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Example 13.1 : Page 400
Reinforced Concrete - Design Theory and Examples
Second Edition
T.J. Macginley & B.S. Choo

4 - appVersion 0.99.7822.147

4 appVersion 0.99.7822.147

t.0 0 time :

https://www.academia.edu/34894173/Reinforced_Concrete_Design_Theory_and_Examples

Example 13.1 Simplified analysis of concrete framed building—vertical load
The application of the various simplified methods of analysis given in section 3.2 of the code is shown in the following example.
(a) Specification
The cross-section of a reinforced concrete building is shown in Fig. 13.5(a). The frames are at 4.5 m centres, the length of the building is 36 m and the column bases are fixed. Preliminary sections for the beams and columns are shown in Fig. 13.5(b). The floor and roof slabs are designed to span one way between the frames. Longitudinal beams are provided between external columns of the roof and floor levels only. :

Frames are spaced at 4.5 m

This worked example shows only the analysis of the sub-frame shown. Hence, only Dead & Imposed loads on Beams BF & FK are considered.

S.frame 4.5 m * : L.building 36 m * : γ.cu 24 kN m 3^/ * : 31 line

Iterations -----> While LOOP based
on a given threshold value, to be defined at MAIN PROGRAM level

For column GF

col.FE_b 400 mm * : col.FE_h 500 mm * : 12 mat col.GF_b 300 mm * : col.GF_h 400 mm * : beam.FK_b 400 mm * : beam.FK_h 600 mm * : 22 mat 21 line

Define Number of Member Moments to evaluate N.members 16 :

L.BF 6 m * : L.FK 8 m * : 21 line

Loadings

Following load cases are required for the beam BFK for the braced frame.
But, only Case 1 is shown in this example.

For the floors

Floors D/L g.k_floor 6.2 kN m 2^/ * :

Case 1

1.4 G.k * 1.6 Q.k * +

on the whole beam

Case 2

1.4 G.k * 1.6 Q.k * +

on BF and

1.6 G.k *

on FK

Floors I/L q.k_floor 3.0 kN m 2^/ * :

Case 3

1.0 G.k *

on BF and

1.4 G.k * 1.6 Q.k * +

on FK

Loads on Beans

Dead loads on Beams

Imposed loads on Beams

G.k_floor g.k_floor S.frame * : kN m / 27.9

Q.k_floor q.k_floor S.frame * : kN m / 13.5

Calculation of Moments of Inertia

I.Col_FE 112 / col.FE_b * col.FE_h (3^* : mm 4^4.1667 109^*

I.Col_FG 112 / col.GF_b * col.GF_h (3^* : mm 4^1.6 109^*

I.beam_FK 112 / beam.FK_b * beam.FK_h (3^* : mm 4^7.2 109^*

I.beam_FB I.beam_FK : mm 4^7.2 109^*

Beam BF and FK : same cross section

Moment of Inertia I I.Col_FE I.Col_FG I.beam_FK I.beam_FB 41 mat : mm 4^4.1667 109^* 1.6 109^* 7.2 109^* 7.2 109^* 41 mat

LengthCols & Beams L 550040008000600041 mat mm * :

Stiffness K.FE K.FG K.FK K.FB 41 mat I L / vectorize : 105^mm 3^* 7.5758 491241 mat

(d) : Subframe analysis for braced frames

The subframe consists of the beams at first floor level and the columns above and below that level with ends fixed. The frame is analysed for the dead and imposed load cases given..
Member Stiffness of the subframe at B, F, K are as follows

Stiffnesses (Arrangement of K is very important)

Joint B

Joint F

Joint K

Group A, B, C, D

K.B K.FE K.FB K.FG 31 mat :

K.F K.FE K.FB K.FG K.FK 41 mat :

K.K K.FE K.FK K.FG 31 mat :

K K.B K.F K.K 31 mat :

Fixed End Moments

Case I

Case II (Not considered in this example)

The FEMs are for case1 (F1) fon both beams

F1 1.4 G.k_floor * 1.6 Q.k_floor * + : kN m / 60.66

F2 1.0 G.k_floor * : kN m / 27.9

FEM for span BF: Case I FEM.BF F1 L.BF 2^12 / * - : kN m * 181.98 -

FEM for span FK: Case I FEM.FK F1 L.FK 2^12 / * - : kN m * 323.52 -

FEM for FB: Case I FEM.FB FEM.BF - : kN m * 181.98

FEM for span KF: Case I FEM.KF FEM.FK - : kN m * 323.52

These programs (Yellow Color) are applicable to any Method

PROGRAM 1 : Find Nested DF for ALL JOINTS. Using vectorize function k Find_DF df k k sum / (vectorize : 11 line :

PROGRAM 2 : To replace zeros with blank spaces.(To improve readability in TABLE). Lots of zeros will clutter the appearance. M# format j 1 M# rows range : k 1 M# cols range : BB j k el M# j k el 0 ≡ M# j k el cases : BB 41 line :

PROGRAM 3 : To find column totals. (Last row in Mom. Dist Table) M# tot_BMM j 1 M# cols range : A# j el M# j col sum : 21 line :

PROGRAM 4 : To stack a nested matrix M Stack_M A M 1 el : j 2 M rows range A A M j el stack : 11 line for A eval 31 line :

PROGRAM 5: To Group LOAD, DF, FEM, BMM & BMM from MAIN PROGRAM PROGRAM 4 stacks nested 'd'. PROGRAM 3 calc all Col Totals (A3).A1 removes the last CO row of output from MAIN PROGRAM. ld d fem# df Group d d Stack_M kN m * / : A1 d 1 d rows 1 - (range 1 d cols range el : A2 fem# kN m * / A1 stack : A3 A2 tot_BMM transpose : ld df transpose A2 A3 stack eval 51 line :

Define variables for use in PROGRAM 6

P1 Bal CO 21 mat : P2 DF FEM 21 mat : P3 Final BM 11 mat : 31 line

PROGRAM 6 : Create Left Stub for Table n Create_LS B P1 : j 1 n 1 - (range B B P1 stack : 11 line for B B 1 B rows 1 - (range 1 B cols range el : Load P2 B P3 stack 41 line :

Following PROGRAMS apply ONLY to the type of Example used

Header for Results Table Hd AB BA BF BC CB EF FE FB FG FK GF JK KJ KF KL LK 116 mat :

Hd cols 16

Should be equal to

N.members 16

Dist. Factors

PROGRAM 7: Call PROGRAM 1 to find DF Array (Nested)K = Stiffness DF K Find_DF : 0.32 0.51 0.17 31 mat 0.23 0.37 0.12 0.28 41 mat 0.37 0.44 0.19 31 mat 31 mat

PROGRAM 8DFV vector; ONLY to display in the Final Result & Table DFV 0 DF 1 el 00 DF 2 el 00 DF 3 el 0 stack :

DFV transpose cols 16

DFV transpose 0 0.32 0.51 0.17 00 0.23 0.37 0.12 0.28 00 0.37 0.44 0.19 0116 mat

PROGRAM 9 : Balancing Moments M# n# Bal_M Z 1 n# el 0 : Z 2 el Z 3 el Z 4 el 13 mat M# - 1 el DF 1 el transpose * : Z 7 el Z 8 el Z 9 el Z 10 el 14 mat M# 2 el M# 3 el + (DF 2 el transpose * - : Z 13 el Z 14 el Z 15 el 13 mat M# - 4 el DF 3 el transpose * : Z eval 51 line :

PROGRAM 10 : Carry Over Moments M BL CO_M M 1 el 0.5 BL 2 el * : M 3 el 0.5 BL 8 el * : M 5 el 0.5 BL 4 el * : M 6 el 0.5 BL 7 el * : M 8 el 0.5 BL 3 el * : M 10 el 0.5 BL 14 el * : M 11 el 0.5 BL 9 el * : M 12 el 0.5 BL 13 el * : M 14 el 0.5 BL 10 el * : M 16 el 0.5 BL 15 el * : M eval 111 line :

PROGRAM 11 : Load arrangement M Find_LD M 3 el F1 : M 8 el F1 : M 10 el F1 : M 14 el F1 : M kN m / / eval 51 line :

MAIN PROGRAM : Moment Distribution. mem = Number of members;  Δba = Threshold value mem Δba BMM fem 1 mem el 0 : co 1 mem el 0 : BA 1 mem el 0 : LD 1 mem el 0 : 41 mat LDD LD Find_LD : fem 3 el FEM.BF : fem 8 el fem 3 el - : fem 10 el FEM.FK : fem 14 el fem 10 el - : ΔBal Δba kN * m * : --------------- j 1 : BA j el fem 3 el fem 8 el fem 10 el fem 14 el 41 mat mem Bal_M : CO j el co BA j el CO_M : D j el BA j el CO j el stack : tmp BA j el : tmp abs vectorize Max ΔBal ≥ ------------------ j j 1 + : BA j el co 3 el co 8 el co 10 el co 14 el 41 mat mem Bal_M : CO j el co BA j el CO_M : tmp BA j el : D j el BA j el CO j el stack : 61 line while A LDD D fem DFV Group : B j Create_LS : A B j 31 mat eval 171 line :

MAIN PROGRAM

Initialize input arrays

Load arrangement : Program 11

FEM Assign

Assign Unit to Δba (Threshold Value)

Ex: Largest Numerical Value of an Array

j = 1 Iteration starts

XX 3 - 51 - 10 - 14 mat :

XX Max 5

Bal Program 9

XX abs vectorize 3511014 mat

XX abs vectorize Max 10

CO Program 10

Stack Bal & CO

Assign BA array to tmp.
Can't vectorize Max(BAj) directly.

Vectorize abs (tmp) to find NUMERICALLY LARGEST value of each Bal array

j : Iteration continue

Bal Program 9

CO Program 10

Assign BA array to tmp. Can't use BAj directly.

Group Loads, DF, FEM, Bal, CO, Totals Program 5

Left Stub for TABLE Program 6

j = N umber of Iterations

Already defined N.members 16

Define Threshold value ΔVal 0.1 :

Results from MAIN PROGRAM Ans N.members ΔVal BMM :

Moment Distrb. MD Ans 1 el :

Left Stub for Table LS Ans 2 el :

Iterations Iter Ans 3 el : 6

Format "MD" for Table CC MD format :

MD 00 60.66 0000 60.66 0 60.66 000 60.66 000 0.32 0.51 0.17 00 0.23 0.37 0.12 0.28 00 0.37 0.44 0.19 000 181.98 - 0000 181.98 0 323.52 - 000 323.52 000 58.48 92.63 30.88 00 32.92 52.14 17.38 39.1 00 119.12 - 141.51 - 62.89 - 0 29.24 0 26.07 0 15.44 16.46 0 46.31 0 70.76 - 8.69 59.56 - 0 19.55 0 31.45 - 0 8.38 - 13.27 - 4.42 - 00 5.68 93 6.75 00 7.2 - 8.55 - 3.8 - 0 4.19 - 0 4.5 0 2.21 - 2.84 0 6.63 - 0 4.28 - 1.5 3.6 - 0 3.38 0 1.9 - 0 1.45 - 2.29 - 0.76 - 00 2.54 4.02 1.34 3.01 00 1.24 - 1.48 - 0.66 - 0 0.72 - 0 2.01 0 0.38 - 1.27 0 1.15 - 0 0.74 - 0.67 0.62 - 0 1.51 0 0.33 - 0 0.65 - 1.02 - 0.34 - 00 0.44 0.69 0.23 0.52 00 0.55 - 0.66 - 0.29 - 0 0.32 - 0 0.35 0 0.17 - 0.22 0 0.51 - 0 0.33 - 0.12 0.28 - 0 0.26 0 0.15 - 0 0.11 - 0.18 - 0.06 - 00 0.2 0.31 0.1 0.23 00 0.1 - 0.11 - 0.05 - 0 0.06 - 0 0.15 0 0.03 - 0.1 0 0.09 - 0 0.06 - 0.05 0.05 - 0 0.12 0 0.03 - 0 0.05 - 0.08 - 0.03 - 00 0.03 0.05 0.02 0.04 00 0.04 - 0.05 - 0.02 - 0 23.95 47.85 73.11 - 25.26 12.64 20.89 41.81 286.13 22.07 350.01 - 11.03 64.1 - 128.25 - 195.97 67.72 - 33.85 - 1516 mat

0 time t.0 - 0.14 s *

CC LS MD ΔVal XX BMM Find_LD CO_M Bal_M Hd Create_LS P3 Group Stack_M tot_BMM format Find_DF FEM.KF FEM.FB FEM.FK FEM.BF F2 F1 K.FB K.FK K.FG K.FE I.beam_FB I.beam_FK I.Col_FG I.Col_FE Q.k_floor G.k_floor q.k_floor g.k_floor L.FK L.BF beam.FK_h beam.FK_b col.GF_h col.GF_b col.FE_h col.FE_b γ.cu L.building S.frame t.0 CC LS MD Ans ΔVal BMM Find_LD CO_M Bal_M DFV DF Hd Create_LS P3 P2 P1 Group Stack_M tot_BMM format Find_DF FEM.KF FEM.FB FEM.FK FEM.BF F2 F1 K K.K K.F K.B K.FB K.FK K.FG K.FE L I I.beam_FB I.beam_FK I.Col_FG I.Col_FE Q.k_floor G.k_floor q.k_floor g.k_floor L.FK L.BF beam.FK_h beam.FK_b col.GF_h col.GF_b col.FE_h col.FE_b γ.cu L.building S.frame t.0 ΔVal FEM.KF FEM.FB FEM.FK FEM.BF F2 F1 K.FB K.FK K.FG K.FE I.beam_FB I.beam_FK I.Col_FG I.Col_FE Q.k_floor G.k_floor q.k_floor g.k_floor L.FK L.BF beam.FK_h beam.FK_b col.GF_h col.GF_b col.FE_h col.FE_b γ.cu L.building S.frame t.0

CC
		60.66					60.66		60.66				60.66
	0.32	0.51	0.17			0.23	0.37	0.12	0.28			0.37	0.44	0.19
		-181.98					181.98		-323.52				323.52
	58.48	92.63	30.88			32.92	52.14	17.38	39.1			-119.12	-141.51	-62.89
29.24		26.07		15.44	16.46		46.31		-70.76	8.69	-59.56		19.55		-31.45
	-8.38	-13.27	-4.42			5.68	9	3	6.75			-7.2	-8.55	-3.8
-4.19		4.5		-2.21	2.84		-6.63		-4.28	1.5	-3.6		3.38		-1.9
	-1.45	-2.29	-0.76			2.54	4.02	1.34	3.01			-1.24	-1.48	-0.66
-0.72		2.01		-0.38	1.27		-1.15		-0.74	0.67	-0.62		1.51		-0.33
	-0.65	-1.02	-0.34			0.44	0.69	0.23	0.52			-0.55	-0.66	-0.29
-0.32		0.35		-0.17	0.22		-0.51		-0.33	0.12	-0.28		0.26		-0.15
	-0.11	-0.18	-0.06			0.2	0.31	0.1	0.23			-0.1	-0.11	-0.05
-0.06		0.15		-0.03	0.1		-0.09		-0.06	0.05	-0.05		0.12		-0.03
	-0.05	-0.08	-0.03			0.03	0.05	0.02	0.04			-0.04	-0.05	-0.02
23.95	47.85	-73.11	25.26	12.64	20.89	41.81	286.13	22.07	-350.01	11.03	-64.1	-128.25	195.97	-67.72	-33.85
#
AB	BA	BF	BC	CB	EF	FE	FB	FG	FK	GF	JK	KJ	KF	KL	LK
Load
DF
FEM
Bal
CO
Bal
CO
Bal
CO
Bal
CO
Bal
CO
Bal
Final BM
#
#

Iter 6

0 time t.0 - 0.16 s *

Checking Balacing Moments of While Loop Calculations

Balancing Rows in MD : 4, 6, 8, 10, 12, 14.

MD rows 1 - 14

Last row of MD is Final BM

NUMERICALLY LARGEST VALUE of each BAL row of "MD" M# Check_Max_Bal A M# 4 M# rows 1 - (6 range 1 M# cols range el : k 1 A rows range B A k 1 A cols range el : C k el B abs vectorize Max : 21 line for C 31 line :

Extract Bal rows from MD matrix

Largest Value of each row of Bal matrix

M_Bal MD Check_Max_Bal : 141.51 13.27 4.02 1.02 0.31 0.08 61 mat

≤

ΔVal 0.1

Iter 6

0 time t.0 - 0.16 s *

---------------------------------------------------------------------------------------------------------------------------------------------------------------

Check for a Different Threshold value

New Threshold value ΔVal2 0.05 :

Results from MAIN PROGRAM Ans2 N.members ΔVal2 BMM :

Moment Distrb. MD2 Ans2 1 el :

Iterations Iter2 Ans2 3 el : 7

M_Bal2 MD2 Check_Max_Bal : 141.51 13.27 4.02 1.02 0.31 0.08 0.02 71 mat

≤

ΔVal2 0.05

Iter2 7

0 time t.0 - 0.27 s *