Table B4|www.BuildingHow.com

VOLUME A
The art of construction and detailing
Introduction
The structural frame
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  - Beams
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  - The behavior and reinforcement of a slab
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Tables
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Drawings
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Model (exemplary) construction

« Table B3 Table B5_1 »

Table B4

Limit stress resultants of continuous slab [*] Note This table also applies to continuous beams, although rarely met at earthquake resistant structures. as a function of the ratio g_d/p_d

g_d/p_d	g_d=0 p_d=q_d	Two-span beam									g_d=p_d q_d=0
g_d/p_d	0.0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.0
m₁	10.45	10.75	11.07	11.41	11.75	12.12	12.50	12.90	13.32	13.76	14.22
m_B	-8.00	-8.00	-8.00	-8.00	-8.00	-8.00	-8.00	-8.00	-8.00	-8.00	-8.00
p_1A	2.29	2.32	2.35	2.39	2.42	2.46	2.50	2.54	2.58	2.62	2.67
p_1B	-1.60	-1.60	-1.60	-8.00	-1.60	-1.60	-1.60	-1.60	-1.60	-1.60	-1.60

g_d/p_d	g_d=0 p_d=q_d	Three-span beam									g_d=p_d q_d=0
g_d/p_d	0.0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.0
m₁	9.88	10.10	10.33	10.57	10.82	11.07	11.34	11.61	11.9	12.19	12.50
m_B	-8.57	-8.70	-8.82	-8.96	-9.09	-9.23	-9.37	-9.52	-9.68	-9.84	-10.00
m₂	13.33	14.29	15.38	16.67	18.18	20.00	22.22	25.00	28.57	33.33	40.00
p_1A	2.22	2.25	2.27	2.30	2.33	2.35	2.38	2.41	2.44	2.47	2.50
p_1B	-1.62	-1.63	-1.63	-1.63	-1.64	-1.64	-1.65	-1.65	-1.66	-1.66	-1.67
p_2B	1.71	1.74	1.76	1.79	1.82	1.85	1.87	1.90	1.94	1.97	2.00

g_d/p_d	g_d=0 p_d=q_d	Four-span beam									g_d=p_d q_d=0
g_d/p_d	0.0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.0
m₁	10.04	10.28	10.53	10.80	11.07	11.36	11.65	11.96	12.28	12.61	12.96
m_B	-8.30	-8.39	-8.48	-8.58	-8.68	-8.78	-8.89	-9.00	-9.11	-9.22	-9.33
m₂	12.42	13.14	13.96	14.87	15.92	17.12	18.53	20.17	22.14	24.54	27.51
m_C	-9.33	-9.66	-10.00	-10.37	-10.77	-11.20	-11.67	-12.17	-12.73	-13.33	-14.00
p_1A	2.24	2.27	2.30	2.32	2.35	2.38	2.41	2.45	2.48	2.51	2.55
p_1B	-1.61	-1.61	-1.62	-1.62	-1.63	-1.63	-1.63	-1.64	-1.64	-1.64	-1.65
p_2B	1.66	1.68	1.70	1.72	1.74	1.76	1.78	1.80	1.82	1.84	1.87
p_2C	-1.75	-1.78	-1.82	-1.85	-1.89	-1.93	-1.97	-2.01	-2.06	-2.11	-2.15

g_d/p_d	g_d=0 p_d=q_d	Five-span beam									g_d=p_d q_d=0
g_d/p_d	0.0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.0
m₁	9.99	10.23	10.48	10.74	11.00	11.28	11.57	11.87	12.18	12.5	12.84
m_B	-8.36	-8.46	-8.57	-8.67	-8.78	-8. 89	-9.01	-9.13	-9.25	-9.37	-9.50
m₂	12.65	13.43	14.31	15.32	16.48	17.82	19.40	21.29	23.59	26.45	30.08
m_C	-8.99	-9.26	-9.54	-9.85	-10.17	-10.52	-10.89	-11.28	-11.71	-12.17	-12.67
m₃	11.69	12.26	12.88	13.57	14.34	15.20	16.17	17.27	18.54	20.00	21.71
p_1A	2.24	2.26	2.29	2.32	2.35	2.37	2.41	2.44	2.47	2.50	2.53
p_1B	-1.61	-1.62	-1.62	-1.63	-1.63	-1.63	-1.64	-1.64	-1.64	-1.65	-1.65
p_2B	1.67	1.69	1.71	1.73	1.76	1.78	1.80	1.83	1.85	1.87	1.90
p_2C	-1.73	-1.77	-1.80	-1.83	-1.87	-1.90	-1.94	-1.98	-2.02	-2.07	-2.11
p_3C	1.69	1.72	1.75	1.77	1.80	1.83	1.86	1.90	1.93	1.96	2.00

g_d/p_d	g_d=0 p_d=q_d	∞ -span beam									g_d=p_d q_d=0
g_d/p_d	0.0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.0
m_F	9.99	10.23	10.48	10.74	11.00	11.28	11.57	11.87	12.18	12.50	12.84
m_S	-8.36	-8.46	-8.57	-8.67	-8.78	-8.89	-9.01	-9.13	-9.25	-9.37	-9.50
p_FS	12.65	13.43	14.31	15.32	16.48	17.82	19.4	21.29	23.59	26.45	30.08

Example:

Given: Three-span continuous slab with L=5.0 m, with nominal loads g=6.5 kN/m and q=5.0 kN/m

Question: Calculate the maximum absolute values of bending moments and shear forces for ULS, at supports and spans, forγ_g =1.35 and γ_q=1.50.

Solution:

The design loads for ULS are equal to:

g_d=1.0×g, q_d=(1.00-γ_g)×g + γ_q×q = 0.35×g + 1.35×q
and p_d= g_d + q_d= γ_g×g + γ_q×q = 1.35×g + 1.50×q →

g_d=1.0×6.5=6.5 kN/m, q_d=0.35×6.5+1.5×5.0=9.775 kN/m p_d=1.35g+1.50q=1.35×6.5+1.50×5.0=16.275 kN/m

Using the table for three-span slabs yields:

g_d/p_d=6.5/16.275=0.40 → m₁=10.82, m_B=-9.09, m₂=18.18, p_1A=2.33, p _1B=-1.64, p_2B=1.82

V_01,max=p_d×L/p_1A=16.275×5.0/2.33=34.9 kN

V_10,min=p_d×L/p_1B=-16.275×5.0/1.64=-49.6 kN

V_12,max=p_d×L/p_2B=16.275×5.0/1.82=44.7 kN

M_01,max=p_d×L²/m₁=16.275×5.0²/10.82=37.6 kNm

M_1,min=p_d×L²/m₁=-16.275×5.0²/9.09=-44.8 kNm

M_12,max=p_d×L²/m₂=16.275×5.0²/18.18=22.4 kNm

« Table B3 Table B5_1 »