Suspension Cable Systems

Data: span $L = 200$ m, two cables and two three-hinged stiffening girders, dip $d = 20$ m. Dead load of the bridge 50 kN/m (total); three point loads of 800 kN each on the centre line at 50, 100 and 150 m. The loads are shared equally by the two cables, so for one cable: $w = 25$ kN/m and three loads of 400 kN. Cable: $y = \dfrac{4d\,x(L-x)}{L^2}$, $\tan\alpha_s = \dfrac{4d}{L} = 0.4$.

Horizontal pull (one cable)

Dead load: $H_d = \dfrac{wL^2}{8d} = \dfrac{25\times 200^2}{8\times 20} = 6250\ \text{kN}$

Point loads: the simple-beam reaction is $R_A = 1.5\times 400 = 600$ kN (symmetrical), and the moment at the middle hinge:

$$M_C = 600\times 100 - 400\times 50 = 40000\ \text{kN m}, \qquad H_p = \frac{M_C}{d} = \frac{40000}{20} = 2000\ \text{kN}$$ $$H = H_d + H_p = 8250\ \text{kN}$$

Maximum and minimum tension in the cable

• Minimum tension: at mid-span (the lowest point of the cable), where the cable is horizontal: $T_{min} = H = 8250$ kN.
• Maximum tension: at the supports, where $\tan\alpha_s = 0.4$ and the vertical component is $V_s = H\tan\alpha_s = 3300$ kN:

$$T_{max} = \sqrt{H^2 + V_s^2} = H\sqrt{1+0.16} = 8885.5\ \text{kN}$$

Length of the cable

$$S = L\left[1 + \frac83\left(\frac{d}{L}\right)^2 - \frac{32}{5}\left(\frac{d}{L}\right)^4\right] = 200\left[1+\frac83(0.1)^2 - \frac{32}{5}(0.1)^4\right] = 205.21\ \text{m}$$

Answer: $T_{max} = 8885.5$ kN (at the supports), $T_{min} = 8250$ kN (at mid-span), length of each cable $= 205.21$ m.

Assumptions. The loads given are for the whole bridge, so each of the two girders and cables carries half: $w$ = 40/2 = 20 kN/m and $P$ = 200/2 = 100 kN (if the data are taken for a single cable, double all forces below; the method does not change). Span $L$ = 100 m, dip $d$ = 10 m.

Cable pull $H$

A three-hinged girder gives the cable a uniform load, so the cable stays parabolic.

• Dead load part: $H_d = \dfrac{wL^2}{8d} = \dfrac{20(100)^2}{8(10)} = 2500.0$ kN
• Point loads: $R_A = 1.5P = 150.0$ kN; $M_{0C} = 150.0(50) - 100(25) = 5000.0$ kN m, so $H_p = \dfrac{M_{0C}}{d} = 500.0$ kN

$$H = H_d + H_p = 2500.0 + 500.0 = 3000.0\ \text{kN}$$

Maximum and minimum tension

End slope: $\tan\theta = 4d/L = 0.40$, $\sec\theta = 1.0770$.

• Maximum tension at the supports A and B: $T_{max} = H\sec\theta = 3000.0(1.0770) = 3231.10$ kN
• Minimum tension at mid-span (crown, where the cable is horizontal): $T_{min} = H = 3000.0$ kN

Length of the cable

Using $l = L\left[1+\dfrac{8}{3}\left(\dfrac{d}{L}\right)^2-\dfrac{32}{5}\left(\dfrac{d}{L}\right)^4\right]$ with $d/L = 0.1$:

$$l = 100\left[1 + \frac{8}{3}(0.1)^2 - \frac{32}{5}(0.1)^4\right] = 102.60\ \text{m}$$

Girder shear force and bending moment (per girder)

The dead load is carried wholly by the cable, so it produces no SF or BM in the girder. For the three loads, with $y = 0.004x(100-x)$ and $H_p$ = 500 kN:

$$M = M_0 - H_p\,y,\qquad S = S_0 - H_p\,\frac{dy}{dx}$$

• $0 \le x \le 25$: $M = 2x^2 - 50x$, $S = 4x - 50$
• $25 \le x \le 50$: $M = 2x^2 - 150x + 2500$, $S = 4x - 150$
• The right half is the mirror image (SF with opposite sign).

So $M = 0$ at the load points (0, 25, 50, 75, 100 m), and $M = -312.5$ kN m at 12.5 m, 37.5 m, 62.5 m and 87.5 m (the greatest hogging value). Shear force is $-50$ to $+50$ kN in each quarter, changing sign at the mid-point of each panel (a saw-tooth).

SFD (kN), +-50 in each quarter span
            *           *           *           *
          **          **          **          **
         *           *           *           *
       **          **          **          **
------*-----------*-----------*-----------*------
    **          **          **          **
   *           *           *           *
 **          **          **          **
*
(top = 50.0, bottom = -50.0; '-' is zero line)

BMD (kN m), all hogging, zero at the loads
*-----------*-----------*-----------*-----------*

 *         * *         * *         * *         *

  *       *   *       *   *       *   *       *

   *     *     *     *     *     *     *     *
    *   *       *   *       *   *       *   *
     ***         ***         ***         ***
(top = 0.0, bottom = -312.5; '-' is zero line)

Answer: $T_{max} = 3231.10$ kN at the supports; $T_{min} = 3000.0$ kN at the crown (mid-span); cable length = 102.60 m; girder $M_{max} = -312.5$ kN m (hogging) at 12.5, 37.5, 62.5 and 87.5 m; $S_{max} = 50$ kN.

Assumptions. The data are for the whole bridge; two girders and two cables share the load equally, so per cable $w$ = 30/2 = 15 kN/m and $P$ = 300/2 = 150 kN. (If a single cable is intended, all forces below double: $H$ = 22000 kN, $T_{max}$ = 22968.7 kN. The length is unchanged.) Span 400 m, dip 30 m, point loads at 100, 200 and 300 m.

Horizontal pull $H$

• Dead load: $H_d = \dfrac{wL^2}{8d} = \dfrac{15(400)^2}{8(30)} = 10000.0$ kN
• Point loads: $R_A = 1.5P = 225.0$ kN, $M_{0C} = 225.0(200) - 150(100) = 30000.0$ kN m, so $H_p = \dfrac{30000}{30} = 1000.0$ kN

$$H = 10000.0 + 1000.0 = 11000.0\ \text{kN}$$

Maximum and minimum tension

$\tan\theta = 4d/L = 0.300$ at the supports, $\sec\theta = 1.04403$.

• Maximum (at the two supports): $T_{max} = H\sec\theta = 11000.0(1.04403) = 11484.34$ kN
• Minimum (at mid-span, where the cable is horizontal): $T_{min} = H = 11000.0$ kN

Length of the cable

Using $l = L\left[1+\dfrac{8}{3}\left(\dfrac{d}{L}\right)^2-\dfrac{32}{5}\left(\dfrac{d}{L}\right)^4\right]$ with $d/L = 0.075$:

$$l = 400\left[1+\frac{8}{3}(0.075)^2-\frac{32}{5}(0.075)^4\right] = 405.92\ \text{m}$$

Answer: $T_{max} = 11484.34$ kN at the supports; $T_{min} = 11000.0$ kN at mid-span; length of cable = 405.92 m (per cable, loads shared equally by the two cables).

Data: three-hinged stiffening girder, span $L = 150$ m, cable dip $d = 15$ m. Point loads 180 kN at 50 m and 150 kN at 100 m from the left support. Dead load of the girder $w = 7.5$ kN/m, taken wholly by the cable (so the girder carries no dead-load moment). Cable: $y = \dfrac{4d\,x(L-x)}{L^2} = \dfrac{x(150-x)}{375}$.

Horizontal pull

Dead load: $H_d = \dfrac{wL^2}{8d} = \dfrac{7.5\times 150^2}{8\times 15} = 1406.25\ \text{kN}$

Point loads (hinge at mid-span):

$$R_A = \frac{180\times 100 + 150\times 50}{150} = 170.0\ \text{kN}, \qquad M_C = 170\times 75 - 180\times 25 = 8250\ \text{kN m}$$ $$H_p = \frac{M_C}{d} = \frac{8250}{15} = 550\ \text{kN}$$

Total: $H = 1406.25 + 550 = 1956.25$ kN.

SF and BM in the girder at 30 m from the left

Cable ordinate: $y = \dfrac{30\times 120}{375} = 9.60$ m; slope $\tan\alpha = \dfrac{4\times 15\,(150-60)}{22500} = 0.24$. Only the point loads act on the girder (the dead load is carried by the cable):

$$\begin{aligned} M &= R_A\times 30 - H_p\,y = 170\times 30 - 550\times 9.60 = -180.0\ \text{kN m}\\ V &= R_A - H_p\tan\alpha = 170 - 550\times 0.24 = 38.0\ \text{kN} \end{aligned}$$

Maximum tension in the cable

At the supports, $\tan\alpha_s = \dfrac{4d}{L} = 0.4$:

$$V_s = H\tan\alpha_s = 782.50\ \text{kN}, \qquad T_{max} = \sqrt{H^2 + V_s^2} = 2106.9\ \text{kN}$$

Length of the cable

$$S = L\left[1 + \frac83\left(\frac{d}{L}\right)^2 - \frac{32}{5}\left(\frac{d}{L}\right)^4\right] = 150\left[1 + \frac83(0.1)^2 - \frac{32}{5}(0.1)^4\right] = 153.90\ \text{m}$$

Answer: at 30 m from the left support, $V = 38.0$ kN and $M = -180.0$ kN m; maximum cable tension $= 2106.9$ kN; length of cable $= 153.90$ m.

Data and assumptions: span $L = 100$ m, two cables with two three-hinged stiffening girders, dip $d = 10$ m, roadway width 5 m. Dead load $5$ kN/m² and live load $10$ kN/m² (right half only). Load per metre run of the bridge: dead $5\times 5 = 25$ kN/m, live $10\times 5 = 50$ kN/m. Each of the two cables and girders carries half: per cable/girder dead load $w_d = 12.5$ kN/m (whole span, taken by the cable) and live load $w_l = 25$ kN/m on the right half ($x = 50$ to 100 m). Cable: $y = 0.004\,x(100-x)$.

Horizontal pull (per cable)

$$H_d = \frac{w_dL^2}{8d} = \frac{12.5\times 10^4}{80} = 1562.5\ \text{kN}$$

Live load: total $25\times 50 = 1250$ kN acting at $x = 75$ m.

$$R_A = \frac{1250\times 25}{100} = 312.5\ \text{kN}, \qquad M_C = R_A\times 50 = 15625.0\ \text{kN m}, \qquad H_l = \frac{M_C}{d} = 1562.5\ \text{kN}$$ $$H = H_d + H_l = 3125.0\ \text{kN}$$

SF and BM in the girder at 20 m from the left support

The dead load is carried by the cable, so only the live load acts on the girder. At $x = 20$ m: $y = 0.004\times 20\times 80 = 6.40$ m, $\tan\alpha = 0.004\,(100-40) = 0.24$. For the section (left of the load): $M_b = R_A\times 20 = 6250.0$ kN m, $V_b = R_A = 312.5$ kN.

$$\begin{aligned} M &= M_b - H_l\,y = 6250.0 - 1562.5\times 6.40 = -3750.0\ \text{kN m}\\ V &= V_b - H_l\tan\alpha = 312.5 - 1562.5\times 0.24 = -62.5\ \text{kN} \end{aligned}$$

(per girder; the sagging/hogging sign: negative moment means hogging.)

Maximum tension in the cable

At the supports, $\tan\alpha_s = \dfrac{4d}{L} = 0.4$, $V_s = H\tan\alpha_s = 1250.0$ kN:

$$T_{max} = \sqrt{H^2 + V_s^2} = \sqrt{3125.0^2 + 1250.0^2} = 3365.7\ \text{kN (each cable)}$$

Answer (per girder): $V = -62.5$ kN, $M = -3750.0$ kN m at 20 m from the left support; maximum tension in each cable $= 3365.7$ kN.

Set-up. The cable carries a UDL $w$ = 15 kN/m on the horizontal span, so it is a parabola with its lowest point O as the origin. Let the left support A be higher. The lower (right) support B is 10 m above O, so A is $10+5 = 15$ m above O. Let $x_1$, $x_2$ be the horizontal distances of A and B from O.

Position of the lowest point and $H$

For a parabola, $y = \dfrac{w x^2}{2H}$, hence $\dfrac{x_1^2}{x_2^2} = \dfrac{y_1}{y_2} = \dfrac{15}{10}$, so $x_1 = 1.2247\,x_2$. With $x_1 + x_2 = 180$ m:

$$\begin{aligned} x_2 &= \frac{180}{1+1.2247} = 80.908\ \text{m}, \qquad x_1 = 99.092\ \text{m} \\ H &= \frac{w x_2^2}{2 y_2} = \frac{15(80.908)^2}{2(10)} = 4909.60\ \text{kN} \end{aligned}$$

Check with the left side: $w x_1^2/(2\times 15) = 4909.60$ kN (same).

Maximum tension

Vertical components at the supports: $V_A = w x_1 = 1486.38$ kN, $V_B = w x_2 = 1213.62$ kN. The tension is largest at the higher support A:

$$T_{max} = \sqrt{H^2 + V_A^2} = \sqrt{4909.60^2 + 1486.38^2} = 5129.66\ \text{kN}$$

(At B: $T_B = 5057.37$ kN.)

Length of the cable

For a parabolic arc from the lowest point, with end slope $s = 2y/x$:

$$l = \frac{x}{2}\left[\sqrt{1+s^2} + \frac{\sinh^{-1}s}{s}\right]$$

Total length of cable between the saddles = 182.31 m.

Thrust on the piers

The saddles are on rollers, so the horizontal pull $H$ of the main cable is balanced only by the horizontal component of the straight backstay. A backstay at 60° to the vertical has $T_b \sin 60^\circ = H$:

$$T_b = \frac{H}{\sin 60^\circ} = 5669.11\ \text{kN}, \qquad V_b = T_b\cos 60^\circ = 2834.56\ \text{kN}$$

The horizontal forces cancel at each saddle, so the pier carries only a vertical thrust (cable vertical component plus backstay vertical component) and no bending moment:

• Left pier: $P_A = 1486.38 + 2834.56 = 4320.93$ kN
• Right pier: $P_B = 1213.62 + 2834.56 = 4048.18$ kN

Answer: (i) $T_{max} = 5129.66$ kN (at the higher support); (ii) cable length = 182.31 m; (iii) maximum thrust on the pier = 4320.93 kN (vertical, at the higher pier).

Set-up. The cable carries a UDL $w$ = 15 kN/m on the horizontal span, so it is a parabola with its lowest point O as the origin. Let the left support A be higher. The lower (right) support B is 10 m above O, so A is $10+5 = 15$ m above O. Let $x_1$, $x_2$ be the horizontal distances of A and B from O.

Position of the lowest point and $H$

For a parabola, $y = \dfrac{w x^2}{2H}$, hence $\dfrac{x_1^2}{x_2^2} = \dfrac{y_1}{y_2} = \dfrac{15}{10}$, so $x_1 = 1.2247\,x_2$. With $x_1 + x_2 = 200$ m:

$$\begin{aligned} x_2 &= \frac{200}{1+1.2247} = 89.898\ \text{m}, \qquad x_1 = 110.102\ \text{m} \\ H &= \frac{w x_2^2}{2 y_2} = \frac{15(89.898)^2}{2(10)} = 6061.23\ \text{kN} \end{aligned}$$

Check with the left side: $w x_1^2/(2\times 15) = 6061.23$ kN (same).

Maximum tension

Vertical components at the supports: $V_A = w x_1 = 1651.53$ kN, $V_B = w x_2 = 1348.47$ kN. The tension is largest at the higher support A:

$$T_{max} = \sqrt{H^2 + V_A^2} = \sqrt{6061.23^2 + 1651.53^2} = 6282.20\ \text{kN}$$

(At B: $T_B = 6209.42$ kN.)

Length of the cable

For a parabolic arc from the lowest point, with end slope $s = 2y/x$:

$$l = \frac{x}{2}\left[\sqrt{1+s^2} + \frac{\sinh^{-1}s}{s}\right]$$

Total length of cable between the saddles = 202.08 m.

Thrust on the piers

The saddles are on rollers, so the horizontal pull $H$ of the main cable is balanced only by the horizontal component of the straight backstay. A backstay at 60° to the vertical has $T_b \sin 60^\circ = H$:

$$T_b = \frac{H}{\sin 60^\circ} = 6998.91\ \text{kN}, \qquad V_b = T_b\cos 60^\circ = 3499.45\ \text{kN}$$

The horizontal forces cancel at each saddle, so the pier carries only a vertical thrust (cable vertical component plus backstay vertical component) and no bending moment:

• Left pier: $P_A = 1651.53 + 3499.45 = 5150.98$ kN
• Right pier: $P_B = 1348.47 + 3499.45 = 4847.92$ kN

Answer: (i) cable length = 202.08 m; (ii) $T_{max} = 6282.20$ kN at the higher support; (iii) maximum thrust on pier = 5150.98 kN (vertical), and moment on the pier = 0 because the rollers transmit no horizontal force.

Single cable and three-hinged stiffening girder: span $L$ = 150 m, dip $d$ = 15 m, dead load $w$ = 5 kN/m, loads 180 kN at 50 m from A and 120 kN at 30 m from B (that is, 120 m from A).

Cable pull $H$

$$\begin{aligned} R_A &= \frac{180(100)+120(30)}{150} = 144.0\ \text{kN} \\ M_{0C} &= 144.0(75) - 180(25) = 6300.0\ \text{kN m} \\ H_p &= \frac{M_{0C}}{d} = 420.0\ \text{kN}, \qquad H_d = \frac{wL^2}{8d} = \frac{5(150)^2}{8(15)} = 937.5\ \text{kN} \\ H &= H_d + H_p = 1357.5\ \text{kN} \end{aligned}$$

BM and SF at 30 m from the left support

The dead load is carried entirely by the cable (parabolic), so it gives no moment or shear in the girder; only the point loads act on it.

$$y_{30} = \frac{4(15)(30)(120)}{150^2} = 9.60\ \text{m},\qquad \frac{dy}{dx} = \frac{4d(L-2x)}{L^2} = \frac{4(15)(90)}{150^2} = 0.24$$ $$\begin{aligned} M_{30} &= R_A(30) - H_p\,y_{30} = 4320.0 - 420(9.60) = 288.0\ \text{kN m} \\ S_{30} &= R_A - H_p\,\frac{dy}{dx} = 144.0 - 420(0.24) = 43.2\ \text{kN} \end{aligned}$$

Maximum tension in the cable

At the supports $\tan\theta = 4d/L = 0.40$, so $\sec\theta = 1.0770$:

$$T_{max} = H\sec\theta = 1357.5(1.0770) = 1462.07\ \text{kN}$$

Length of the cable

Using $l = L\left[1+\dfrac{8}{3}\left(\dfrac{d}{L}\right)^2-\dfrac{32}{5}\left(\dfrac{d}{L}\right)^4\right]$ with $d/L = 0.1$:

$$l = 150\left[1+\frac{8}{3}(0.1)^2-\frac{32}{5}(0.1)^4\right] = 153.90\ \text{m}$$

Answer: $M_{30} = 288.0$ kN m (sagging), $S_{30} = 43.2$ kN; $T_{max} = 1462.07$ kN; cable length = 153.90 m.

Loads per girder. The roadway is 6 m wide and is carried by two girders and two cables, each taking half:

• Dead load: $5\times 6 = 30$ kN/m on the bridge, so $w_d$ = 15 kN/m per girder
• Live load on the left half (0 to 60 m): $10\times 6 = 60$ kN/m, so $w_l$ = 30 kN/m per girder

Span $L$ = 120 m, dip $d$ = 12 m.

Cable pull $H$

• Dead load: $H_d = \dfrac{w_d L^2}{8d} = \dfrac{15(120)^2}{8(12)} = 2250.0$ kN
• Live load on the left half: $R_A = \dfrac{30(60)(90)}{120} = 1350.0$ kN, then $M_{0C} = 1350.0(60) - 30(60)(30) = 27000.0$ kN m
• $H_l = \dfrac{M_{0C}}{d} = \dfrac{27000}{12} = 2250.0$ kN

$$H = H_d + H_l = 2250.0 + 2250.0 = 4500.0\ \text{kN}$$

SF and BM in the girder at 30 m from the left end

The dead load causes no girder BM or SF (the cable carries it). For the live load:

$$y_{30} = \frac{4(12)(30)(90)}{120^2} = 9.00\ \text{m},\qquad \frac{dy}{dx} = \frac{4(12)(120-60)}{120^2} = 0.20$$ $$\begin{aligned} M_{0,30} &= 1350(30) - 30(30)(15) = 27000.0\ \text{kN m} \\ M_{30} &= M_{0,30} - H_l\,y_{30} = 27000.0 - 2250.0(9.00) = 6750.0\ \text{kN m} \\ S_{30} &= \left[R_A - 30(30)\right] - H_l\,\frac{dy}{dx} = 450.0 - 2250.0(0.20) = 0.0\ \text{kN} \end{aligned}$$

Maximum tension in the cable for this loading

$\tan\theta = 4d/L = 0.40$ at the supports, so $\sec\theta = 1.0770$:

$$T_{max} = H\sec\theta = 4500.0(1.0770) = 4846.65\ \text{kN}$$

Answer: At 30 m from the left end, $M = 6750.0$ kN m (sagging) and $S = 0.0$ kN (the shear is zero, so this is the section of maximum BM in the loaded half); maximum cable tension = 4846.65 kN in each cable.

The main parts of a suspension bridge are:

1. Main cables: two or more parabolic (or catenary) cables of high-tensile steel wire that carry the whole load in tension and pass over the towers to the anchorages.
2. Towers (pylons): steel or reinforced concrete columns that support the cables through saddles at their tops and transfer the vertical load to the foundations.
3. Saddles: castings at the tower tops (and at the splay points) on which the cable rests; they allow the cable to slide or roll and so keep the tower free from bending.
4. Suspenders (hangers): vertical ropes or rods that transfer the load from the deck or stiffening girder to the main cable.
5. Stiffening girder (or truss): a beam or truss along the deck that distributes concentrated loads, reduces local deflection and prevents the cable from changing shape under moving loads (three-hinged, two-hinged or unhinged).
6. Deck (roadway) and floor system: floor beams and slab that carry the traffic.
7. Anchorages (anchor blocks): massive concrete blocks or rock anchors at the ends of the cables that resist the horizontal pull and any uplift.
8. Side spans and back stays: cable portions from the tower to the anchorage.
9. Wind cables, wind ties and bracing: lateral cables and bracing that resist wind forces.
10. Bearings and expansion joints: at the towers and abutments for movement of the deck.

        tower                         tower
         ||        main cable          ||
 anchor  ||  .--------.______.--------.||  anchor
 block<--||-'   |  |  |  |  |  |   | '-||-->block
         ||     hangers (suspenders)    ||
   ======||=== stiffening girder + deck ===||======
         ||                              ||

Tower structures

Towers carry the main cables over a saddle and transfer the vertical component of the cable pull to the foundation. They may be of steel or reinforced concrete and are built as:

• Portal frame towers: two legs connected by cross beams (struts) with diagonal bracing; used for road bridges, and stiff against wind across the bridge.
• Single-column or A-frame towers: for narrow bridges or where the cables are in one plane.
• Rocker towers (hinged at the base): allow the tower top to move with the cable.

Design points: the tower is designed mainly for axial compression, with some bending due to the unequal horizontal pull of the main and side spans; the cable sits on a saddle (rollers or a sliding plate) at the top so that the horizontal pull on both sides is the same and the tower is not bent.

     saddle
   ___|___|___        top strut
  |    X    |        cross bracing
  |   / \   |
  |  /   \  |        legs (compression)
  |_/_____\_|
   foundation

Wind cables and ties

Suspension bridges are light and flexible, so they are prone to swing sideways and vibrate under wind. The lateral stability is increased by:

• Wind ties (lateral bracing): horizontal bracing between the two stiffening girders or between the hangers in the plane of the deck, forming a horizontal truss that carries the wind load to the towers and abutments.
• Wind cables: additional cables placed horizontally on each side of the deck (or inclined from the deck level to the piers or the ground) and anchored on the abutments; they limit the lateral sway of the deck.
• Inclined suspenders and stays: hangers arranged in a triangular form, or cross ties, to damp vertical oscillation.
• Streamlined box girders: reduce wind forces and flutter.

 plan view:   =====girder=====        wind
              \   /  \   /  \          ->
              wind ties (horizontal truss)
              =====girder=====

In a suspension bridge with a three-hinged stiffening girder (hinges at the two ends and at mid-span $C$), the hangers carry a point load to the cable as a uniform load.

Assumptions

Hangers are closely spaced and inextensible, so the cable transmits an upward uniform load $p$ per unit length to the girder. The cable is a parabola of dip $d$.

Derivation

Let a point load $W$ act on the girder (span $L$) at a distance $x$ from the left end, $x\le L/2$.

        W
   A    |        C(hinge)       B
   o----+---------o-------------o
   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^   p per unit length (from the cable)

The bending moment at the middle hinge is zero. Taking the girder as a simple beam under $W$ and the upward uniform load $p$:

$$M_C = \frac{W x}{2} - \frac{p L^2}{8} = 0 \Rightarrow p = \frac{4Wx}{L^2}$$

Here $\dfrac{Wx}{2}$ is the moment at mid-span of the simple beam under $W$ (reaction $R_B = Wx/L$ acting at $L/2$). For a load on the other half, put $L - x$ in place of $x$:

$$p = \frac{4W\,(L-x)}{L^2} \quad (x\ge L/2)$$

Result

The load transferred to the main cable is a uniformly distributed load of intensity

$$p = \frac{4Wx}{L^2}\ \ (x\le L/2), \qquad p = \frac{4W(L-x)}{L^2}\ \ (x\ge L/2)$$

The total load on the cable is $pL = \dfrac{4Wx}{L}$, which is $2\times$ the reaction of the simple beam. It is largest ($p = \dfrac{2W}{L}$) when $W$ is at mid-span. The horizontal pull in the cable is

$$H = \frac{pL^2}{8d} = \frac{Wx}{2d}$$

Example: $W = 100$ kN at $x = 10$ m on a girder of span $L = 50$ m: $p = \dfrac{4\times 100\times 10}{2500} = 1.6$ kN/m.

Data: a transmission line hangs between two towers at the same level, span $\ell$, with weight $\gamma$ per unit length of cable (so the load is uniform along the cable length, not along the span). Let the sag (dip) at mid-span be $d$.

Shape: catenary

Take the origin at the lowest point $O$. For the portion of the cable of length $s$ from $O$ to a point $P(x, y)$, the forces are the horizontal tension $H$ at $O$, the weight $\gamma s$ and the tension $T$ at $P$:

$$\tan\psi = \frac{dy}{dx} = \frac{\gamma s}{H}$$

Let $c = H/\gamma$. Then $\dfrac{dy}{dx} = \dfrac{s}{c}$ and $ds = \sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx$, which gives $s = c\sinh\dfrac{x}{c}$. Integrating,

$$y = c\left(\cosh\frac{x}{c} - 1\right)$$

This is the equation of a catenary.

Sag and horizontal tension

At the tower, $x = \ell/2$:

$$d = c\left(\cosh\frac{\ell}{2c} - 1\right), \qquad c = \frac{H}{\gamma}$$

Length of the cable between the towers

$$S = 2c\,\sinh\frac{\ell}{2c}$$

Maximum tension at the towers: $T_{max} = \gamma\,(c + d) = H + \gamma d$.

Parabolic approximation (small sag, $d/\ell \le 1/10$)

When the sag is small, the weight per unit length can be taken as acting per unit horizontal span ($w = \gamma$). The cable is then a parabola:

$$y = \frac{\gamma x^2}{2H}, \qquad H = \frac{\gamma \ell^2}{8d}$$ $$S = \ell\left[1 + \frac{8}{3}\left(\frac{d}{\ell}\right)^2 - \frac{32}{5}\left(\frac{d}{\ell}\right)^4 + \dots\right]$$

so, to a good approximation, $S \approx \ell + \dfrac{8d^2}{3\ell}$, and the tension at the tower is $T = \sqrt{H^2 + (\gamma\ell/2)^2}$.

Example: $\ell = 300$ m, $d = 6$ m: $S \approx 300 + \dfrac{8\times 36}{900} = 300.32$ m.

Data: cable of span $L$ carrying a UDL $w$ per unit horizontal length; the supports are at heights $h_1$ and $h_2$ above the lowest point of the cable. Let $x_1$ and $x_2$ be the horizontal distances from the lowest point to the supports, so $x_1 + x_2 = L$.

 support 1 o                       o support 2
            \                     /  ^
          h1 \                   /   | h2
             |\___.___lowest___.'     
             |<x1->|<---x2---->|

Equation of the cable

Take the origin at the lowest point $O$, where the tension is horizontal and equal to $H$. For a point at horizontal distance $x$ from $O$, take moments about the point of the portion of the cable from $O$:

$$H\,y = \frac{w\,x^2}{2} \Rightarrow y = \frac{w\,x^2}{2H}$$

Apply to the two supports

$$h_1 = \frac{w\,x_1^2}{2H} \Rightarrow x_1 = \sqrt{\frac{2H\,h_1}{w}}, \qquad h_2 = \frac{w\,x_2^2}{2H} \Rightarrow x_2 = \sqrt{\frac{2H\,h_2}{w}}$$

Since $x_1 + x_2 = L$:

$$\sqrt{\frac{2H}{w}}\,\left(\sqrt{h_1}+\sqrt{h_2}\right) = L$$

Squaring both sides:

$$\frac{2H}{w}\left(\sqrt{h_1}+\sqrt{h_2}\right)^2 = L^2$$ $$\boxed{H = \frac{w\,L^2}{2\left(\sqrt{h_1}+\sqrt{h_2}\right)^2}}$$

This is the horizontal reaction at each end, as the cable carries no horizontal external load. Check: for $h_1 = h_2 = h$, $H = \dfrac{wL^2}{2\times 4h} = \dfrac{wL^2}{8h}$, the usual result for level supports.

Data: span $L = 300$ m, central dip $d = 1.5$ m (supports at the same level), uniformly distributed load $w = 300$ kN/m on the horizontal span (the value is as read from the scan; the method is the same for any $w$). The cable is a parabola.

Horizontal tension

$$H = \frac{wL^2}{8d} = \frac{300\times 300^2}{8\times 1.5} = 2250000\ \text{kN}$$

Vertical reaction at each support

$$V = \frac{wL}{2} = \frac{300\times 300}{2} = 45000\ \text{kN}$$

Maximum tension (at the supports)

$$T_{max} = \sqrt{H^2 + V^2} = \sqrt{(2250000)^2 + (45000)^2} = 2250450.0\ \text{kN}$$

Slope at the supports: $\tan\alpha = \dfrac{4d}{L} = \dfrac{6}{300} = 0.02$, $\alpha = 1.15^\circ$.

Length of the cable

$$S = L\left[1 + \frac{8}{3}\left(\frac{d}{L}\right)^2 - \frac{32}{5}\left(\frac{d}{L}\right)^4\right] = 300\left[1 + \frac{8}{3}(0.005)^2 - \frac{32}{5}(0.005)^4\right] = 300.020\ \text{m}$$

Answer: $T_{max} = 2250450.0$ kN (at the supports); length of the cable $\approx 300.020$ m.

Data: span $L = 50$ m, dip $d = 5$ m, three-hinged stiffening girder (hinges at both ends and mid-span). Dead load $w = 8$ kN/m carried entirely by the cable. A moving load $W = 80$ kN crosses the girder. The cable is a parabola, $y = \dfrac{4d\,x(L-x)}{L^2} = 0.008\,x(50-x)$; $\tan\alpha_{support} = \dfrac{4d}{L} = 0.4$.

Horizontal pull in the cable

Dead load: $H_d = \dfrac{wL^2}{8d} = \dfrac{8\times 2500}{40} = 500$ kN.

Live load $W$ at distance $a$ from the left end (hinge at mid-span: girder moment at $C$ is zero):

$$H_L = \frac{W a}{2d} \ (a\le 25),\qquad \frac{W(L-a)}{2d}\ (a\ge 25)$$

The maximum is with $W$ at mid-span: $H_L = \dfrac{80\times 25}{10} = 200$ kN.

$$H_{max} = 500 + 200 = 700\ \text{kN}$$

Maximum tension in the cable

It occurs at the supports, where the cable is steepest:

$$T_{max} = H_{max}\sqrt{1 + \left(\frac{4d}{L}\right)^2} = 700\sqrt{1+0.16} = 753.9\ \text{kN}$$

(vertical component at the support $= 700\times 0.4 = 280$ kN).

Maximum bending moment at 10 m from the left end

Section $x_s = 10$ m, cable ordinate $y_s = 0.008\times 10\times 40 = 3.2$ m. For a unit load at $a$, $M = M_b - H\,y_s$:

• $a\le 10$: $M = 0.8a - \dfrac{a}{10}(3.2) = 0.48a$ (peak $+4.8$ m at $a=10$)
• $10\le a\le 25$: $M = 0.2(50-a) - 0.32a = 10 - 0.52a$ ($-3.0$ m at $a = 25$)
• $a\ge 25$: $M = -0.12\,(50-a)$

The ILD is $0 \to +4.8$ (at 10 m) $\to -3.0$ (at 25 m) $\to 0$.

$$M_{max}^{+} = 80\times 4.8 = 384.0\ \text{kN m (load at the section)}, \qquad M_{max}^{-} = 80\times(-3.0) = -240.0\ \text{kN m (load at mid-span)}$$

BMD and SFD (load at 10 m, giving the maximum positive moment)

With $W$ at 10 m: $R_A = 64$ kN, $R_B = 16$ kN, $H_L = \dfrac{80\times 10}{10} = 80$ kN.

$$M(x) = 64x - 80\times 0.008x(50-x) = 32x + 0.64x^2 \ (x\le 10), \qquad M(x) = 800 - 48x + 0.64x^2 \ (x\ge 10)$$ $$V(x) = V_b - H_L\tan\alpha_x,\quad \tan\alpha_x = 0.4 - 0.016x$$

 BMD: A(0) -> +384 at x=10 -> 0 at hinge C (x=25) -> -100 at x=37.5 -> 0 at B
      sagging from A to C, hogging from C to B (parabolic segments)
 SFD: +32 at A -> +44.8 at 10-  | jump of -80 |  -35.2 at 10+ -> -16 at C -> +16 at B

Answer: $T_{max} = 753.9$ kN; maximum BM at 10 m = $+384.0$ kN m (load at the section); at the same section the minimum is $-240.0$ kN m with the load at mid-span.

Data: span $L = 50$ m; the left support is 3 m below the right one; the lowest point of the cable is 5 m below the lower (left) support. So the heights above the lowest point are $h_1 = 5$ m (left) and $h_2 = 5 + 3 = 8$ m (right). UDL $w = 15$ kN/m on the horizontal span. The cable is a parabola with its vertex at the lowest point.

Horizontal thrust

With $y = \dfrac{w x^2}{2H}$ from the lowest point and $x_1 + x_2 = L$:

$$H = \frac{wL^2}{2\left(\sqrt{h_1}+\sqrt{h_2}\right)^2} = \frac{15\times 50^2}{2\left(\sqrt5+\sqrt8\right)^2} = \frac{37500}{2\times(5.0645)^2} = 731.0\ \text{kN}$$

Position of the lowest point

$$x_1 = \frac{L\sqrt{h_1}}{\sqrt{h_1}+\sqrt{h_2}} = 22.08\ \text{m (from the left support)}, \qquad x_2 = 50 - x_1 = 27.92\ \text{m}$$

Check: $\dfrac{w x_1^2}{2H} = 5.000$ m and $\dfrac{w x_2^2}{2H} = 8.000$ m.

Maximum tension

The tension is greatest at the higher (right) support:

$$T_{2} = \sqrt{H^2 + (w x_2)^2} = \sqrt{731.0^2 + (15\times 27.92)^2} = 842.5\ \text{kN}$$

(At the left support, $T_1 = \sqrt{H^2 + (w x_1)^2} = 802.5$ kN.)

Length of the cable

For each portion from the lowest point, with $u = \dfrac{w x}{H}$:

$$s = \frac{H}{w}\cdot\frac{u\sqrt{1+u^2} + \sinh^{-1}u}{2}$$

Left portion: $u_1 = 0.4530$, $s_1 = 22.81$ m. Right portion: $u_2 = 0.5730$, $s_2 = 29.38$ m.

$$S = s_1 + s_2 = 52.19\ \text{m}$$

(Approximation: $s \approx x\left[1 + \tfrac23\left(\tfrac{h}{x}\right)^2\right]$ gives 52.28 m.)

Answer: $H = 731.0$ kN; maximum tension $= 842.5$ kN (at the higher support); length of the cable $= 52.19$ m.

Data: three-hinged stiffening girder (hinges at both ends and at mid-span), span $L = 100$ m, loads 200 kN at 25 m and 300 kN at 50 m from the left end; cable dip $d = 10$ m. The cable is parabolic: $y = \dfrac{4d\,x(L-x)}{L^2} = 0.004\,x\,(100-x)$, slope $\tan\alpha = \dfrac{4d\,(L-2x)}{L^2}$.

Horizontal pull in the cable

Reaction of the simple beam (taking moments about the right end):

$$R_A = \frac{200\times 75 + 300\times 50}{100} = 300.0\ \text{kN}$$

Moment at the mid-span hinge $C$ of the simple beam:

$$M_C = 300\times 50 - 200\times 25 = 10000\ \text{kN m}$$

The girder has a hinge at $C$, so the cable thrust makes the net moment there zero:

$$H\,d = M_C \Rightarrow H = \frac{10000}{10} = 1000\ \text{kN}$$

Shear force and bending moment at 30 m from the left

At $x = 30$ m: $y = 0.004\times 30\times 70 = 8.40$ m; $\tan\alpha = 0.004\,(100-60) = 0.16$.

$$M_b = 300\times 30 - 200\times 5 = 8000\ \text{kN m}, \qquad V_b = 300 - 200 = 100\ \text{kN}$$ $$\begin{aligned} M &= M_b - H\,y = 8000 - 1000\times 8.40 = -400\ \text{kN m}\\ V &= V_b - H\tan\alpha = 100 - 1000\times 0.16 = -60\ \text{kN} \end{aligned}$$

Maximum tension in the cable and its slope

The tension is greatest at the supports, where the slope is $\tan\alpha_s = \dfrac{4d}{L} = 0.4$:

$$\alpha_s = \tan^{-1}0.4 = 21.80^\circ, \qquad V_s = H\tan\alpha_s = 400\ \text{kN}$$ $$T_{max} = \sqrt{H^2 + V_s^2} = \sqrt{1000^2 + 400^2} = 1077.0\ \text{kN}$$

Answer: at 30 m from the left, $V = -60$ kN and $M = -400$ kN m (hogging); maximum cable tension $= 1077.0$ kN at the supports, inclined at 21.80° to the horizontal.