Dr. Mohammed E. Haque, P.E. Lecture Notes

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1 nalysis of Selected Determinate Structural Systems Planar Trusses Method of Joints Planar Trusses Method of Sections Pinned Frames with Multi-force Members Retaining Walls OS321Haque 1

2 Planar Trusses Method of Joints Dr. Mohammed E. Haque, P.E. Example 1: alculate support reactions and member forces of the truss as shown. 3k D E 8 8k x y y First alculate the support reactions F x = 0 x=3k (Left) M = 0 y(24) 3(8) 8(12) = 0 y = 5k (up) M = 0 -y(24) -3(8) +8(12) = 0 y = 3k (up) Proof: F y = 0 y + y - 8 =? = 0 Take Joints FD, and solve member forces rrow directed towards a joint represents ompression member force, and arrow directed away from the joint represents Tension member force. Dy D Dx F y = 0 Dy =3 D(4/5) = 3 D=15/4=3.75k () F x = 0 = Dx = 3.75(3/5) = 2.25k (T) FD of Joint y = 3k OS321Haque 2

3 E Ex Ey x = 3k F y = 0 Ey =5 E(4/5) = 5 E=25/4=6.25k () F x = Ex 3 = (3/5) 3 =0 = 0.75(T) y = 5k FD of Joint 3k D DE F y = 0 Dy = Dy D(4/5) = 3.75(4/5) D=3.75k (T) D =3.75 D F x = 0 -DE+3+Dx + Dx =0 DE= (3/5) (3/5) = 7.5k() FD of Joint D D=3.75 E F y = 0 Ey +Dy 8 =0 E(4/5) +3.75(4/5) 8 =0 E(4/5) =0 E = 6.25k (T) D=0.75 =2.25 FD of Joint 8k NSWER: Reactions: y=3k (up); x=3k (Left), y=5k(up) Member Forces: = 2.25(T); = 0.75k (T); DE=7.5k() D=3.75k( ) ; D=3.75k(T); E=6.25k(T); E=6.25k() OS321Haque 3

4 Example 2: alculate support reactions and member forces, DE, D and of the truss as shown. 3k D a E 8 y 6 a 6 8k 6 6 y x First alculate the support reactions F x = 0 x=3k (Left) M = 0 y(24) 3(8) 8(12) = 0 y = 5k (up) M = 0 -y(24) -3(8) +8(12) = 0 y = 3k (up) Proof: F y = 0 y + y - 8 =? = 0 Take a section along a-a, and solve for member forces 3k D DE E F y = 0 -Dy +3 =0 -D(4/5) +3 = 0 D=15/4=3.75k (T) D M = 0 DE(8) -3(8) - 3(12) =0 DE= 7.5k( ) F x = 0 +3+Dx-DE = (3/5) 7.5 =0 = 2.25k (T) y=3k OS321Haque 4

5 Example 3: alculate reactions, x, y and x. Find member forces, E, E, and D using section method. y a x E 2m 2.5kN 2m D 2.0kN a 2m x Find out support Reactions: F y = 0 y= = 4.5 kn (up) M = 0 -x (2) +2.5(4) + 2(2) =0 x= 7 kn(right) F x = 0 x= 7 kn (Left) OS321Haque 5

6 a E E 1 2 E 2.5kN D 2.0kN D a 2m Take a section along a-a M = 0 -Ex(1) - Ey(2) +2.5(4) + 2(2) =0 E(2/ 5)(1) + E(1/ 5)(2) = 14 E = kn (T ) M = 0 Ex( 1) + Ey (2) 2(2) = 0 E(2/ 5)( 1) + E(1/ 5)(2) =4 E = kn ( ) M E = 0 D(1) + 2.5(2) =0 D = -5 kn () (arrow should be toward joint D) OS321Haque 6

7 Pinned Frames with Multi-force Members Example 1: alculate the support reactions at and. lso calculate internal pin force at. 4 ft 4 ft 4 ft 200 lb 6ft 250 lb D 4 ft 400 lb 6ft x y 4 ft 4 ft 4 ft 200 lb 250 lb x y D 4 ft 400 lb OS321Haque 7

8 Find Support Reactions at &, and Pin force at : M = 0 x(6) 200(4) 400(4) 250(12) = 0 x=900 lb (Left) Fx = 0 x = 900 lb (Right) x y 4 ft 4 ft 4 ft 200 lb FD-1 x x y y x 250 lb D 4 ft 400 lb FD-2 From FD-2: M = 0 y(8) + 250(4) 400(4) = 0 y = 75 lb (up) F Y = 0 y y = y = 0 y = 575 lb From FD-1: M = 0 x(6) y(8) +200(4) = 0 900(6) y(8) +200(4) = 0 y=775 lb (up) F X = 0 x x =0 x=900 lb Results: x=900 lb (Left); y=775 lb (Up); x=900 lb (Right); y=75 lb (Up); x=900 lb; y=575 lb OS321Haque 8

9 Example 2: wind force of 0.12k/ft is applied to the vertical projection of a building as shown in fig. alculate the support reactions at and. lso calculate internal pin force at k/ft OS321Haque 9

10 Dr. Mohammed E. Haque, P.E F = 0.12x35 =4.2 k 20 35/2=17.5 x x FD-1 y y x y From FD-1 M = 0 y(40) 4.2(17.5) = 0 y=1.838 k (Up) Fy = 0 y = k (down) From FD-2 Fy = 0 y = 1.838k M = 0 x(35) = 1.838(20) x=1.05k (Left) F X = 0 x x =0 x = 1.05k 35 From FD-1 F X = 0 -x x +4.2 =0 -x = 0 x=3.15k (Left) FD-2 x y=1.838k RESULTS: x=3.15k (Left); y=1.838k (Down) x=1.05k (Left); y=1.383k (Up) x=1.05k; y=1.838k 20 OS321Haque 10

11 Retaining Walls ommon Retaining Walls Ground Level ack Fill Ground Level ack Fill Stem Ground Level Ground Level Toe Heel Gravity or Semi-gravity Retaining wall Toe Footing antilever Retaining wall Heel Ground Level ack Fill Ground Level ack Fill uttress Stem Stem ounterfort Ground Level Ground Level Toe Footing Heel Toe Footing Heel uttress Retaining wall ounterfort Retaining wall OS321Haque 11

12 First Floor Ground Level Stem asement Floor Footing asement Wall Foundation OS321Haque 12

13 ack-wall earing Plate ridge Girder ack Fill Ground Level Pile ap Piles ridge butment OS321Haque 13

14 Lateral Soil Pressure on Retaining Walls ackfill γ soil h Fig. 1: Soil Pressure on the back of wall (No surcharge) P max = K a γ soil h K a = oefficient of ctive Soil Pressure K a = tan 2 ( φ/2) φ = ngle of Internal Friction for backfill soil γ soil = Unit weight of Soil (pcf) P max = K a γ soil h P max = w h Where w = equivalent fluid pressure of soil w = K a γ soil Typical ngle of Internal Friction for backfill soil Soil Type φ (Degree) Gravel and coarse sandy backfill soil Medium to fine sandy backfill soil Silty sand OS321Haque 14

15 β ackfill γ soil h K a = cos β cos β - (cos 2 β - cos 2 φ) cos β + (cos 2 β - cos 2 φ) P max = K a γ soil h Fig. 2: Soil Pressure on the back of wall (Sloping ackfill) OS321Haque 15

16 Surcharge, w sc ackfill γ soil + h P sc = K a w sc P max = K a γ soil h Fig. 3: Soil Pressure on the back of wall (with uniform surcharge) OS321Haque 16

17 Q1(a): nalyze the stability of the reinforced cantilever retaining wall as shown in Figure. Use the following values: oncrete unit weight = 150 pcf Soil unit weight, γ soil =110 pcf oefficient of ctive Soil Pressure, K a = 0.33 (Neglect oefficient of passive Soil Pressure, K p ) oefficient of friction between the bottom of footing and soil, µ = OS321Haque 17

18 Solution: W soil 12 W wall F H 2 1/3(12 )=4 2 W Footing Toe P max Step 1: alculate lateral soil pressure and overturning moment P max = K a γ soil h = 0.33 (110)(12) = psf F H = ½ P max h = ½ (435.6)(12)= lb/ft of wall Sliding Force, F H = lb/ft of wall Overturning Moment, M OT about toe = x 4 = lb-ft /ft of wall. OS321Haque 18

19 Step 2: alculate weights, Resisting Moment, Sliding Resisting Force; W soil = (4 x 10)(110) = 4400 lb./ft of wall W wall = (1 x 10)(150) = 1500 lb / ft of wall W Footing = (2 x 7)(150) = 2100 lb / ft of wall W Total = = 8000 lb / ft of wall Resisting moment, M R about toe = (4400 x 5) + (1500 x 2.5) + (2100 x 3.5) = lb-ft/ft of wall Sliding Resisting Force, F R = µ X W Total = 0.5 (8000) = 4000 lb Step 3: determining the Factor of Safety (FS) against overturning and sliding. FS OT = M R / M OT = / = 3.17 > 2.0 OK FS Sliding = F R / F H = 4000 / = 1.53 > 1.5 OK OS321Haque 19

20 Soil Pressure under the footing of the Retaining Wall c= /2.L. c > /3.L. e</6 c /3.L. e /6 P= W Total P= W Total P= W Total f = f max = P/ = P/(x1) ase I: Load, P with zero eccentricity (e=0) f max f max = P/ + (P e /S m ) f min = P/ - (P e /S m ) where = 1x S m = 1x( 2 )/6 f min f max f max = (2/3) P/c ase III: Load, P with large eccentricity (e /6) ase II: Load, P with small eccentricity (e</6) OS321Haque 20

21 Q1(b) alculate the soil pressure under the footing of the retaining wall of Q1(a). e =7.L. of Footing 2 c= W Total Toe f max f min Determine the resultant vertical force, W Total intersects the bottom of the footing: c = (M R M OT )/ W Total = ( ) / 8000 = ft. from the Toe. Eccentricity, e = /2 c = (7/2) = < /6 (=7/6= ) Therefore, ase II is applicable. alculate soil pressures, f max and f min under the footing: = 1x = 7 sqft. S m = 1x( 2 )/6 = 7 2 / 6 = ft 3 P= W Total = 8000 lb. f max = P/ + (P e /S m ) = (8000/7) +(8000x0.6693/8.1667) = = psf f min = P/ - (P e /S m ) = = = psf OS321Haque 21

Compute the lateral force per linear foot with sloping backfill and inclined wall. Use Equation No. 51, page 93. Press ENTER.

Compute the lateral force per linear foot with sloping backfill and inclined wall. Use Equation No. 51, page 93. Press ENTER. Sample Problems Problem 5.1 A gravity retaining wall is supporting a cohesionless soil. The active lateral force per linear foot of the retaining wall is most nearly (A) 5,000 lb/ft (B) 6,000 lb/ft (C)