Deflection

The elastic line is calculated, considering separately the uniform load p and the concentrated load P. The left end of the bar is considered to be the origin of coordinate z (local coordinate system xyz, where z is the bar axis).

For uniform load p along entire length L :

The basic equation of elastic line is solved in two steps:

Step 1

Step 2

Maximum deflection occurs at position z=L and is equal to:

For concentrated load P applied at distance c :

For 0≤z≤c:

Step 1

Step 2

For c≤z≤L:

Step 1

Step 2

Maximum deflection occurs at position z=L and is equal to:

In the special case where the concentrated load P is applied at the tip of the cantilever, e.g. c=L, maximum deflection is

Notice:

· When the cantilever slab is in continuity with an adjacent slab, its deflection increases by the linear deflection due to the rotation of their common edge.

Example

Cantilever slab with thickness h=200 mm, length L=2.00 m and concrete C40/50, is subjected to covering load g_επ=1.0 kN/m2, live load q=5.0 kN/m2 and concentrated load G=4.0 kN/m applied at the end.

Calculate bending moments, shear forces and deflections in ultimate limit state.

Solution:

Loads:

Consider strip width b=1.00 m.

Self-weight:

g_o=γ (b (h=25. 0kN/m3 (1.00m (0.20m=5.0 kN/m

Covering load:

g_e=1.0kN/m2·1.00m=1.0 kN/m

Total distributed dead load:

g=g_o+g_e=5.0+1.0=6.0 kN/m

Dead concentrated load:

G=4.0kN/m·1.00m=4.0 kN

Live load:

q=5.0kN/m2·1.00m=5.0 kN/m

Design loads

p=γ_g·g+γ_q·q=1.35x6.0+1.50x5.0=15.6 kN/m

P=γ_g·G=1.35x4.0kN=5.4 kN

Design shear forces

V_A=P+p·L=5.4+15.6x2.00=36.6 kN

V_k=V_A-p·L=36.6-15.6x2.00=5.4 kN

Design bending moments

M_A=-P·L-p·L²/2=-5.4x2.00-15.6x2.002/2=-42.0 kNm

M_k=0

Deflections

For C40/50 it is E=35.2x10⁹ Pa [EC2, 3.1.2] and from volume C' §1.1 a

I=b·h³/12=1.00x0.203/12=6.67x10-4 m⁴

E·I=35.2x10⁹N/m²x6.67x10^-4m⁴=234.8x10⁵ N·m²

Due to uniformly distributed load p:

Due to concentrated load P:

In total: y_max=y_max,1+y_max,2=1.33+0.61=1.94 mm

Note:

The calculated deflection corresponds to the ultimate limit state. The deflection for serviceability limit state, would result considering the respective serviceability loads with values e.g. p_s=1.0g+1.0q=1.0x6.0+1.0x5.0=11.0 kN/m and P_s=1.0G= 1.0x4.0=4.0 kN. Thus, the total deflection would be y_s=1.33x(11.0/15.6)+0.61x(4.0/5.4)= 1.39 mm, of course, ignoring creep effects.

One-way slabs