Suspension & Portal Frames

Cables as structural members

A cable is a flexible member that can carry only tension — it has no bending or compression resistance. Under any load it takes the shape in which every part is in pure tension (the funicular shape). Cables are used in suspension and cable-stayed bridges, ropeways, transmission lines, guyed masts and cable roofs.

Assumptions in elementary analysis: the cable is perfectly flexible and inextensible (or its elongation is small), self-weight is either neglected or treated as a UDL, and loads are vertical.

Cable under point loads

Between loads the cable is straight. At each load point, equilibrium gives a polygon. Because loads are vertical, the horizontal component of tension $H$ is the same everywhere in the cable.

For supports at the same level, the reactions follow from treating the cable like a beam, and $H$ follows from one known sag.

General cable theorem

It is the arch relationship with $M=0$: a cable is an arch turned upside down that can only be in tension.

Cable under a UDL (parabolic cable)

A UDL $w$ per unit horizontal length (e.g. the deck of a suspension bridge, which is much heavier than the cable) makes the cable a parabola. With span $L$ and central sag $h$ (supports at the same level):

Tension is minimum at the lowest point ($=H$) and maximum at the supports.

A cable hanging under its own weight per unit length of cable forms a catenary; for small sag–span ratios the parabola is an excellent approximation.

Supports at different levels

Let the lowest point be at horizontal distances $x_1$ and $x_2$ from the supports, with sags $h_1$ and $h_2$ below them ($x_1+x_2=L$). Each side behaves like half a parabola:

Vertical reactions: $V_1=w{x}_1$, $V_2=w{x}_2$.

Length of a parabolic cable

For supports at the same level:

Each half of a cable with unequal supports: ${\ell }_i\approx x_i+\frac{2h_i^2}{3x_i}$.

Change of sag with length or temperature

From $\ell =L+8h^2/3L$: $\delta \ell =\frac{16h}{3L}\delta h$ → $\delta h=\frac{3L}{16h}\delta \ell$.

A temperature rise $T$ lengthens the cable by $\delta \ell =\alpha T\ell \approx \alpha TL$, which increases the sag and reduces $H$. Elastic stretch under tension does the same.

Why approximate analysis

Exact analysis of a multi-storey, multi-bay frame is highly indeterminate. Approximate methods are still valuable to:

• make preliminary sizing before computer analysis;
• check computer output for gross errors;
• understand how loads flow through a frame.

Each method introduces enough assumptions (usually locations of zero moment, or distribution of shear or axial force) to make the structure determinate.

Gravity (vertical) loads

Assumed points of inflexion

For a beam in a building frame under UDL:

• If the ends were fixed, the points of contraflexure would be at $0.211L$ from each end.
• If the ends were pinned, the moment would be zero at the ends.
• Actual beam ends are partly restrained, so a common assumption is inflexion points at about $0.1L$ from each end, and zero axial force in the beam.

The beam then becomes a simply supported central segment of length $0.8L$ on two cantilevers of $0.1L$:

Substitute frame method

To find beam moments at one floor, analyse only that floor's beams together with the columns immediately above and below, with the far ends of the columns fixed. Use moment distribution with two or three cycles. This is the approach suggested in codes for regular frames under gravity load, where pattern loading (live load on alternate or adjacent spans) is also considered for maximum span and support moments.

Two-cycle method

A shortened moment distribution — two cycles of balancing and carry-over — gives beam moments within engineering accuracy for regular frames.

Lateral loads (wind, earthquake)

Under horizontal loads at floor levels, columns and beams of a regular frame bend in double curvature, with points of inflexion near mid-height of columns and mid-span of beams.

Portal method

Suitable for low-rise frames (height smaller than width), where behaviour is dominated by shear (frame racking).

Steps

1. Storey shear $V$ = sum of lateral loads above the mid-height of that storey. With $n$ bays (so $n-1$ interior columns): exterior column shear $=V/2n$; interior column shear $=V/n$.
2. Column end moments $=$ shear $\times h/2$ (both ends, opposite sense).
3. Beam end moments from joint equilibrium: at each joint, beam moments balance the column moments above and below.
4. Beam shears $=2M_{beamend}/L$ (inflexion at mid-span).
5. Column axial forces from beam shears; exterior columns carry the axial force, interior columns carry only the difference (zero for equal bays).

Cantilever method

Suitable for tall, slender frames (height large compared with width), which bend like a vertical cantilever.

Steps

1. Locate the centroid of column areas at each storey.
2. Take moments of lateral loads above a section through column inflexion points about the centroid; this equals $\sum F_i{x}_i$ where $F_i\propto A_i{x}_i$. Solve for column axial forces.
3. Beam shears from vertical equilibrium of joints (difference of column axial forces).
4. Beam end moments $=$ beam shear $\times L/2$.
5. Column shears and moments from joint equilibrium.

Factor method (outline)

The factor method (Wilbur) uses relative stiffnesses of beams and columns at each joint to estimate column and girder moments. It is more accurate than the portal and cantilever methods for frames with varying member sizes but more laborious.