Structural Steel Design Project

Size: px

Start display at page:

Download "Structural Steel Design Project"

Janis Cole
5 years ago
Views:

1 Job No: Sheet 1 of 6 Rev Worked Example - 1 Made by Date Checked by PU Date Analyse the building frame shown in Fig. A using portal method. 15 kn C F I L 4 m 0 kn B E H K 6 m A D G J 4 m 6 m 4 m Top storey: Fig. A (i) Column Shears: Shear in columns of the top storey is obtained by considering the free body diagram shown in Fig. A.1 15 kn C Hinge F I L m V 1 V 1 V 1 V 1 Plane containing Fig. A.1 points of contraflexure Version II 38 - {PAGE }

2 Job No: Sheet of 6 Rev Worked Example 1 Made by Date Checked by PU Date V 1 + V 1 + V 1 + V 1 15 kn [Assumption 3] Shear in end column, V 1.5 kn Shear in middle columns, V kn Thus, shear in columns are: Column Shear (kn) CB.5 FE 5.0 IH 5.0 LK.5 (ii) Column moments: Column moments are found by multiplying column shear and half the height of column as shown below: Column Shear (kn) Moment (kn-m) CB.5.5 * 5.0 FE * 10.0 IH * 10.0 LK.5.5 * 5.0 (iii) Girder Moments: At any joint, sum of the girder moments is equal to the sum of the column moments. Starting from left corner of the frame, C Joint C: M CB M CF 5 kn-m Version II 38 - {PAGE }

3 Job No: Sheet 3 of 6 Rev Worked Example - 1 Made by Date Checked by PU Date Joint F: Joint I: M FC + M FI M FE M FI M FE - M FC kn-m M IF + M IL M IH M IL M IH - M IF kn-m (iv) Girder shears: Girder shear Girder Moment / (span/) Span Span/ (m) Moment (kn-m) Shear (kn) CF FI IL (v) Column axial forces: (See Fig. A.) Axial force on a column is determined by summing up the girder shears and other axial forces at each joint. Starting from the left corner of the frame, we have Joint C: Σ F y 0 F CB V CF.5 kn Joint F: Σ F y 0 F EF + V FC V FI F EF V FI - V FC kn Version II 38 - {PAGE }

4 Job No: Sheet 4 of 6 Rev Worked Example - 1 Made by Date Checked by-pu Date Joint I : Σ F y 0 V IF F IH + V IL F IH V IF - V IL kn Joint L : Σ F y 0 F KL V LI.5 kn Ground storey: (i) Column shears: Shear in the columns of ground storey is calculated in similar way to the top storey. The plane containing points of contraflexure will pass through the half the height of the ground storey. Then shear in columns is calculated as ΣF H 0 6V 1 35 kn. V 1 35/6 5.8 kn. Column Shears; V AB 5.8 kn ; V DE 11.7 kn V JK 5.8 kn ; V GH 11.7 kn. (ii) Column moments: Column Length/ (m) Shear (kn) Moment (kn-m) AB DE GH JK

5 Job No: Sheet 5 of 6 Rev Worked Example - 1 Made by Date Checked by PU Date (iii) Girder moments: Joint B: M BE M BC + M BA kn-m Joint E: M EH M EF + M ED - M EB Joint H: kn-m M HK M IH + M HG - M HE kn-m Joint K: M KH M KL + M JK kn-m (iv) Girder shears: Beam Span/ (m) Moment (kn-m) Shear (kn) BE EH HK (v) Column axial forces: (Consider the Fig. A.) Joint B: Σ F V 0 F AB F BC + V BE kn

6 Job No: Sheet 6 of 6 Rev Worked Example - 1 Made by Date Checked by PU Date Joint E: F ED F EF + V EH - V EB kn Joint H: F GH V EH + F IH - V HK kn Joint K: F JK V KH + F LK kn 15 kn 0 kn C M 5.0 V.5 F.5 B F M 5.0 V.5 M 10.0 V 5.0 F M.5 V 11.3 E M 5.0 V 1.67 M.5 V 7.5 I M 10.0 V 5.0 F H M 5.0 V.5 M 5.0 V.5 F.5 M.5 V 11.3 L K 4 m M 17.5 V 5.8 F 13.8 M 35.0 V 11.7 F -4.6 M 35.0 V 11.7 F -4.6 M 17.5 V 5.8 F m A D G J 4 m 6 m 4 m Fig. A. Fig. A.. Axial forces in columns and shear forces in members M Moment in kn-m V Shear in kn F Axial force in kn

7 Job No: Sheet 1 of 7 Rev Worked Example - Made by Date Checked by PU Date Problem : Analyse the building frame shown in previous example (Fig. A) using Cantilever method. Assume cross-sectional areas of all the columns as equal. Top storey: (1) Location of centroidal line of columns of the storey: Let the area of each column be A and x be distance to the centre of gavity of columns shown in Fig. B.1 15 kn C Girder shears F I L A x A A A m X 1 Axial forces in columns 4 m 6 m 4 m Fig. B.1 Take moments about column BC 0 A + 4 A + 10 A+ 14 A 8A x 7m 4A 4A () Column axial forces: (See Fig. B.1) In cantilever method, it is assumed that the axial forces in the columns are proportional to the horizontal distance from the center of gravity of the columns in the storey. Say, F BC F

8 Job No: Sheet of 7 Rev Worked Example Made by Date Checked by PU Date F BC 7 F EF 3 3 F HI F 7 ; F EF 3 F BC 3 / 7 F 7 F KL F Take moments about X 1, 15 + F EF 4-10 F HI - 14 F KL F 4-10 F -14 F F - 14F F 7 10 F 1.81kN 116 F BC 1.81 kn F HI 0.78 kn F EF 0.78 kn F KL 1.81kN (3) Shear forces at the ends of beams: (See Fig. B.1) Joint C: F y 0 V CF F BC 1.81 kn Joint F: V FC + F EF V FI V FI k N ( Q V FC V CF )

9 Job No: Sheet 3 of 7 Rev Worked Example - Made by Date Checked by PU Date Joint I: V IL V IF - F HI kn (V IF V FI ) Joint L: V LI F KL 1.81 kn (4) Girder moments: Girder moment Girder shear * Span/ Girder Shear (kn) Span / (m) Moment (kn-m) CF FI IL (5) Column moments: At each joint, sum of girder moments equals to sum of column moments. Consider joints from left corner of the floor. Joint C: M CB M CF 3.6 kn-m Joint F: M FE M FC + M FI kn-m Joint I: M IH M IF + M IL kn-m Joint L: M LI M LK 3.6 kn-m (6) Column Shears: Column Shear Column moment / (Length/) Column Moment (kn-m) Length/ (m) Shear (kn) BC EF HI KL

10 Job No: Sheet 4 of 7 Rev Worked Example - Made by Date Checked by PU Date Ground storey: (i) Location of centroidal line of columns of the storey: Consider the following free body diagram shown Fig. B. 15 kn C F I L x 0 kn B E H K A A A A X 4 m 6 m 4 m F AB F DE F GH F JK Fig. B. Let the area of each column be 'A' and x be the distance to the centre of gravity of columns as shown in Fig. B. 0 A + 4 A + 10 A+ 14 A x 7 m 4A (ii) Column axial forces: Say, F AB F Then, F DE 3/7 F ; F GA 3/7 F and F JK F

11 Job No: Sheet 5 of 7 Rev Worked Example - Made by Date Checked by PU Date Taking moments about X, 15 (4 + 6/ ) + 0 (6/ ) + F DE 4 F GH F JK F F 14F F kN Axial forces in columns are, F AB 10.0 kn ; F DE 4.3 kn F GH 4.3 kn ; F JK 10.0 kn. (iii) Beam shears: [See Fig B.3] Joint B: Joint E: Joint H: F AB V BE + F BC V BE V EB + F DE V EH + F EF V EH kn (V EB V BE ) V HK V HE - F GH + F HI V HK kn (V HE V EH ) Joint K: V KH F JK - F KL kn

12 Job No: Sheet 6 of 7 Rev Worked Example - Made by Date Checked by PU Date (iv) Girder Moments: Span Length/ (m) Shear (kn) Moment (kn-m) BE EH HK (v) Column moments: At each joint sum of column moments equals to sum of girder moments Joint B: Joint E: Joint H: Joint K: M BC + M BA M BE M BA M BE - M BC kn-m M EB + M EH M EF + M ED M ED M EB + M EH - M EF kn-m M HE + M HK M HI + M HG M HG M HE + M HK - M HI kn-m M KH M KL + M JK M JK M KH - M KL kn-m (vi) Column shears: Span Length/ (m) Moment (kn-m) Shear (kn) AB DE GH JK

13 Job No: Sheet 7 of 7 Rev Worked Example Made by Date Checked by PU Date kn 0 kn C M 3.6 V 1.8 F 1.8 B F M 3.6 V 1.8 M11.4 V 5.7 F 0.78 M 16.4 V 8. E M 7.77 V.59 M 35.1 V 11.7 I M 11.4 V 5.7 F 0.78 H M 3.6 V 1.8 M 3.6 V 1.8 F 1.8 M 16.4 V 8. L K 4 m M 1.8 V 4.3 F 10.0 M 40.1 V 13.3 F 4.3 M 40.1 V 13.3 F 4.3 M 1.8 V 4.3 F m A D G J 4 m 6 m 4 m Fig. A. Fig. B. 3. Axial forces in columns and shear forces in members M Moment in kn-m V Shear in kn F Axial force in kn

14 Job No: Sheet 1 of 8 Rev Worked Example 3 Made by Date Checked by PU Date Problem 3: Determine the moments at the ends of columns and beams in the building frame shown in Fig. C by factor method. The relative stiffness factors (k) are mentioned in figure. 15 kn C 4 F I 3 L 4 m 0 kn B 4 E H 3 K 6 m Relative stiffness factors A D G J 4 m 6 m 4 m Fig. C (1) Girder factors: Girder factor, g Sum of column relative stiffness factors at the joint Sum of total relative stiffness factors at that joint. Joint C: g C Joint F: g F 4+ +

15 Job No: Sheet of 8 Rev Worked Example - 3 Made by Date Checked by PU Date Joint I: g I Joint L: g L Joint B: g B Joint E: g E Joint H: g H Joint K: g K () Column factors: Column factor, c 1-g Joint g c (1-g) g/ c/ C F I L B

16 Job No: Sheet 3 of 8 Rev Worked Example - 3 Made by Date Checked by PU Date Joint g c (1-g) g/ c/ E H K A D G J (3) Column and girder moment factors (C & G): Joint Members c or g Half values (3) + (4) k C or G of factors of opposite end (1) () (3) (4) (5) (6) (5) * (6) C CF F I L B E CB FE FI FC IF IH IL LI LK BE BC BA EF EB EH ED

17 Job No: Sheet 4 of 8 Rev Worked Example - 3 Made by Date Checked by PU Date Joint Members c or g Half values of factors of opposite end (3) + (4) k C or G (1) () (3) (4) (5) (6) (5)* (6) H HI HE HG HK K KL KH KJ A AB D DE G GH J JK (4) Storey Constants: For ground storey, Let A 1 be the storey constant for determination of moments at the ends of columns of the ground storey. Then Total horizontal shear of ground storey kn Height of ground storey 6 m C (C AB + C BA ) + (C ED + C DE ) + (C HG + C GH ) + (C KJ + C JK ) A (.5 +.0) + (. +.6) + ( ) + ( )

18 Job No: Sheet 5 of 8 Rev Worked Example - 3 Made by Date Checked by PU Date For top storey, Let A be the storey constant for determination of moments at the ends of columns of the top storey, then A Total horizontal Shear of top storey ΣC X Height of top storey where, C Sum of the column end moment factors of the storey. Total horizontal shear of top storey 15 kn. Height of top storey 4m. C (C CB + C BC ) + (C FE + C EF ) + (C IH + C HI ) + (C LK + C KL ) ( ) + ( ) + ( ) + ( ) A (5) Moments at the ends of columns: Ground storey moments: Moment at end of the column Column moment factor at that end *A 1 M AB.5 * kn-m M DE.6 * kn-m M GH.556 * kn-m M JK.43 * kn-m M BA.0 * kn-m M ED. * kn-m M HG.11 * kn-m M KJ * kn-m

19 Job No: Sheet 6 of 8 Rev Worked Example - 3 Made by Date Checked by PU Date Top storey moments: Moment at end of the column Column moment factor at that end * A M BC 1.67 * kn-m M CB 1.84 * kn-m M EF 1.95 * kn-m M FE.1 * kn-m M HI 1.86 * kn-m M KL * kn-m M IH * kn-m M LK 1.63 * kn-m (6) Joint Constants: Joint constant (B) { EMBED Equation.3 } For ground storey, B B M + M G BC BE BA B E M G EF EB + M + G ED EH B H M G HI HE + M + G HG HK B K M KL G + KH M KJ

20 Job No: Sheet 7 of 8 Rev Worked Example - 3 Made by Date Checked by PU Date For top storey, B C M CB G CF B F M FE G FC + G FI B I M IH G IF + G IL B L M LK G LI (7) Moments at the ends of beams: Moment at the end of beam equals to Girder moment factor at that end multiplied by respective joint constant. M CF 1.8 * kn-m ; M FC 1.66 * kn-m M FI * kn-m ; M IF 0.8* kn-m M IL * kn-m ; M LI 1.69 * kn-m M BE.8 * kn-m ; M EB.6 * kn-m M EH 1.44 * kn-m ; M HE 1.88 * kn-m M HK 1.33 * kn-m ; M KH.187 * kn-m The values of girder factors and column factors are shown in Fig. C.

21 Job No: Sheet 8 of 8 Rev Worked Example - 3 Made by Date Checked by PU Date kn 0 kn C B F E I H L K A D G J m 6 m 4 m Fig. C Fig. C. Girder factors and column factors of the frame Factors presented in rectangular boxes are girder factors and column factors are presented simply without any rectangular box.

22 Job No: Sheet 1 of Rev Worked Example - 4 Made by Date Checked by PU Date Problem 4: Determine the moments at the ends of columns and beams of the rigidly jointed building frame shown in Fig. D for the gravity load applied. 4 m 0 kn/ m 4 m 4 m 5 m 5 m 5 m Fig. D Consider the following approximate model. 0. L 0.6 L 0. L L 5 m Approximate model

23 Job No: Sheet of Rev Worked Example - 4 Made by Date Checked by PU Date Maximum + ve B.M. at mid-span wl 8 0* 3 /8.5 kn-m End reaction wl/ 0 * 3/ 30 kn Maximum negative B.M. at end column 30 * 1 + (0 * 1 * 1) / 40 kn-m Bending moment in the interior column B.M. diagram for the frame:

3.4 Analysis for lateral loads

3.4 Analysis for lateral loads 3.4.1 Braced frames In this section, simple hand methods for the analysis of statically determinate or certain low-redundant braced structures is reviewed. Member Force Analysis