DESIGN OF BASE PLATE.Tower Leg section ( D ) = 27.3 cm dia. Tube.= 200= 8.15( I !"5#!1$78)= 25"8= 1#81.#8 %g&m2FOUNDATION LOADS:'(timate )om*ression ( +uc ) = 718"7 %g'(timate Tension ( +ut ) = #$3"# %g'(timate s,ear ( -!dir ). +s/ = "7"8 %g'(timate s,ear ( 0!dir ). +s1 = "7"8 %g= #71".#$ %gFOUNDATION BOLT DETAILS:Dia. 23 3oundation bo(t ( d ) = 3.# cm4o. o3 3oundation bo(ts ( n ) = 8 no.sLengt, o3 embeddment ( L ) = 100 cm5o(t edge distance ( ed ) = 7 cm5o(t to bo(t distance ( g ) = $ cmX+oundation bo(t5ase *(ate.52.538.5ti33ener).6.(ine 153X7 52.5Base plate 6rade o3 concrete !7!20 ( 3 c% )%g&cm25ond stress o3 concrete tbd%g&cm26rade o3 5ase 8(ate tee( ( 39 )%g&m2:((owab(e bending stress ( 3b. a((ow = 0.## 39 );esu(tant s,ear *er (eg ( +s )+s = ided ( : = 5 / 5 ) = 275#.25= 2#.07 ? 50HENCE. O.K.CHECK FOR BENDING STRESS IN PLATE ( LEG IN COMPRESSION 7a/. bending moment about -!- a/ is. 7 /!/ = 1751$0." %g!cm.).6. o3 5ase *(ate @ sti33ner! section!-!-. / = 2."2 cm 3rom bottom o3 base *(ate. -ma/. = 15.58 cm7oment o3 Inertia about t,e a/is -!-. ( I // ) = 17#".0$ cm"ection modu(us. ( 0 /!/ (min) = I/!/&-ma/ ) = 113.23 cm35ending stress 3b = 7/!/&0/!/( min ) = 15"7.21 %g&cm2 ?1#81.#8 HENCE. SAFE.CHECK FOR BENDING STRESS IN PLATE ( LEG IN TENSION :7a/. bending moment about -!-. 7 /!/A+orce in eac, bo(t ( + = +ut & n ) = 8##8.25 %g.Distance o3 outer bo(ts to tube = 5.# cm7a/. moment. 7 /!/ = $708"." %g!cm.ection modu(us. 0 /!/ = 113.23 cm3'(timate bending stress ( 3 ) = 857."088 %g&cm2 ? 1#81.#8 HENCE. SAFE.%g&cm2cm27a/. bearing *ressure on *(ate ( * = +uc&: )%g&cm2%g&cm2CHECK FOR STRESSES IN FOUNDATION BOLTS:Dia o3 3oundation bo(t. d = 3.# cm4o. o3 3oundation bo(ts. n = 8 no.s.6ross area ( :g ) = 10.18= 8.#5Tension on eac, bo(t = 8##8.25 %g,ear on eac, bo(t ( s ) = 83$.3" %g)a(cu(ated bo(t stress in tension ( 3bt.ca( ) = 1002.11)a(cu(ated bo(t stress in s,ear ( 3bs.ca( ) = 82."58ermssib(e bo(t stress in a/ia( tension( 3bt) = 1223)(ause 8.$.".1. Tab(e 8.1 o3 I!800!1$8"8ermssib(e bo(t stress in s,ear( 3bs) = 815Interaction 3or combined bo(t tension and s,earA3bt. ca(&3bt = 3bs. ca(&3bs = 0.$2 ? 1 He!"e #$lts a%e $.&.+oundation bo(t embeddment (engt,( L ) = 100 cm5ond strengt, o3 eac, bo(t = $212.7# %g B 8##8.25 He!"e' p%$()*e e+#e**+e!t le!,t- )s $.&.cm24et area ( :net )cm2%g&cm2%g&cm2%g&cm2%g&cm2DESIGN OF BASE PLATE.Tower Leg section ( D ) = 27.3 cm dia. Tube.= 200= 8.15( I !"5#!1$78)= 25"8= 1#81.#8 %g&m2FOUNDATION LOADS:'(timate )om*ression ( +uc ) = "133$ %g'(timate Tension ( +ut ) = ""3"" %g'(timate s,ear ( -!dir ). +s/ = 1271 %g'(timate s,ear ( 0!dir ). +s1 = 1271 %g= 17$7."7 %gFOUNDATION BOLT DETAILS:Dia. 23 3oundation bo(t ( d ) = 3 cm4o. o3 3oundation bo(ts ( n ) = 8 no.sLengt, o3 embeddment ( L ) = $0 cm5o(t edge distance ( ed ) = 7 cm5o(t to bo(t distance ( g ) = $ cmX+oundation bo(t5ase *(ate."02#ti33ener).6.(ine 152X7 "0BASE PLATE DETAILS:i1e o3 5ase 8(ate ( 5 ) = "0 / "0 cmT,ic%ness o3 5ase 8(ate ( t ) = 2 cmi1e o3 ti33ner= 15 / 1.2 cmCHECK FOR BEARING PRESSURE:8ermissib(e bearing stress in concrete = 500.25 3c% ( )(ause 33." o3 I!"5#!1$78 )5earing area *ro>ided ( : = 5 / 5 ) = 1#00= 25.8" ? 506rade o3 concrete !7!20 ( 3 c% )%g&cm25ond stress o3 concrete tbd%g&cm26rade o3 5ase 8(ate tee( ( 39 )%g&m2:((owab(e bending stress ( 3b. a((ow = 0.## 39 );esu(tant s,ear *er (eg ( +s )+s =