SOLUTION; DEAD LOAD SELF WEIGHT OF SLAB=0.12*25=3 KN/M^2 FLOOR FINISH=3  KN/M^2 TOTAL DEAD LOAD= 6 KN/M^2 6 KN/M^2 *(3*9) M^2=162 KN 162KN/9M =18 KN/M SELF WEIGHT OF BEAM =0.5*0.75*25=9.375 KN/M TOTAL DEAD LOAD = 27.375 KN/M LIVE LOAD =5 KN/M^2*(3*9) M^2=135 KN 135KN/9M=15 KN/M TOTAL LIVE LOAD = 15KN/M TOTAL DEAD…

SOLUTION 1; Truss type reasoning   PRATT TRUSS 1.  IT CONSIST  OF VERTICAL AND DIAGONAL MEMBERS THAT FORM AN "N" SHAPE OF PATTERN. 2.  THE DIAGONAL MEMBERS ARE ARRANGED TO  DEVELOP TENSILE FORCES 3.  ALSO THEY CAN BE DESIGNED TO RESIST AXIAL TENSION FORCES ONLY. 4.  THE VERTICAL MEMBER…

   SOLUTION 1 A....   CHAIN BOLTING= PLATE SIZE =260*20 MM FOR A 20 MM BOLT BOLT DIAMETER= 22 MM NET AREA =(B-N*D@0)*T NET AREA =(260-4*22)*20=3440 MM^2       SOLUTION 1 B..... ZIG ZAG BOLTING   PLATE SIZE =260*20 MM FOR A 20 MM BOLT BOLT DIAMETER= 22 MM         PATH…

 SOLUTION 1 A ; PROBLEM Z DIRECTION K FACTOR =0.8 L Y DIRECTION K FACTOR =2.0 L B ; PROBLEM PICTURE 1  FRAME =BRACED, ASSUMING BASE TO BE PINNED SINCE NOT MENTIONED  K=1.0. PICTURE 2 K=2.0 PICTURE 3  K= 1.2.     SOLUTION 2; GIVEN DATA ; HOT ROLLED SECTION=ISMB 300 FLANGE COVER PLATE =200*12…

SOLUTION 1 A. EPISILON VALUE FOR ALL SECTIONS=SQUAREROOT(250/F@Y)=SQUAREROOT(250/250)=1 ISMB 500 D=500 T@F==17.2 B@F=180 T@W=10.2 R=17 d=D-2(t@f+R)=500-2(17.2+17)=431.6 B@F/T@F=180/17.2=10.46<10.5*EPISILON=10.5*1=10.5 d/T@W=431.6/10.2=42.31<84*EPISILON=84*1=84 HENCE THE SECTION IS COMPACT   B. ISMC-200 D=200…

SOLUTION 1 P@U=1000 KN F@Y =250 MPA D=1200 MM d=1150 MM T@F=25 MM T@W=10 MM TAU@CR,E=K@V*PI^2*E/(12*(1-MU^2)*(d/T@W)^2)=5.35*PI^2*2*10^5/(12*(1-0.3^2)*(1150/10)^2 TAU@CR,E=73.12 N/MM^2 LAMBDA=SQRT(F@YW/(SQRT(3)*TAU@CR,E))=SQRT(250/(SQRT(3)*73.12))=1.404 TAU@B=F@YW/SQRT(3)*)(LAMBDA@W)^2=250/SQRT(3)*)(1.404)^2=73.12 N/MM^2…

SOLUTION 3 C= COMPRESSION T= TENSION ALL SUPPORT ARE PINNED  CASE A 1. SINGLE DIAGONAL BRACING (N SHAPE)  WITH LATERAL LOAD 2. SINGLE DIAGONAL BRACING N SHAPE WITH ALTERNATE DIRECTION (V SHAPE) WITH LATERAL LOAD 3. X BRACING WITH LATERAL LOAD 4. ECCENTRIC BRACING WITH LATERAL LOAD CASE B. 1. SINGLE…

Modelling of Structure in STAAD Pro   OPEN STAAD PRO SELECT MODEL TYPE AS  ANALYTICAL MODEL AND UNIT AS METRIC CALCULATION OF RIDGE HEIGHT 15*TAN(20)DEGREE=5.46 M         STEP 1. SELECT GEOMETRY AND ENTER NODE VALUES WITH ORIGIN AS X= 0, Y=0 Z=0 ALONG THE WIDTH OF 30 M   LIKE WISE AS…

OPEN STAAD PRO THE STAAD MODEL 1-7LOAD .STD HAS BEEN UPDATED WITH LOADING AND DESIGN PARAMETERS AND ADDED AS AN ATTACHMENT     STEP 1. GO TO LOADING IN RIBBON TOOL BAR CLICK LOAD CASES DETAILS CLICK PRIMARY DEFINE LOAD 1 WITH LOADING TYPE DEAD LOAD AND TITLE AS = DL   STEP 2. DEFINE LOAD 2 WITH LOADING TYPE…

  SOLUTION 1   REFER STAAD MODEL STRUCTURE 50     GIVEN DATA Bay spacing as 5 m  Frame width is 20 m Slope of the roof is 1 in 10 LOADING DATA Roof is accessible for maintenance purpose (assume 75 kg/m2) Weight of roofing sheet = 5kg/m2 Weight of purlins + tertiary members = 25 kg/m2 DESIGN CODE…

SOLUTION 1.   PURLIN NUMBER DETERMINATION  . RAFTER SPAN =50M BAY SPACING =8 M ROOF SLOPE =1 IN 10 HEIGHT OF COLUMN= 12 M BRICK WALL HEIGHT = 2 M PURLIN SPACING =1.2 M GIRT SPACING =1.5 M X=SQRT(2.5^2+25^2)=25.125 M NUMBER OF PURLIN =RAFTER LENGTH/PURLIN SPACING +1=25.125/1.2+1=22 NUMBER HENCE NUMBER OF PURLIN…

SOLUTION 1. DESIGN CODE Design code as per IS 800- 2007 RIDGE HEIGHT TAN^(1/10)=5.71  15*TAN(5.71)= 1.50 M Determining Wind load for certain category and condition as per IS 875 Part 3 :2015 PLAN DIMENSION = 140 m x 30 m HEIGHT OF BUILDING = 11.1 M BAY SPACING =  10 m. THE BASIC WIND SPEED IN CHENNAI FROM WIND…

SOLUTION 1. 1. Friction load determination VERTICAL FORCE =125 KN TAKE FRICTION COEFFICIENT =0.3 ASSUMPTION ALONG THE PIPE -RESTRAINED IS THERE TRANSVERSE-NO RESTRAINT ONLY 10 PERCENT IS CONSIDERED FOR TRANSVERSE LONGITUDINAL =125*0.3=37.5 KN TRANSVERSE =125*0.1=12.5 KN     If TEFLON pad is used between shoe…

  SOLUTION 1. ISOLATED FOOTING REFER MODEL WEEK 12 OPEN STAAD FOUNDATION GO TO FOUNDATION PLAN LINEAR GRID SET UP ENTER THE FOLLOWING INPUT     GO TO COLUMN POSITION. ENTER THE FOLLOWING INPUT         GO TO COLUMN DIMENSION . ENTER THE FOLLOWING INPUT SHAPE =RECTANGULAR COLUMN LENGTH…

SOLUTION 1. DESIGN CODE Design code as per IS 800- 2007 RIDGE HEIGHT TAN^(1/20)=2.862 29.745*TAN(2.862)= 1.487 M Determining Wind load for certain category and condition as per IS 875 Part 3 :2015 PLAN DIMENSION = 169.080 m x 59.49 m HEIGHT OF BUILDING = 9.7 M BAY SPACING =  8 m. THE BASIC WIND SPEED FROM WIND ZONE…

  SOLUTION 1 REFER STAAD MODEL PROJECT 1 CALCULATION FOR M 18 DIAMETER BOLTS CONSIDER DIAMETER OF BOLT D@B= 18 MM DIAMETER OF BOLT HOLE D@H= 20+2 = 20 MM BOLTS G.8.8 F@UB = 800 N/MM^2 F@Y= 640 N/MM^2 ASSUMING PLATE PROPERTY  F@UP= 410 N/MM^2 F@Y= 250 N/MM^2 MINIMUM EDGE DISTANCE=1.5*D@0=1.5*20=30…

SOLUTION 1 REFER STAAD MODEL PROJECT 1 CALCULATION FOR M 18 DIAMETER BOLTS CONSIDER DIAMETER OF BOLT D@B= 18 MM DIAMETER OF BOLT HOLE D@H= 20+2 = 20 MM BOLTS G.8.8 F@UB = 800 N/MM^2 F@Y= 640 N/MM^2 ASSUMING PLATE PROPERTY  F@UP= 410 N/MM^2 F@Y= 250 N/MM^2 MINIMUM EDGE DISTANCE=1.5*D@0=1.5*20=30…