beamanal (metric) copie

7/28/2019 BEAMANAL (Metric) Copie

1/19

"BEAMANAL" --- SINGLE-SPAN and CONTINUOUS-SPAN BEAM ANALYSIS

Program Description:

"BEAMANAL" is a spreadsheet program written in MS-Excel for the purpose of analysis of either single-span or

continuous-span beams subjected to virtually any type of loading configuration. Four (4) types of single-span beams

and two (2) through (5) span, continuous-span beams, considered. Specifically, beam end reactions as well as themaximum moments and deflections are calculated. Plots of both the shear and moment diagrams are produced,

as well as a tabulation of the shear, moment, slope, and deflection for the beam or each individual span.

Note: this is a metric units version of the original "BEAMANAL.xls" spreadsheet workbook.

This program is a workbook consisting of three (3) worksheets, described as follows:

Worksheet Name Description

Doc This documentation sheet

Single-Span Beam Single-span beam analysis for simple, propped, fixed, & cantilever beams

Continuous-Span Beam Continuous-span beam analysis for 2 through 5 span beams

Program Assumptions and Limitations:

1. The following reference was used in the development of this program (see below):

"Modern Formulas for Statics and Dynamics, A Stress-and-Strain Approach"

by Walter D. Pilkey and Pin Yu Chang, McGraw-Hill Book Company (1978), pages 11 to 21.

2. This program uses the three (3) following assumptions as a basis for analysis:

a. Beams must be of constant cross section (E and I are constant for entire span length).

b. Deflections must not significantly alter the geometry of the problem.

c. Stress must remain within the "elastic" region.

3. On the beam or each individual span, this program will handle a full length uniform load and up to eight (8) partial

uniform, triangular, or trapezoidal loads, up to fifteen (15) point loads, and up to four (4) applied moments.

4. For single-span beams, this program always assumes a particular orientation for two (2) of the the four (4)

different types. Specifically, the fixed end of either a "propped" or "cantilever" beam is always assumed to be on

the right end of the beam.

5. This program will calculate the beam end vertical reactions and moment reactions (if applicable),the maximum positive moment and negative moment (if applicable), and the maximum negative deflection

and positive deflection (if applicable). The calculated values for the end reactions and maximum moments

and deflections are determined from dividing the beam into fifty (50) equal segments with fifty-one (51) points,

and including all of the point load and applied moment locations as well. (Note: the actual point of maximum

moment occurs where the shear = 0, or passes through zero, while the actual point of maximum deflection is

where the slope = 0.)

6. The user is given the ability to input two (2) specific locations from the left end of the beam to calculate the

shear, moment, slope, and deflection.

7. The user is also given the ability to select an AISC W, S, C, MC, or HSS (rectangular tube) shape to aide in

obtaining the X-axis moment of inertia for input for the purely analysis worksheets.

8. The plots of the shear and moment diagrams as well as the displayed tabulation of shear, moment, slope,

and deflection are based on the beam (or each individual span) being divided up into fifty (50) equal segments

with fifty-one (51) points.

9. For continuous-span beam of from two (2) through five (5) spans, this program utilizes the "Three-Moment

Equation Theory" and solves a system simultaneous equations to determine the support moments

10. This program contains numerous comment boxes which contain a wide variety of information including

explanations of input or output items, equations used, data tables, etc. (Note: presence of a comment box

is denoted by a red triangle in the upper right-hand corner of a cell. Merely move the mouse pointer to the

desired cell to view the contents of that particular "comment box".)


2/19

Formulas Used to Determine Shear, Moment, Slope, and Deflection in Single-Span Beams

For Uniform or Distributed Loads:

Loading functions for each uniform or distributed load evaluated at distance x = L from left end of beam:

FvL = -wb*(L-b-(L-e)) + -1/2*(we-wb)/(e-b)*((L-b)^2-(L-e)^2)+(we-wb)*(L-e)FmL = -wb/2*((L-b)^2-(L-e)^2) + -1/6*(we-wb)/(e-b)*((L-b)^3-(L-e)^3)+(we-wb)/2*(L-e) 2

FqL = -wb/(6*E*I)*((L-b)^3-(L-e)^3) + -1/(24*E*I)*(we-wb)/(e-b)*((L-b)^4-(L-e)^4)+(we-wb)/(6*E*I)*(L-e)^3

FDL = -wb/(24*E*I)*((L-b)^4-(L-e)^4) + -1/(120*E*I)*(we-wb)/(e-b)*((L-b)^5-(L-e)^5)+(we-wb)/(24*E*I)*(L-e)^4

Loading functions for each uniform or distributed load evaluated at distance = x from left end of beam:

If x >= e:

Fvx = -wb*(x-b-(x-e)) + -1/2*(we-wb)/(e-b)*((x-b)^2-(x-e)^2)+(we-wb)*(x-e)

Fmx = -wb/2*((x-b)^2-(x-e)^2) + -1/6*(we-wb)/(e-b)*((x-b)^3-(x-e)^3)+(we-wb)/2*(x-e) 2

Fqx = -wb/(6*E*I)*((x-b)^3-(x-e)^3) + -1/(24*E*I)*(we-wb)/(e-b)*((x-b)^4-(x-e)^4)+(we-wb)/(6*E*I)*(x-e)^3

FDx = -wb/(24*E*I)*((x-b)^4-(x-e)^4) + -1/(120*E*I)*(we-wb)/(e-b)*((x-b)^5-(x-e)^5)+(we-wb)/(24*E*I)*(x-e)^4

else if x >= b:

Fvx = -wb*(x-b) + -1/2*(we-wb)/(e-b)*(x-b)^2 else: Fvx = 0

Fmx = -wb/2*(x-b)^2 + -1/6*(we-wb)/(e-b)*(x-b)^3-(x-e)^3 else: Fmx = 0

Fqx = -wb/(6*E*I)*(x-b)^3 + -1/(24*E*I)*(we-wb)/(e-b)*(x-b)^4 else: Fqx = 0

FDx = -wb/(24*E*I)*(x-b)^4 + -1/(120*E*I)*(we-wb)/(e-b)*(x-b)^5 else: FDx = 0

For Point Loads:

Loading functions for each point load evaluated at distance x = L from left end of beam:

FvL = -P

FmL = -P*(L-a)

FqL = -P*(L-a)^2/(2*E*I)

FDL = P*(L-a)^3/(6*E*I)

Loading functions for each point load evaluated at distance = x from left end of beam:

If x > a:

Fvx = -P else: Fvx = 0

Fmx = -P*(x-a) else: Fmx = 0

Fqx = -P*(x-a)^2/(2*E*I) else: Fqx = 0

FDx = P*(x-a)^3/(6*E*I) else: FDx = 0

For Applied Moments:

Loading functions for each applied moment evaluated at distance x = L from left end of beam:

FvL = 0

FmL = -M

FqL = -M*(L-c)/(E*I)

FDL = M*(L-c)^2/(2*E*I)

Loading functions for each applied moment evaluated at distance = x from left end of beam:

If x >= c:

Fvx = 0 else: Fvx = 0

Fmx = -M else: Fmx = 0

Fqx = -M*(x-c)/(E*I) else: Fqx = 0

FDx = M*(x-c)^2/(2*E*I) else: FDx = 0

(continued)


3/19

Formulas Used to Determine Shear, Moment, Slope, and Deflection (continued)

Initial summation values at left end (x = 0) for shear, moment, slope, and deflection:

Simple beam:

Vo =-1/L*S(FmL)

Mo = 0

qo = 1/L*S(FDL)+L/(6*E*I)*S(FmL)

Do = 0

Propped beam:

Vo = -3*E*I/L^3*S(FDL)-3*E*I/L^2*S(FqL)

Mo = 0

qo = 3/(2*L)*S(FDL)+1/2*S(FqL)

Do = 0

Fixed beam:

Vo = -12*E*I/L^3*S(FDL)-6*E*I/L^2*S(FqL)

Mo = 6*E*I/L^2*S(FDL)+2*E*I/L*S(FqL)

qo = 0

Do = 0

Cantilever beam:

Vo = 0

Mo = 0

qo = -S(FqL)

Do = -S(FDL)-L*S(FqL)

Summations of shear, moment, slope, and deflection at distance = x from left end of beam:

Shear: Vx = Vo+S(Fvx)

Moment: Mx = Mo+Vo*x+S(Fmx)

Slope: qx = qo+Mo*x/(E*I)+Vo*x^2/(2*E*I)+ S(Fqx)

Deflection: Dx = -(Do-qo*x-Mo*x^2/(2*E*I)-Vo*x^3/(6*E*I)+ S(FDx)

Reference:

"Modern Formulas for Statics and Dynamics, A Stress-and-Strain Approach"

by Walter D. Pilkey and Pin Yu Chang, McGraw-Hill Book Company (1978)


4/19

"Three-Moment Theory" Used for Continuous-Span Beam Analysis:

The "Three-Moment" Equation is valid for any two (2) consecutive spans as follows:

Ma*L1/I1+2*(Mb)*(L1/I1+L2/I2)+Mc*L2/I2

= -6*(FEMab*L1/(6*I1)+FEMba*L1/(3*I1))-6*(FEMbc*L2/(3*I2)+FEMcb*L2/(6*I2))=-(FEMab+2*FEMba)*L1/I1-2*(FEMbc+FEMcb)*L2/I2

where: Ma = internal moment at left support

Mb = internal moment at center support

Mc = internal moment at right support

L1 = length of left span

I1 = moment of inertia for left span

L2 = length of right span

I2 = moment of inertia for right span

FEMab = total Fixed-End-Moment for left end of left span

FEMba = total Fixed-End-Moment for right end of left span

FEMbc = total Fixed-End-Moment for left end of right span

FEMcb = total Fixed-End-Moment for right end of right span

N = actual number of beam spans

Note: "Dummy" spans are used to model the left end and right end support conditions for the beam. A pinned

end is modeled as a very flexible span (very long length and very small inertia). A fixed end is modeled

as a very stiff span (very short length and very large inertia). Thus, the theoretical number of spans used

is = N + 2.

By writing an equation for each pair of consecutive spans and introducing the known values (usually zero)

of end moments, a system of (N+1) x (N+1) simultaneous equations can be set up to solve for the

unknown support moments.

Reference:

AISC Manual of Steel Construction - Allowable Stress Design (ASD) - 9th Edition (1989), page 2-294


5/19

Student: Ivancu Aurel Profil

Curs:

SE CERE

Dimensionarea unui dig de ritinere cunoscand:

1. deschiderea de calcul : h = 3.80 ml

pa = 315.00 kN/mp

Din considerente constructive se alege o armare rigida cu profile laminate, dispuse la distanta de:

b = 0.60 ml

REZOLVARE

NOTA :

q = pa x b = 189.00 kN /ml

Pentru calculul de rezistenta se considera o grinda incastrata la ambele capete, de latime egala cu dinstant

dintre armatura rigida , solicitata de o incarcare uniform distribuita de valoare :

Constructii miniere

2. presiune orizontala:

3. distanta intre armatura rigida :

TEMA DE CASADimensionarea unui dig de retinere

I.P.C.M. - Master anul I

5 of 19 6/28/2013 7:52 AM


6/19

A . Calculul eforturilor:

Versiunea 1

Introducere date: c

d

Date despre grinda: Rez & Rez b

Tip grinda Inc & Inc a

eschidere, L = 3.8000 m Rez & Inc +P +M +

Mod. el, E = 200000 MPa + qb

Inertie, I = 605.00 cm^4 Inc & Inc

E,I L

Incarcari: Liber & Inc RL x RR

Permanente:

p = -189.00 kN/m

Utile Start End

Distribuite: b (m) qb(kN/m) d (m) qd(kN/m)

#1: 0.0000 0.0000 0.0000 0.0000 RL = -359.10 RR = -359.10

#2: ML = 227.43 MR = 227.43

#3:

#4: +M(max) = 227.43 @ x = 0.000

#5: -M(max) = -113.72 @ x = 1.900#6:

#7: -D(max) = 0.000 @ x = 0.000

#8: +D(max) = 84.816 @ x = 1.900

D(ratio) = L/45

Concentrate: a (m) P (kN)

#1: 0.0000 0.00

#2:

#3:

#4:

#5:

#6:

#7:

#8:

#9:

#10:

#11:

#12:

#13:

#14:

#15:

Momente: c (m) M (kN-m)

#1: 0.0000 0.00

#2:

#3:

#4:

Momentul Max. [ kN*m ] / pozitia [m]:

Reactiuni / Forta taietoare [ kN ]:

Sageata Max. [mm] / pozitia [m]:

REZULTATE:

-400.0

-300.0

-200.0

-100.0

0.0

100.0

200.0

300.0

400.0

0.0

0

0.2

3

0.4

6

0.6

8

0.9

1

1.1

4

1.3

7

1.6

0

1.8

2

2.0

5

2.2

8

2.5

1

2.7

4

2.9

6

3.1

9

3.4

2

Forta(kN)

x (m)

Diagrama fortei taietoare

-150.0

-100.0

-50.0

0.0

50.0

100.0

150.0

200.0

250.0

0.0

0

0.2

3

0.4

6

0.6

8

0.9

1

1.1

4

1.3

7

1.6

0

1.8

2

2.0

5

2.2

8

2.5

1

2.7

4

2.9

6

3.1

9

3.4

2

Moment(kN-m)

x (m)

Digrama de moment

6 of 19 6/28/2013 7:52 AM


7/19

B. DIMENSIONARE

INCOVOIERE Sectiuni dreptunghiulare simplu armate

B.1.

C12/15 b h Rc Ra M Aa nec

OB 37 600 800 9.50 210.00 227,430,000 1546

(mm) (mm) (N/mm2) (N/mm

2) (N*mm) (mm

2)

a= 70 ho= 730 m= 0.075

DA 0

z= 0.078

Aa= 1546 p= 0.35

p 1,546

SE ALEGE: nr.buc f / A Aa ef

5 14

1 U 12Armat.rezistenta

DIMENSIONARE ARMATURI

TOTAL(arm.rez.)

mb=0,42ptr.OB37

mb=0,40ptr.PC

m


8/19

1.2 INCOVOIERE Sectiuni dreptunghiulare simplu armate

B. 2.

b h Rc Ra Aa / Aef Mcap

600 800 9.50 210.00 1,701.00 261,227,580


2) (mm

2) (N*mm)

a= 70 ho= 730 p= 0.39 % z= 0.09

zb=

DA

m= 0.086

Mcap= (N*mm)

p


9/19

1.1 INCOVOIERE Sectiuni dreptunghiulare s implu armate

C12/15 b h Rc Ra M Aa nec

OB37 600 800 9.26 210.00 227,430,000.36 1545


2) (N*mm) (mm

2)

a= 70 ho= 730 m= 0.077

DA 0

0

z= 0.08

Aa= 1545 p= 0.35

p 1,545

nr.buc f / A Aa ef

5 14

1 U 12

1.2 INCOVOIERE Sectiuni dreptunghiulare simplu armate

b h Rc Ra Aa / Aef

600 800 9.26 210.00 1,701.00


2) (mm

2) (N*mm)

a= 70 ho= 730 p= 0.39 % z= 0.09

zb=

DA

m= 0.086

Mcap= (N*mm)

p


10/19

Ra Ea

OB37 210 210000 Rc*= 210

PC52 300 210000

PC60 350 210000

j0 Rck/gbc Rtk/gtcC12/15 3.3 24000 9.26 0.79 Rc

*= 9.259259

C16/20 3 27000 12.30 0.95 Rt*= 0.793333

C20/25 2.8 30000 15.19 1.10

Rt N/mm2 Rc N/mm2 Rt N/mm2 Rc N/mm2

b


11/19


12/19

Incovoiere sectiuni dreptunghiulare dublu armate

C12/15 Rc N/mm2 9.26

OB37 Rt N/mm2 0.793333333

b= 600 mm

h= 800 mm

Aa'= 0 mm2

Aa= 785 mm2 0 F 8

Ea= 210000 N/mm2 0 F 12

Eb= 24000 N/mm2

ME= 24590000 Nmm = 24.59 kNm 10 F 10

ME

ld= 24590000 Nmm = 24.59 kNm 0 F 12

a= 32.5 mm

h0= 767.5 mm

a'= 32.5 mm

j= 3.0 pt. cond normale de lucru si BC20 tabel 4

p=Aa/bh0*100= 0.17 % peconomic={0,61,2)

p'=Aa'/bh0*100= 0.00 %

sb max=ME/Ibi*xh0= 0.58 N/mm2 < 9.26 N/mm2 Supradimensionare

sa=ne*ME/Ibi*(1-x)h0= 44.7 N/mm2 < 210 N/mm2 Supradimensionare

sa=ne*sbmax*(1-x)/x= 44.7 N/mm2 < 210 N/mm2 Supradimensionare

sa'=sa*(x-a'/h0)/(1-x)= 13.3 N/mm2 < 210 N/mm2 Supradimensionarev=M

Eld/M

E= 1.000

Eb'=0,8Eb/(1+0,5vj)= 7680 N/mm2

ne=Ea/Eb'= 27.34

a=nep/100= 0.047

a'=nep'/100= 0.000

x=(a+a'){[1+2(a+a'a'/h0)/(a+a')2]0,5

-1}= 0.262

Ibi=[x3/3+a(1-x)

2+a'(x-a'/h0)

2]bh0

3= 8.516E+09 mm

4

Aa'=

Aa=

nr.B

are

diam

etru


13/19

Ra Ea

OB37 210 210000

PC52 300 210000

PC60 350 210000

j0 Rck/gbc Rtk/gtcC12/15 3.3 24000 9.26 0.79 Rc

*= 9.259259 N/mm

2

C16/20 3 27000 12.30 0.95 Rt*= 0.793333 N/mm

3

C20/25 2.8 30000 15.19 1.10


b


14/19

Dimensionare la forta taietoare pentru incovoiere sectiuni dreptunghiulare dublu armateSectiunea se afla intr-o zona potential plastica ? DA/NU da

Ra N/mm2 210.00

C20/25 Rc N/mm2 15.19

OB37 Rt N/mm2 0.94

b= 250 mm

h= 400 mmAa'= 2454 mm

2

Aa= 2454 mm2 3 F 25

QE= 300000 N = 300 kN 2 F 25

a'= 32.5 mm

a= 32.5 mm 3 F 25

h0= 367.5 mm 2 F 25

ne= 3.00 nr. bare intersectate ne= 3.00

Ae= 78.54 mm2 3 F 10

0.00 F 8ae= 100 mm

Qeb=qeb*bh0Rt redus= 364240.24 N

Qb=q*bh0Rt redus= 3000

petr=ne*Ae/(aeb)*100= 0.94 %

p'=Aa'/bh0*100= 2.67 % peconomic={0,61,2)

p=Aa/bh0*100= 2.67 %

si/h0=(100p

0,5

/petr*Rt redus/(0,8Ra))

0,5

= 14.72si/h02,5 qeb=2(petrp

0,50,8Ra/(100Rt redus))

0,5= 33.27

si/h0>2,5 qeb=p0,5

/2,5+petrRa/(50Rt redus)= 424.01

q=Q/bh0Rt= 3.49

daca q>1,0

ms=(3-q)/2= 0.01

Rt redus=msRt= 0.01

qrecalculat=Q/(bh0Rt redus)= 349.23 F

q4?

nr.B

are

diam

etru

Aa'=

Aa=

Ae=

OK


15/19

Ra Ea

OB37 210 210000

PC52 300 210000

PC60 350 210000j0 Rck/gbc Rtk/gtc

C12/15 3.3 24000 9.26 0.79 Rc*= 15.18519 N/mm

2

C16/20 3 27000 12.30 0.95 Rt*= 1.1 N/mm

3

C20/25 2.8 30000 15.19 1.10


b etrieri cu minim 4 ramuri

8 50.26548 mm2

Placi (orice grosime)

Beton Armat Beton Simplu

Stalpi, diafragme cu h


16/19

( maxim 5 deschideri )

Den. Investitie: Cod Inv.:

Den.Lucrare : Intocmit: V

Introducere date:

Tip material : Metal Ix = 12778.30 cm^4

Sectiune : W12x40 = > Iy = 1835.58 cm^4 b

Date despre grinda: a

Nr. Deschid., N = 3 +P

Tip sport - stg = Rez.simp. Reazem # 1 Deschd. #1 Deschd. #2 Deschd. #3 Deschd. #4 Deschd. #5 +

Tip sport - dr. = Rez.simp. Reazem # 4 5.00 8.50 4.00 [ m ]

Modul elast., E = 210,000 MPa E,I

Moment inertie I= 12,778 (cm^4) - Se alege Ix sau Iy VL x

Date despre lungimea deschiderilor:

Deschidere, L(m) =

Inertie, I(cm^4) =

Distrib. - w(kN/m) =

Distribuite: b (m) wb(kN/m) e (m) we(kN/m) b (m) wb (kN/m) e (m) we (kN/m) b (m) wb (kN/m)

#1:

#2:

#3:

Concentrate:

#1:

#2:

Momente:

#1:

#2:

Forta.taiet. cosola_ Stanga = 0.00 kN Moment. cosola_ Stanga = 0.00 kN-m

-14.60 -14.6

c (m) M (kN-m) M (kN-m) c (m)c (m)

End

P (kN) a (m)a (m) P (kN) a (m)

Start

Deschd. #1 Deschd. #2 Deschd

5.00

12,778.30

4.00

12,778

5.00

12,778.30

-14.60

Start End Start

Traversare retea canalizare _parau " Km 0.0000

Reabilitare .

CALCULUL - G R I N Z I L O R C O N T I N U E

Ing. Ivancu Aurel

Incarcarii - Utile

Incarcarii - Permanente:

Schita grinzi

1 2 3 4 5 6

16 of 19


17/19

R E Z U L T A T E :

F. taietoare (kN): -28.75 44.25 -38.73 34.27 -36.09

Reactiuni ( kN) : -28.75

Deschd. #1 Deschd. #2 De

1 2 3Momente (kN*m):

0.00 27.57 --28.31 38.73 -12.65

-82.98 -70.36

-50

-40

-30

-20

-10

0

10

20

30

40

50

0.0

000

0.4

000

0.8

000

1.2

000

1.6

000

2.0

000

2.4

000

2.8

000

3.2

000

3.6

000

4.0

000

4.4

000

4.8

000

5.1

000

5.5

000

5.9

000

6.3

000

6.7

000

7.1

000

7.5

000

7.9

000

8.3

000

8.7

000

9.1

000

9.5

000

9.9

000

1

0.1

600

1

0.4

800

1

0.8

000

1

1.1

200

1

1.4

400

1

1 7 6 0 0

Shear(kN)

-30

-20

-10

0

10

20

30

40

50

0.0

000

0.4

000

0.8

000

1.2

000

1.6

000

2.0

000

2.4

000

2.8

000

3.2

000

3.6

000

4.0

000

4.4

000

4.8

000

5.1

000

5.5

000

5.9

000

6.3

000

6.7

000

7.1

000

7.5

000

7.9

000

8.3

000

8.7

000

9.1

000

9.5

000

9.9

000

10.1

600

10.4

800

10.8

000

11.1

200

11.4

400

Moment(kN-m)


18/19

ri:

s:

kN

kN

kN

kN

kN

kN

m (Deschd. #1)

m (Deschd. #1)

m (Deschd. #2)

m (Deschd. #1)

we (kN/m)

kN-m

kN)

N-m)

d

18 of 19


19/19

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19 f 19

beamanal (metric) copie

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