11.3 Cables, Trusses, and Internal Forces in Members

11.3 Cables, Trusses, and Internal Forces in Members

Civil engineering systems rely on skeletal structural frameworks and tensile elements to span significant distances efficiently. Trusses, cables, and frames are core topics on the PRC Civil Engineering Licensure Examination (CELE). Understanding how applied loads translate into internal axial thrusts, cable tensions, and multi-force frame reactions is essential for structural analysis, bridge design, and temporary falsework engineering.

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Planar Trusses and Ideal Truss Assumptions

A truss is a structural framework composed of slender members joined together at their ends to form a rigid assembly of triangles. The classical ideal truss assumptions are:

1. Frictionless Pin Connections: All members are connected at their ends by smooth, frictionless spherical or cylindrical pins.
2. Joint Loading Only: All external design loads and support reactions are applied exclusively at the panel joints (pin nodes).
3. Centroidal Concurrency: The centroidal axes of all framing members at any given joint intersect at a single concurrent point.
4. Weightless Members: The self-weight of members is negligible compared to live and dead service loads (or half of each member's weight is allocated to its two end nodes).

Under these ideal assumptions, bending moments and transverse shear forces are identically zero. Every member behaves as a two-force member, carrying purely axial forces: tension ($T > 0$) pulling away from the joint, or compression ($C < 0$) pushing toward the joint.

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Classification of Coplanar Trusses

A planar truss containing $m$ members, $j$ joints, and supported by $r$ independent external reaction components is classified by comparing total unknowns to equilibrium equations:

$\text{Total Unknowns} = m + r, \qquad \text{Total Equations} = 2j$

For a standard simple planar truss with three external determinate reactions ($r = 3$), the internal determinacy criterion simplifies to: $m = 2j - 3$

Stability Traps

Satisfying $m + r = 2j$ is a necessary but not sufficient condition for stability:

• External Instability: Occurs if support reactions are all parallel or concurrent.
• Internal Instability: Occurs if members are arranged improperly (e.g., forming an unbraced four-bar rectangular mechanism in one panel while over-stiffening another panel).

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Visual Identification of Zero-Force Members

Identifying zero-force members by inspection prior to calculation dramatically reduces arithmetic and prevents algebraic errors on board exams.

Case 1: Two Non-Collinear Members at an Unloaded Joint

If only two non-collinear members meet at a joint and no external load or reaction acts at that joint, both members are zero-force members. $\sum F_y' = 0 \implies F_1 \sin \theta = 0 \implies F_1 = 0$ $\sum F_x' = 0 \implies F_2 = 0$

          Member 1
         / 
        /  θ
   (Joint A)-------- Member 2      ===>  F_1 = 0 and F_2 = 0
    (No load)

Case 2: Three Members (Two Collinear) at an Unloaded Joint

If three members meet at a joint with no external load or reaction, and two of the members are collinear, the non-collinear third member is a zero-force member, and the two collinear members carry equal forces. $\sum F_y = 0 \implies F_3 = 0, \qquad \sum F_x = 0 \implies F_1 = F_2$

               Member 3 (F3 = 0)
                     |
                     v
   Member 1 ---- (Joint B) ---- Member 2    ===>  F_3 = 0, F_1 = F_2
                   (No load)

Case 3: Two Non-Collinear Members with Collinear Applied Load

If two non-collinear members meet at a joint subjected to an external load $P$ that is collinear with one of the members, the other member carries zero force ($F_2 = 0$), and the collinear member balances the load ($F_1 = P$).

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Methods of Truss Analysis

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Flexible Suspension Cables

Cables are lightweight, flexible structural elements designed to span long distances in suspension bridges, transmission lines, and guyed masts. The classical flexible cable assumptions dictate:

1. Cables have zero flexural rigidity (zero bending stiffness: $M = 0$) and cannot resist transverse shear ($V = 0$).
2. Cables are in pure, continuous tension directed tangentially along the cable curve.

Cables Carrying Uniform Horizontal Loads (Parabolic Cables)

When a suspension bridge deck transmits a uniformly distributed gravity load $w$ per unit of horizontal span length (e.g., in $\text{kN/m}$ along the $x$-axis):

       y ^                     w (kN/m horizontal)
         |                vvvvvvvvvvvvvvvvvvv
   Support A            Vertex (T0)              Support B
        \                  *                      /
         \_              _/' \_                _/
           \___________/'       '\___________/
         | <------- L/2 -------> | <------ L/2 ------> |

Taking the lowest point (vertex) as the coordinate origin $(0, 0)$ where the cable slope is horizontal:

• Minimum Tension ($T_0$): Occurs at the vertex ($x = 0$) and acts purely horizontal.
• Governing Differential Equation: $\frac{dy}{dx} = \frac{w x}{T_0} \implies y = \frac{w x^2}{2 T_0}$ The cable assumes a parabolic geometry.

For a symmetric cable of total horizontal span $L$ and maximum central sag $h$ at midspan ($x = L/2, y = h$): $h = \frac{w (L/2)^2}{2 T_0} = \frac{w L^2}{8 T_0} \implies T_0 = \frac{w L^2}{8 h}$

• Tension at Any Distance $x$: $T = \sqrt{T_0^2 + (w x)^2} = T_0 \sqrt{1 + \left( \frac{w x}{T_0} \right)^2} = T_0 \sqrt{1 + \left( \frac{2y}{x} \right)^2}$

• Maximum Tension ($T_{\max}$): Occurs at the tower supports ($x = \pm L/2$): $T_{\max} = \sqrt{T_0^2 + \left( \frac{w L}{2} \right)^2} = T_0 \sqrt{1 + \left( \frac{4h}{L} \right)^2}$

• Cable Arc Length ($S$): Evaluated by series expansion for practical flat sags ($h/L \le 0.1$): $S = \int_{-L/2}^{L/2} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx \approx L \left( 1 + \frac{8}{3} \left( \frac{h}{L} \right)^2 - \frac{32}{5} \left( \frac{h}{L} \right)^4 + \dots \right) \approx L + \frac{8 h^2}{3 L}$

Cables Carrying Uniform Self-Weight (Catenary Cables)

When a heavy cable (such as an electrical transmission conductor) hangs freely under its own uniform weight $w$ per unit of arc length (along $s$):

• Governing Equation: $y = c \cosh\left( \frac{x}{c} \right) - c \qquad \left( \text{where } c = \frac{T_0}{w} \right)$
• Fundamental Catenary Relations: $s = c \sinh\left( \frac{x}{c} \right), \qquad y^2 + 2 c y = s^2$ If the vertical axis origin is shifted down by distance $c$ ($y' = y + c = c \cosh(x/c)$): $T = w y' = w (y + c) = T_0 + w y$ The tension at any elevation equals the horizontal tension $T_0$ plus the weight of a vertical column of cable extending to that point!

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Frames and Machines: Multi-Force Members

Unlike ideal trusses, frames and machines are structural assemblies that contain at least one multi-force member (a member acted upon by forces at three or more points, or carrying distributed loads/applied couples).

• Frames: Rigid, stationary structures designed to support external service loads.
• Machines: Structures designed to transmit or modify input forces (contain moving components).

Analysis Strategy for Frames

1. External Equilibrium: Treat the entire assembly as a single rigid body to find external reactions (if determinate).
2. Dismantle the Assembly: Disassemble the frame into individual member free-body diagrams.
3. Apply Newton's Third Law: At every internal connecting pin, the reactive forces exerted between members are equal in magnitude and opposite in direction ($B_{x1} = -B_{x2}, B_{y1} = -B_{y2}$).
4. Apply Equations of Statics: Solve $\sum F_x = 0, \sum F_y = 0, \sum M = 0$ for each individual member.

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Step-by-Step Worked Problem Examples

Worked Example 1: Method of Sections for a Bridge Truss

Problem: A simply supported Warren bridge truss spans $L = 18.0\text{ m}$, consisting of six panels of $3.0\text{ m}$ each with a constant truss height $h = 4.0\text{ m}$. Vertical downward loads of $P = 60.0\text{ kN}$ act at panel joints $L_1, L_2, L_3, L_4, L_5$. The vertical reactions at supports $L_0$ and $L_6$ are each $R = 150.0\text{ kN} \uparrow$. Determine the axial force in the top chord member $U_2 U_3$.

          U1       U2       U3       U4       U5
           *--------*--------*--------*--------*
          / \      / \      / \      / \      / \
         /   \    /   \    /   \    /   \    /   \
        *-----*--*-----*--*-----*--*-----*--*-----*
       L0    L1       L2       L3       L4       L5    L6
       ^  (60kN)    (60kN)   (60kN)   (60kN)   (60kN)  ^
    150 kN                                          150 kN

Solution:

Step 1: Pass an Imaginary Cutting Section Pass a vertical cutting plane through top chord $U_2 U_3$, diagonal $U_2 L_3$, and bottom chord $L_2 L_3$. Isolate the left portion of the truss ($L_0$ to section cut).

Step 2: Identify External Forces on Left Portion

• Upward reaction at $L_0$: $R = 150.0\text{ kN} \uparrow$ at $x = 0\text{ m}$.
• Downward panel load at $L_1$: $P_1 = 60.0\text{ kN} \downarrow$ at $x = 3.0\text{ m}$.
• Downward panel load at $L_2$: $P_2 = 60.0\text{ kN} \downarrow$ at $x = 6.0\text{ m}$.
• Cut member forces: $F_{U2U3}$ (top chord), $F_{U2L3}$ (diagonal), $F_{L2L3}$ (bottom chord).

Step 3: Choose Strategic Moment Center Notice that cut diagonal $F_{U2L3}$ and cut bottom chord $F_{L2L3}$ both pass through joint $L_3$ ($x = 9.0\text{ m}, y = 0\text{ m}$). Taking moments about joint $L_3$ eliminates both forces completely!

Step 4: Formulate and Solve Moment Equilibrium Assume $F_{U2U3}$ is in tension (pulling to the right along $y = 4.0\text{ m}$): $\sum M_{L3} = 0 \quad (\text{CCW } +)$ $-(R \times 9.0) + (P_1 \times 6.0) + (P_2 \times 3.0) - (F_{U2U3} \times 4.0) = 0$ $-(150.0 \times 9.0) + (60.0 \times 6.0) + (60.0 \times 3.0) - 4.0 F_{U2U3} = 0$ $-1350.0 + 360.0 + 180.0 - 4.0 F_{U2U3} = 0$ $-810.0 - 4.0 F_{U2U3} = 0$ $4.0 F_{U2U3} = -810.0$ $F_{U2U3} = -202.5\text{ kN}$ The negative sign indicates compression: $F_{U2U3} = 202.5\text{ kN} \quad (\text{Compression})$

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Worked Example 2: Suspension Cable Analysis

Problem: A main suspension bridge cable spans $L = 120.0\text{ m}$ between tower pylons at equal elevations. The cable supports a uniform deck load of $w = 25.0\text{ kN/m}$ across the horizontal span. The central sag is designed at $h = 10.0\text{ m}$.

1. Calculate the minimum horizontal cable tension $T_0$.
2. Calculate the maximum cable tension $T_{\max}$ at the tower supports.
3. Estimate the total required cable arc length $S$.

Solution:

Step 1: Compute Minimum Horizontal Cable Tension ($T_0$) $T_0 = \frac{w L^2}{8 h} = \frac{25.0\text{ kN/m} \times (120.0\text{ m})^2}{8 \times 10.0\text{ m}} = \frac{25.0 \times 14,400}{80.0} = \frac{360,000}{80.0} = 4,500.0\text{ kN}$

Step 2: Compute Maximum Cable Tension ($T_{\max}$) The vertical reaction at each support is half the total suspended weight: $V_{\max} = \frac{w L}{2} = \frac{25.0 \times 120.0}{2} = 1,500.0\text{ kN}$ The maximum tension occurs at the towers: $T_{\max} = \sqrt{T_0^2 + V_{\max}^2} = \sqrt{(4,500.0)^2 + (1,500.0)^2} = \sqrt{20,250,000 + 2,250,000}$ $T_{\max} = \sqrt{22,500,000} = 4,743.42\text{ kN} \approx 4,743.4\text{ kN}$

Step 3: Estimate Cable Arc Length ($S$) Using the parabolic series approximation: $S \approx L + \frac{8 h^2}{3 L} = 120.0 + \frac{8 \times (10.0)^2}{3 \times 120.0} = 120.0 + \frac{800.0}{360.0} = 120.0 + 2.22\text{ m} = 122.22\text{ m}$

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