BEA DIAGAS AND FOULAS Nomencture E = moduus of esticity of stee t 9,000 ksi I = moment of inerti of em (in. 4 ) L = tot ength of em etween rection points (ft) m = mimum moment (kip-in.) = mimum moment in eft section of em (kip-in.) = mimum moment in right section of em (kip-in.) = mimum positive moment in em with comined end moment conditions (kip-in.) = moment t distnce from end of em (kip-in.) = concentrted od (kips) = concentrted od nerest eft rection (kips) = concentrted od nerest right rection, nd of different mgnitude thn (kips) = end em rection for ny condition of symmetric oding (kips) = eft end em rection (kips) = right end or intermedite em rection (kips) = right end em rection (kips) = mimum vertic sher for ny condition of symmetric oding (kips) = mimum vertic sher in eft section of em (kips) = vertic sher t right rection point, or to eft of intermedite rection point of em (kips) = vertic sher t right rection point, or to right of intermedite rection point of em (kips) = vertic sher t distnce from end of em (kips) W = tot od on em (kips) = mesured distnce ong em (in.) = mesured distnce ong em which my e greter or ess thn (in.) = tot ength of em etween rection points (in.) w = uniformy distriuted od per unit of ength (kips per in.) w = uniformy distriuted od per unit of ength nerest eft rection (kips per in.) w = uniformy distriuted od per unit of ength nerest right rection, nd of different mgnitude thn w (kips per in.) = ny distnce mesured ong em from eft rection (in.) = ny distnce mesured ong overhng section of em from nerest rection point (in.) = mimum defection (in.) m = defection t point of od (in.) = defection t ny point distnce from eft rection (in.) = defection of overhng section of em t ny distnce from nerest rection point (in.)
BEA DIAGAS AND FOULAS Frequenty Used Formus The formus given eow re frequenty required in structur designing. They re incuded herein for the convenience of those engineers who hve infrequent use for such formus nd hence my find reference necessry. BEAS : Feur stress t etreme fier f = c / I = / S :Feur stress t ny fier f = y / I y = distnce from neutr is to fier Averge vertic sher (for mimum see eow): v = / A = / dt (for ems nd girders) Horizont shering stress t ny section A-A: v = Q / I Q = sttic moment out the neutr is of tht portion of the cross section ying outside of section A-A = width t section A-A (Intensity of vertic sher is equ to tht of horizont sher cting norm to it t the sme point nd oth re usuy mimum t mid-height of em.) nd defection t ny point: EI d y d = nd y re sciss nd ordinte respectivey of point on the neutr is, referred to es of rectngur coordintes through seected point of support. (First integrtion gives sopes; second integrtion gives defections. Constnts of integrtion must e determined.) CONTINUOUS BEAS (the theorem of three moments) Uniform od: + I + + I I c = I w 4 + w I I Concentrted ods: + I + + I I c = + p + I I I I Considering ny two consecutive spns in ny continuous structure:,, c = moments t eft, center, nd right supports respectivey, of ny pir of djcent spns nd = ength of eft nd right spns, respectivey, of the pir I nd I = moment of inerti of eft nd right spns, respectivey w nd w = od per unit of ength on eft nd right spns, respectivey nd = concentrted ods on eft nd right spns, respectivey nd = distnce of concentrted ods from eft support, in eft nd right spns, respectivey nd = distnce of concentrted ods from right support, in eft nd right spns, respectivey The ove equtions re for em with moment of inerti constnt in ech spn ut differing in different spns, continuous over three or more supports. By writing such n eqution for ech successive pir of spns nd introducing the known vues (usuy zero) of end moments, other moments cn e found.
BEA DIAGAS AND FOULAS Te of Concentrted Lod Equivents Simpe Bem Bem Fied One End, Supported t Other Bem Fied Both Ends n Loding Coeff. 0.5 0.070 0.04 0.5 0.08 c 0.500 0.75 d 0.65 0.500 e 0.0 0.005 0.00 f.000.000 0.667 g.000 0.45 0.00 0.50 0.56 0.5 0.88 0.5 c 0.500 0. d 0.688 0.500 e 0.0 0.009 0.005 f.000.500.000 g 0.800 0.477 0.400 0. 0. 0. 0. 0. c.000 0.667 d..000 e 0.06 0.05 0.008 f.667.667.778 g.0 0.48 0. 4 0.500 0.66 0.88 0.469 0. c.500.0 d.969.500 e 0.050 0.0 0.00 f 4.000.750.500 g 0.950 0.48 0.0 5 0.600 0.60 0.00 0.600 0.400 c.000.400 d.600.000 e 0.06 0.07 0.0 f 4.800 4.800.00 g.008 0.44 0. imum positive moment (kip-ft): L imum negtive moment (kip-ft): L inned end rection (kips): c Fied end rection (kips): d imum defection (in): e / EI Equivent simpe spn uniform od (kips): f Defection coefficient for equivent simpe spn uniform od: g Numer of equ od spces: n Spn of em (ft): L Spn of em (in):
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions. SILE BEA UNIFOLY DISTIBUTED LOAD m w Tot Equiv. Uniform Lod...... = w =................. = w................. = w m (t center)............ = w 8................. = w ( ) m (t center)............ = 5w4 84EI................. = w 4EI ( + ). SILE BEA LOAD INCEASING UNIFOLY TO ONE END Tot Equiv. Uniform Lod...... = 6W 9 =.064W W =................. = W.5774 m = m............... = W................. = W W m (t = W =.5774 )....... = =.8 W 9................. = W ( ) m (t = 8 W =.59 ).. = 0.04 5 EI W................. = 0 + 7 4 ) 80EI (4. SILE BEA LOAD INCEASING UNIFOLY TO CENTE W Tot Equiv. Uniform Lod...... = 4W =................. = W (when < ).......... = W ( 4 ) m (t center)............ = W 6 (when < ).......... = W m m (t center)............ = W 60EI (when < ).......... = W 480EI (5 4 )
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 4. SILE BEA UNIFOLY LOAD ATIALLY DISTIBUTED c w + w = = (m. when < c)....... = w (c + ) (m. when > c)....... = w ( + ) (when > nd < ( + ))... = w( ) m t = + w........ = + w (when < )......... = m (when > nd < ( + ))... = w ( ) (when > ( + ))....... = ( ) 5. SILE BEA UNIFO LOAD ATIALLY DISTIBUTED AT ONE END = m............... = w ( ) m w w =................ = w (when < )......... = w m t = w.......... = w (when < )......... = w (when > )......... = ( ) w (when < )......... = 4EI ( ( ) ( ) + ) (when > )......... = w ( ) (4 ) 4EI 6. SILE BEA UNIFO LOAD ATIALLY DISTIBUTED AT EACH END m w w c w c =................ = w ( ) + w c =................ = w c( c) + w (when < )......... = w (when > nd < ( + )) = w (when > ( + ))....... = w ( ) m t = when w < w = w m t = when w < w c = w (when < )......... = w (when > nd < ( + ))... = w ( ) (when > ( + ))....... = ( ) w ( )
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 7. SILE BEA CONCENTATED LOAD AT CENTE Tot Equiv. Uniform Lod.......... = =.................... = m (t point of od)........... = 4 when <............. = m m (t point of od)........... = 48EI when <............. = 48EI ( 4 ) m 8. SILE BEA CONCENTATED LOAD AT ANY OINT Tot Equiv. Uniform Lod.......... = 8 = = m m (m when < )........... = (m when > )........... = (t point of od)........... = (when < ).............. = ( + ) t = when >... = (t point of od)........... = ( + ) ( + 7EI EI (when < ).............. = 6EI ( ) m 9. SILE BEA TWO EQUAL CONCENTATED LOADS SYETICALLY LACED Tot Equiv. Uniform Lod.......... = 8 =.................... = m m (etween ods)............ = (when < ).............. = (t center)............... = 4EI ( 4 ) (when < ).............. = 6EI ( ) (when > nd < ( ))....... = 6EI ( )
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 0. SILE BEA TWO EQUAL CONCENTATED LOADS UNSYETICALLY LACED = (m. when < )......... = ( + ) = (m. when > )......... = ( + ) (when > nd < ( ))..... = ( ) (m. when > )......... = (m. when < )......... = (when < )............ = (when > nd < ( ))..... = ( ). SILE BEA TWO UNEQUAL CONCENTATED LOADS UNSYETICALLY LACED =.................. = ( ) + =.................. = + ( ) (when > nd < ( ))..... = (m. when < )........ = (m. when < )........ = (when < )............ = (when > nd < ( ))..... = ( ). BEA FIXED AT ONE END, SUOTED AT OTHE UNIFOLY DISTIBUTED LOAD w 8 4 m Tot Equiv. Uniform Lod.......... = w =.................. = w 8 = m.................. = 5w 8.................. = w m.................. = w 8 t = 8 9............. = 8 w.................. = w m t = 6 ( + ) =.45 w... = 4 85EI.................. w 48EI ( + )
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions. BEA FIXED AT ONE END, SUOTED AT OTHE CONCENTATED LOAD AT CENTE Tot Equiv. Uniform Lod....... = m =.................. = 5 5 = m................ = 6 m (t fied end)........... = 6 (t point of od).......... = 5 when <............ = 5 6 when >............ = 6 m t = 5 =.447....... = 48EI 5 (t point of od).......... = 7L 768EI =.0097 EI when <............ = 96EI ( 5 ) when >............ = 96EI ( ) ( ) 4. BEA FIXED AT ONE END, SUOTED AT OTHE CONCENTATED LOAD AT ANY OINT =.................. = ( + ) =.................. = ( ) (t point of od).......... = (t fied end)........... = ( + ) (when < )............ = (when > )............ = ( ) m when <.44 t = ( + ) ( ) = ( + ) EI( ) m when >.44 t =... = + 6EI + (t point of od).......... = ( + ) EI (when < )............ = EI ( ) (when > )............ = EI ( ) ( )
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 5. BEA FIXED AT BOTH ENDS UNIFOLY DISTIBUTED LOADS w Tot Equiv. Uniform Lod........ = w =.................. = w m. m.................. = w m (t ends)............. = w (t center)............. = w 4.................. = w (6 6 ) w m (t center)............. = 4 84EI.................. = w ( ) 4EI 6. BEA FIXED AT BOTH ENDS CONCENTATED LOAD AT CENTE m 4 m m Tot Equiv. Uniform Lod........ = =.................. = m (t center nd ends)........ = 8 when <........... = (4 ) 8 m (t center)............. = 9EI when <........... = ( 4) 48EI 7. BEA FIXED AT BOTH ENDS CONCENTATED LOAD AT ANY OINT = (m. when < )........ = ( + ) = (m. when > )........ = ( + ) (m. when < )........ = (m. when > )........ = (t point of od)......... = (when < )........... = m when > t =.... = + EI( + ) (t point of od)......... = EI (when < )........... = ( ) 6EI
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 8. CANTILEE BEA LOAD INCEASING UNIFOLY TO FIXED END Tot Equiv. Uniform Lod........ = 8 W W =................... = W................... = W m (t fied end)............ = W................... = W m m (t free end)............. = W 5EI W................... = 60EI (5 5 4 + 4 5 ) 9. CANTILEE BEA UNIFOLY DISTIBUTED LOAD Tot Equiv. Uniform Lod........ = 4w w =................... = w................... = w m (t fied end)............ = w m................... = w m (t free end)............. = w4 8EI w................... = 4EI (4 4 + 4 ) 0. BEA FIXED AT ONE END, FEE TO DEFLECT ETICALLY BUT NOT OTATE AT OTHE UNIFOLY DISTIBUTED LOAD Tot Equiv. Uniform Lod........ = 8 w w =................... = w................... = w m (t fied end)............ = w.47 m................... = w 6 ( ) m (t defected end).......... = w4 4EI................... = w( ) 4EI
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions. CANTILEE BEA CONCENTATED LOAD AT ANY OINT Tot Equiv. Uniform Lod........ = 8 =................... = m (t fied end)............ = (when > )............ = ( ) m (t free end)............. = ( ) 6EI (t point of od).......... = EI m (when < )............ = ( ) 6EI (when > )............ = ( ) ( + ) 6EI. CANTILEE BEA CONCENTATED LOAD AT FEE END Tot Equiv. Uniform Lod........ = 8 =................... = m (t fied end)............ =................... = m (t free end)............. = EI m................... = 6EI ( + ). BEA FIXED AT ONE END, FEE TO DEFLECT ETICALLY BUT NOT OTATE AT OTHE CONCENTATED LOAD AT DEFLECTED END Tot Equiv. Uniform Lod........ = 4 =................... = m (t oth ends)............ =................... = m m (t defected end).......... = p EI m................... = ( ) ( + ) EI
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 4. BEA OEHANGING ONE SUOT UNIFOLY DISTIBUTED LOAD =.............. = w ( ) = +........... = w ( + ) w( +)............... = w............... = w ( + ) ( ) ( ) (etween supports)..... = w (for overhng)....... = w( ) t =..... = w 8 ( + ) ( ) (t )............ = w (etween supports)..... = w ( ) (for overhng)....... = w ( ) w (etween supports)..... = 4EI (4 + + ) (for overhng)....... = w 4EI (4 + 6 4 + ) 5. BEA OEHANGING ONE SUOT UNIFOLY DISTIBUTED LOAD ON OEHANG w m =.............. = w +............ = w ( + )............... = w (for overhng)....... = w( ) m (t )........... = w (etween supports)..... = w (for overhng)....... = w ( ) m etween supports t = w = 8 EI = 0.008 w EI m (for overhng t = )... = w (4 + ) 4EI (etween supports)..... = w EI ( ) (for overhng)....... = w 4EI (4 + 6 4 + )
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 6. BEA OEHANGING ONE SUOT CONCENTATED LOAD AT END OF OEHANG =..................... = = +................... = ( + )..................... = m (t )................. = (etween supports).......... = (for overhng)............. = ( ) m etween supports t =..... = =.0645 9 EI EI m m (for overhng t = )........ = ( + ) EI (etween supports).......... = 6EI ( ) (for overhng)............. = 6EI ( + ) m w 7. BEA OEHANGING ONE SUOT UNIFOLY DISTIBUTED LOAD BETWEEN SUOTS Tot Equiv. Uniform Lod.......... = w =..................... = w..................... = w m (t center)............... = w 8..................... = w ( ) m (t center)............... = 5w4 84EI..................... = w 4EI ( + )..................... = w 4EI 8. BEA OEHANGING ONE SUOT CONCENTATED LOAD AT ANY OINT BETWEEN SUOTS Tot Equiv. Uniform Lod.......... = 8 = (m. when < )........... = m = (m. when > )........... = m m (t point of od)............ = (when < ).............. = ( + ) t = when >... = ( + ) ( + 7EI (t point of od)............ = EI (when < ).............. = 6EI ( ) (when > ).............. = ( ) ( ) 6EI..................... = ( + ) 6EI
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 9. CONTINUOUS BEA TWO EQUAL SANS UNIFO LOAD ON ONE SAN w Tot Equiv. Uniform Lod = 49 64 w =........... = 7 6 w = +......... = 5 8 w 7 6 m =........... = 6 w............ = 9 6 w m t = 7 6..... = 49 5 w (t support ).... = 6 w (when < )..... = w (7 8) 6 m (t 0.47 from ).. =.009 w 4 / EI 0. CONTINUOUS BEA TWO EQUAL SANS CONCENTATED LOAD AT CENTE OF ONE SAN m Tot Equiv. Uniform Lod = 8 =........... = = +......... = 6 =........... =............ = 9 m (t point of od)... = 64 (t support ).... = m (t 0.480 from ).. =.05 / EI. CONTINUOUS BEA TWO EQUAL SANS CONCENTATED LOAD AT ANY OINT =........... = 4 (4 ( + )) m = +......... = ( + ( + )) =........... = ( + ) 4............ = 4 (4 + ( + )) m (t point of od)... = 4 (4 ( + )) (t support ).... = ( + ) 4
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions. BEA UNIFOLY DISTIBUTED LOAD AND AIABLE END OENTS =........... = w + w > =........... = w............. = w + t = + w.. = w 8 + + ( ) w............. = w ( ) + (to octe infection points) = 4 + + w w = w 4EI + 4 w 4 w + w + 8 w 4 w. BEA CONCENTATED LOAD AT CENTE AND AIABLE END OENTS =........... = + > =........... = (t center)........ = 4 + when <...... = + when >...... = ( ) + ( ) when < = 48EI 4 8( ) [ ( ) + ( + )]
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 4. CONTINUOUS BEA THEE EQUAL SANS ONE END SAN UNLOADED w w A B C D A = 0.8 w B =.0 w C = 0.450 w 0.8 w 0.58 w 0.0 w 0.67 w +0.075 w 0.67 w 0.47 w +0.054 w 0.0 w D = 0.0 w 0.0w 0.8 0.58 (0.40 from A) = 0.0059 w 4 m / E 5. CONTINUOUS BEA THEE EQUAL SANS END SANS LOADED w w A B C D A = 0.450 w B = 0.550 w C = 0.550 w 0.450 w +0.0 w 0.550 w 0.050 w 0.550 w +0.0 w D = 0.450 w 0.450 w 0.450 0.450 m (0.479 from A or D) = 0.0099 w 4 / E 6. CONTINUOUS BEA THEE EQUAL SANS ALL SANS LOADED w w w A B C D A = 0.400 w B =.0 w C=.0 w 0.400 w 0.500 w 0.600 w 0.600 w 0.500 w 0.00 w 0.00 w +0.080 w +0.05w +0.080 w D = 0.400 w 0.400 w 0.400 0.500 0.500 (0.446 from A or D) = 0.0069 w 4 m / E 0.400
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 7. CONTINUOUS BEA FOU EQUAL SANS THID SAN UNLOADED w w w A B C D E A= 0.80 w B =. w C = 0.57 w 0.80w 0.60 w 0.60 w 0.97 w 0.05 w 0.079 w 0.558 w D = 0.598 w 0.040 w 0.058 w +0.07 w +0.06 w +0.0977 w E = 0.44 w 0.44 w 0.80 0.60 0.44 (0.475 from E) = 0.0094 w 4 m / E 8. CONTINUOUS BEA FOU EQUAL SANS LOAD FIST AND THID SANS w w A B C D E A= 0.446 w B = 0.57 w C = 0.464 w 0.446w 0.08 w 0.48 w 0.554 w 0.056 w 0.057 w +0.0996 w +0.0805 w 0.054 w D = 0.57 w 0.58 w 0.056w E = 0.054w 0.054 w 0.446 (0.477 from A) = 0.0097 w 4 m / E 0.58 9. CONTINUOUS BEA FOU EQUAL SANS ALL SANS LOADED w w w w A B C D E A= 0.9 w B =.4 w C = 0.98 w D =.4 w E = 0.9 w 0.9 w 0.56 w 0.607 w 0.464 w 0.464 w 0.607 w 0.56 w 0.07w 0.074 w 0.07 w +0.077 w +0.064 w +0.064 w +0.077 w 0.9 w 0.9 0.56 0.56 m (0.440 from A nd D) = 0.0065 w 4 / E 0.9
BEA DIAGAS AND FOULAS For rious Sttic Loding Conditions 40. SILE BEA ONE CONCENTATED OING LOAD m = m (t = 0)........... = m t point of od, when =..... = 4 4. SILE BEA TWO EQUAL CONCENTATED OING LOADS m = m (t = 0)........... = when <, ( )..... =.586 m under od t.. = when > ( )..... =.586 with one od t center of spn = 4 (Cse 40) 4. SILE BEA TWO UNEQUAL CONCENTATED OING LOADS m = m (t = 0)........... = + > m under, t = + m my occur with rger od t center of spn nd other od off spn (Cse 40)... = 4 = ( + ) GENEAL ULES FO SILE BEAS CAYING OING CONCENTATED LOADS C.G. The mimum sher due to moving concentrted ods occurs t one support when one of the ods is t tht support. With sever moving ods, the oction tht wi produce mimum sher must e determined y tri. The mimum ending moment produced y moving concentrted ods occurs under one of the ods when tht od is s fr from one support s the center of grvity of the moving ods on the em is from the other support. In the ccompnying digrm, the mimum ending moment occurs under od when =. It shoud so e noted tht this condition occurs when the centerine of the spn is midwy etween the center of grvity of ods nd the nerest concentrted od.
BEA DIAGAS AND FOULAS Design properties of cntievered ems Equ ods, equy spced No. Spns System A B A C D c c D C 4 A E E A 4 d e 5 A F G D C 5 f f C D H H d e e d D C 5 5 A F G G F A 6 (even) f f d e 5 A F G H H D C 7 (odd) C f f f f D H H H H d e f f e d D C 5 5 A F G H H G F A n 4 5 Typic Spn Loding s 4 5 0.086L 0.096L 0.06L 0.09L 0.05L 0.67L 0.88L 0.5L 0.08L 0.04L 0.50L 0.78L 0.67L 0.08L 0.9L 0.L 0.75L 0.50L 0.67L 0.08L 0.49L 0.480L 0.00L 0.7L 0.49L ections A B C D E F G H 0.44.7 0.48.06.086.09 0.977.000 0.8. 0.875.5.67.08.958.000.50.500..67.50..97.000.667 4.667.750 4.50 4. 4.47.97 4.000.07 5.857.00 5.00 5.49 5.557 4.87 5.000 Cntiever Dimensions c d e f 0.7L 0.5L 0.0L 0.04L 0.57L 0.47L 0.50L 0.00L 0.L 0.08L 0.7L 0.50L 0.00L 0.4L 0.50L 0.L 0.8L 0.67L 0.8L 0.4L 0.L 0.L 0.76L 0.67L 0.76L 0.0L 0.9L 0.0L 0.60L 0.50L
BEA DIAGAS AND FOULAS CONTINUOUS BEAS OENT AND SHEA COEFFICIENTS EQUAL SANS, EQUALLY LOADED OENT UNIFO LOAD SHEA in terms of w in terms of w +.07.5 +.07 +.08 +.05 +.08.0.0 +.077 +.06 +.06 +.077.07.07.07 0 5 5 0 8 8 8 0 4 6 5 5 6 4 0 0 0 0 0 +.078 +.078.05.07.07.05 0 7 5 8 8 8 5 7 8 0 8 +.078 +.078.06.077.086.077.06 0 5 0 8 9 9 8 0 8 8 8 8 8 5 0 8 +.078 +.078.06.077.085.085.077.06 0 4 6 55 4 5 04 04 04 5 5 04 5 49 55 6 4 0 04 04 04 0 6 4 86 75 4 67 70 4 7 7 4 7 7 4 70 67 4 75 86 4 6 0 4 OENT in terms of +.56 +.56 +.57 CONCENTATED LOADS t center..69.69. SHEA in terms of +.78.5 +.0.5 +.75.5.65.50.50.65.5 +.7.8 +..9 +..9 +..58 +.7.4.66.54.46.50.50.46.54.66.4 OENT in terms of +. +. +. +.. CONCENTATED LOADS t points.67...67 SHEA in terms of +.44 +.56 +.066 +.066 +.56 +.44.67.67 +.4 +.46 +.076 +.099 +. +. +.099 +.076 +.46 +.4.8...8.7.7.0.0.7.7.7.8.07.9.0.0.9.07.8.7 OENT in terms of +.67 +.67 +.58 +.0 +.0 +.58 -.465 +.4 +.8 +.4 +.8 +.097 +.00 -.7 +.00 +.097 -.7 +.8 +.0 +.55 +.04 +.55 +.0 +.77 +.079 +.006 -.94 +.054 +.079 -.96 +.079 +.054 -.96 +.006 +.079 -.94 +.77 CONCENTATED LOADS t 4 points.0.97.97.0..87.50.50.87. SHEA in terms of..89.60.40.50.50.40.60.89.