11}\ \text{N mm}^2 = 833.3\ \text{kNm}^2$$

 fixed       50kN        40kN
 |A----2m----B-----3m-----C

Use superposition with standard cantilever results (point load $P$ at distance $a$ from the fixed end: $\theta = Pa^2/2EI$, $y = Pa^3/3EI$ at the load point, and the portion beyond the load stays straight).

Slope at C

$$\theta_C = \frac{40\times10^3\times(5000)^2}{2EI} + \frac{50\times10^3\times(2000)^2}{2EI} = 0.600 + 0.120 = 0.72\ \text{rad}$$

Deflection at C

• Due to 40 kN at C: $\dfrac{40\times10^3\times5000^3}{3EI} = 2000$ mm
• Due to 50 kN at B: $\dfrac{50\times10^3\times2000^3}{3EI} = 160$ mm at B, plus slope $0.12$ rad $\times$ 3000 mm $= 360$ mm further to C

$$y_C = 2000 + 160 + 360 = 2520\ \text{mm}$$

Answer: slope at free end = 0.72 rad; deflection = 2520 mm (2.52 m) downward. These values are far beyond the small-deflection limit because the section (50 × 100 mm) is very flexible for these loads; the formulas still give the theoretical result.

Section (as drawn): web 50 mm wide × 120 mm high standing on a flange 200 mm × 50 mm. So the flange is at the bottom, the web top is the top surface, and overall depth is 170 mm.

 8 kN/m                   15 kN
 vvvvvvvvvvvvvvvvvvvvvvv   v
 |#########################
 A            5 m          B

Maximum bending moment (at fixed end A, hogging)

$$M_{max} = \frac{8\times5^2}{2} + 15\times5 = 100 + 75 = 175\ \text{kNm}$$

Top fibres are in tension, bottom fibres in compression.

Section properties (from base, mm)

$$\bar y = \frac{200\times50\times25 + 50\times120\times110}{10000+6000} = 56.875\ \text{mm}$$ $$I = \frac{200\times50^3}{12} + 10000(25-56.875)^2 + \frac{50\times120^3}{12} + 6000(110-56.875)^2 = 36.377\times10^6\ \text{mm}^4$$

Distance to top $= 170 - 56.875 = 113.125$ mm; to bottom $= 56.875$ mm.

Maximum stresses

$$\sigma_{top} = \frac{175\times10^6\times113.125}{36.377\times10^6} = 544.2\ \text{N/mm}^2\ \text{(T)}$$ $$\sigma_{bottom} = \frac{175\times10^6\times56.875}{36.377\times10^6} = 273.6\ \text{N/mm}^2\ \text{(C)}$$

Stress 25 mm below the top surface

$y = 113.125 - 25 = 88.125$ mm above the NA (tension):

$$\sigma = \frac{175\times10^6\times88.125}{36.377\times10^6} = 423.9\ \text{N/mm}^2\ \text{(T)}$$

Answer: maximum bending stress = 544.2 N/mm² (tension, top) at the fixed end; stress 25 mm below top = 423.9 N/mm² (tension). These stresses are well above the strength of steel, so the section as given would fail under these loads.

Section: flange 20 cm × 5 cm on top, stem 5 cm × 12 cm below; overall depth 17 cm.

Maximum bending moment (at fixed end A, hogging)

$$M = \frac{5\times3^2}{2} + 20\times3 = 22.5 + 60 = 82.5\ \text{kNm}$$

Top fibres are in tension.

Neutral axis (from base, cm)

$$\bar y = \frac{20\times5\times14.5 + 5\times12\times6}{100+60} = \frac{1450+360}{160} = 11.31\ \text{cm}$$

Moment of inertia

$$I = \frac{20\times5^3}{12} + 100(14.5-11.31)^2 + \frac{5\times12^3}{12} + 60(6-11.31)^2 = 3637.7\ \text{cm}^4 = 3.638\times10^7\ \text{mm}^4$$

Stress 25 mm below the top

Top fibre is $17 - 11.31 = 5.69$ cm $= 56.9$ mm above the NA. The point 25 mm below the top is $y = 56.9 - 25 = 31.9$ mm above the NA (tension).

$$\sigma = \frac{My}{I} = \frac{82.5\times10^6\times 31.9}{3.638\times10^7} = 72.3\ \text{N/mm}^2\ \text{(T)}$$

For reference, the extreme fibres are 129.0 N/mm² (top, tension) and 256.6 N/mm² (bottom, compression).

Answer: bending stress at 25 mm below the top = 72.3 N/mm² (tensile).

Importance of computing deflection

A beam may be strong enough in bending and shear yet still fail to serve its purpose if it deflects too much. Deflection is computed because:

1. Serviceability: codes limit deflection (for example span/250 for total, span/350 for loads after finishes) so floors do not feel springy or unsafe.
2. Protection of finishes: large deflection cracks plaster, partitions, tiles and glazing.
3. Appearance: sagging beams and slabs look unsafe and spoil the structure.
4. Machine alignment: shafts and machine beds need small deflection for gears, bearings and rails to work properly.
5. Drainage and ponding: sagging roofs collect water and increase the load.
6. Statically indeterminate beams: deflection (compatibility) conditions are needed to find the extra reactions.
7. Vibration control: stiffness affects natural frequency and comfort.

Two examples of pure bending

1. A simply supported beam with two equal loads placed symmetrically at equal distances from the supports: the portion between the loads has constant bending moment and zero shear force.
2. A beam or bar bent into an arc by equal and opposite couples applied at its ends (for example, a bar held in a bending testing machine, four-point bending test).

Given: $P_1 = 20$ kN at $a = 3$ m from the fixed end; $P_2 = 30$ kN at the free end ($L = 5$ m). $E = 2\times10^5$ N/mm², $I = 2\times10^8$ mm⁴.

$$EI = 2\times10^5\times2\times10^8 = 4\times10^{13}\ \text{N mm}^2 = 4\times10^4\ \text{kNm}^2$$

Use superposition. Take $x$ from the fixed end; the moment due to $P_2$ at the free end is $M = P_2(L - x)$.

Effect of $P_1$ (at its own position)

$$\theta_1 = \frac{P_1 a^2}{2EI} = \frac{20\times10^3\times3000^2}{2\times4\times10^{13}} = 0.00225\ \text{rad}$$ $$y_1 = \frac{P_1 a^3}{3EI} = \frac{20\times10^3\times3000^3}{3\times4\times10^{13}} = 4.5\ \text{mm}$$

Effect of $P_2$ (at free end) at the position of $P_1$

Integrate $EI\,y'' = P_2(L-x)$ with $y = y' = 0$ at the fixed end, then put $x = 3$ m:

$$EI\,\theta_2 = P_2\left[Lx - \frac{x^2}{2}\right]_{x=3} = 30\,(15 - 4.5) = 315\ \text{kNm}^2 \Rightarrow \theta_2 = 0.007875\ \text{rad}$$ $$EI\,y_2 = P_2\left[\frac{Lx^2}{2} - \frac{x^3}{6}\right]_{x=3} = 30\,(22.5 - 4.5) = 540\ \text{kNm}^3 \Rightarrow y_2 = 13.5\ \text{mm}$$

Total under $P_1$

$$\theta = 0.00225 + 0.007875 = 0.0101\ \text{rad}$$ $$y = 4.5 + 13.5 = 18.0\ \text{mm}$$

Answer: slope under $P_1$ = 0.0101 rad (about 0.58°); deflection under $P_1$ = 18.0 mm downward.

Section: flange 18 cm × 4 cm on top, web 3 cm × 30 cm; overall depth 34 cm.

Maximum bending moment (mid-span, sagging)

$$M = \frac{wL^2}{8} + \frac{PL}{4} = \frac{30\times4^2}{8} + \frac{40\times4}{4} = 60 + 40 = 100\ \text{kNm}$$

Neutral axis (from base, cm)

$$\bar y = \frac{3\times30\times15 + 18\times4\times32}{90+72} = \frac{1350+2304}{162} = 22.56\ \text{cm}$$

Top fibre distance $y_t = 34 - 22.56 = 11.44$ cm; bottom $y_b = 22.56$ cm.

Moment of inertia

$$I = \frac{3\times30^3}{12} + 90(15-22.56)^2 + \frac{18\times4^3}{12} + 72(32-22.56)^2 = 18406\ \text{cm}^4 = 1.8406\times10^8\ \text{mm}^4$$

Stresses

With the flange at the top and a sagging beam, top = compression, bottom = tension.

$$\sigma_c = \frac{100\times10^6\times114.4}{1.8406\times10^8} = 62.2\ \text{N/mm}^2$$ $$\sigma_t = \frac{100\times10^6\times225.6}{1.8406\times10^8} = 122.5\ \text{N/mm}^2$$

Answer: maximum tensile stress = 122.5 N/mm² (bottom of web); maximum compressive stress = 62.2 N/mm² (top of flange).

Reading of the question: the text states one 20 kN load at 4 m from the left support, so a single load is used. $EI$ is not given, so the answer is in terms of $EI$ (kNm²). $L = 6$ m, $a = 4$ m, $b = 2$ m.

          20 kN
            v
 A ^---------------^---- B
   |<---4 m--->|<2m>|

Reactions

$R_A = 20\times 2/6 = 6.67$ kN, $R_B = 13.33$ kN.

Position of maximum deflection

The maximum deflection lies in the longer segment (A to load), where slope is zero. Measured from A:

$$x = \sqrt{\frac{L^2 - b^2}{3}} = \sqrt{\frac{36-4}{3}} = 3.266\ \text{m}$$

This is less than 4 m, so it lies in the longer segment.

Maximum deflection

$$y_{max} = \frac{P\,b\,x\,(L^2 - b^2 - x^2)}{6EIL} = \frac{20\times2\times3.266\,(36-4-10.667)}{6\times6\,EI} = \frac{77.42}{EI}\ \text{(kNm}^3\text{)}$$

Deflection under the load

$$y_C = \frac{P\,a^2 b^2}{3EIL} = \frac{20\times16\times4}{3\times6\,EI} = \frac{71.11}{EI}\ \text{(kNm}^3\text{)}$$

Answer: (i) maximum deflection $77.42/EI$ at 3.266 m from the left support; (ii) deflection under the load $71.11/EI$ (with $EI$ in kNm², result in m).

If the figure is read as two 20 kN loads at 2 m and 4 m, the beam is symmetrical: maximum deflection at mid-span $= Pa(3L^2-4a^2)/24EI = 153.33/EI$, and under each load $= 133.33/EI$.

Assumptions: cantilever of length $L$, fixed at B, free end A carries a downward load $P$ and a couple $M$ whose sense also tends to bend the beam downward (same direction as $P$). $EI$ constant.

Take $x$ measured from the free end A. Bending moment at $x$ (sagging taken so that the load gives hogging; magnitude used):

$$EI\frac{d^2y}{dx^2} = -(Px + M)$$

Integrate:

$$EI\frac{dy}{dx} = -\left(\frac{Px^2}{2} + Mx\right) + C_1$$

At the fixed end B ($x = L$) the slope is zero:

$$C_1 = \frac{PL^2}{2} + ML$$

Integrate again:

$$EI\,y = -\left(\frac{Px^3}{6} + \frac{Mx^2}{2}\right) + C_1 x + C_2$$

At B ($x=L$), $y = 0$:

$$C_2 = \frac{PL^3}{6} + \frac{ML^2}{2} - \frac{PL^3}{2} - ML^2 = -\frac{PL^3}{3} - \frac{ML^2}{2}$$

At A ($x=0$), $EI\,y_A = C_2$:

$$y_A = \frac{PL^3}{3EI} + \frac{ML^2}{2EI}$$

(Superposition gives the same: $PL^3/3EI$ from the point load plus $ML^2/2EI$ from the couple.)

Answer: deflection at A $= \dfrac{PL^3}{3EI} + \dfrac{ML^2}{2EI}$ downward. If the couple acts in the opposite sense, subtract the second term.

Section (in cm): an inverted T. Flange 5 cm wide × 2 cm thick at the bottom; stem 1 cm wide × 5 cm high above it; overall height 7 cm. $F = 50$ kN.

     +-+   <- top of stem
     | |
     | |  5
     | |
 +---+-+---+
 |         |  2
 +---------+
     5

Neutral axis (from base)

$$\bar y = \frac{5\times2\times1 + 1\times5\times4.5}{10+5} = \frac{32.5}{15} = 2.167\ \text{cm}$$

Moment of inertia

$$I = \frac{5\times2^3}{12} + 10(1-2.167)^2 + \frac{1\times5^3}{12} + 5(4.5-2.167)^2 = 54.58\ \text{cm}^4$$

Shear stress $\tau = \dfrac{F\,A\bar y}{I\,b}$

The top of the stem is 4.833 cm above the NA. The junction of stem and flange (2 cm from the base) lies 0.167 cm below the NA, since the NA is at 2.167 cm.

Example: $\tau_{NA} = \dfrac{50\times10^3\times11.68\times10^3}{54.58\times10^4\times10} = 107.0\ \text{N/mm}^2$.

Variation

• In the stem the stress varies parabolically from 0 at the top to the maximum at the NA, then stays almost constant down to the junction.
• At the junction it drops suddenly from 106.9 to 21.4 N/mm² because the width changes from 1 cm to 5 cm.
• In the flange it falls parabolically (nearly linearly here) to zero at the bottom.

Answer: $\tau_{max} = 107.0\ \text{N/mm}^2$ at the neutral axis; average stress on the stem $= 100\ \text{N/mm}^2$, so $\tau_{max}/\tau_{avg} \approx 1.07$.

Consider a rectangular section of width $b$ and depth $d$ made of an elastic-perfectly plastic material with yield stress $\sigma_y$.

   Elastic limit        Fully plastic
   +--+                  +------+
   | /|                  |      |   sigma_y
 --|/-|-- NA          ---|------|--- NA
   |\ |                  |      |   sigma_y
   +--+                  +------+

Maximum elastic moment

Stress is linear and just reaches $\sigma_y$ at the extreme fibres:

$$M_y = \sigma_y Z = \sigma_y\cdot\frac{bd^2}{6}$$

Fully plastic moment

Every fibre has reached $\sigma_y$. The tension block is the top/bottom half, each of area $bd/2$, acting at the centroid of each half, $d/4$ from the neutral axis:

$$T = C = \sigma_y\frac{bd}{2}$$

The lever arm between the two forces is $d/2$:

$$M_p = T\times\frac{d}{2} = \sigma_y\frac{bd}{2}\times\frac{d}{2} = \sigma_y\frac{bd^2}{4}$$

(The plastic section modulus is $S = bd^2/4$.)

Ratio

$$\frac{M_p}{M_y} = \frac{\sigma_y\,bd^2/4}{\sigma_y\,bd^2/6} = \frac{6}{4} = 1.5$$

Hence $M_p = 1.5\,M_y$. The shape factor of a rectangular section is 1.5.