When the loading is uniformly distributed horizontally the cable is analyzed as.

When the loading is uniformly distributed horizontally the cable is analyzed as.

• A. Parabolic cable
• B. Catenary
• C. Projectile
• D. Rotation

Answer: (A)

When the load is distributed uniformly per horizontal length, the cable assumes a parabolic shape. The key idea is that the horizontal tension along the cable remains essentially constant, so the vertical component of tension must balance a constant load per unit horizontal length. If we denote the horizontal tension by H and the uniform horizontal load by w_h, the vertical equilibrium gives dT_y/dx = w_h. Since T_y = H dy/dx, differentiating gives H d^2y/dx^2 = w_h, a constant. Integrating twice yields a quadratic function of x, i.e., a parabola. For a symmetric span with ends at the same level, the deflection is y(x) = (w_h/(2H)) x (L − x), with maximum sag at midspan y_max = w_h L^2/(8H). Note that if the load were distributed along the actual length of the cable (its own weight), the curve is a catenary instead.

When the load is distributed uniformly per horizontal length, the cable assumes a parabolic shape. The key idea is that the horizontal tension along the cable remains essentially constant, so the vertical component of tension must balance a constant load per unit horizontal length. If we denote the horizontal tension by H and the uniform horizontal load by w_h, the vertical equilibrium gives dT_y/dx = w_h. Since T_y = H dy/dx, differentiating gives H d^2y/dx^2 = w_h, a constant. Integrating twice yields a quadratic function of x, i.e., a parabola. For a symmetric span with ends at the same level, the deflection is y(x) = (w_h/(2H)) x (L − x), with maximum sag at midspan y_max = w_h L^2/(8H). Note that if the load were distributed along the actual length of the cable (its own weight), the curve is a catenary instead.