Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension) B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] 5 [m] 5 [m] 5 [m] 25.6 (tension) 25.6 (compression) 26.3 (tension) 26.3 (compression) Find the force in member CD, in kiloNewtons. [pause] In this problem, --- 60 [kN]

Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension) B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] 5 [m] 5 [m] 5 [m] 25.6 (tension) 25.6 (compression) 26.3 (tension) 26.3 (compression) A truss is subjected to a 60 kiloNewton force, downward, at joint F, and also subjected a ---- 60 [kN]

Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension) B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] 5 [m] 5 [m] 5 [m] 25.6 (tension) 25.6 (compression) 26.3 (tension) 26.3 (compression) 20 kiloNewton force, to the right, at joint B. Externally, the truss is connected --- 60 [kN]

Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension) B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] 5 [m] 5 [m] 5 [m] 25.6 (tension) 25.6 (compression) 26.3 (tension) 26.3 (compression) with a pin connection, at joint A, and a roller connection, at joint E. 60 [kN]

Find: FCD [kN] 25.6 (tension) 25.6 (compression) 26.3 (tension) B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] 5 [m] 5 [m] 5 [m] 25.6 (tension) 25.6 (compression) 26.3 (tension) 26.3 (compression) The dimensions of the truss, are also provided. [pause] The force in member CD can be determined, ---- 60 [kN]

Find: FCD [kN] C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] 5 [m] 5 [m] by cutting an imaginary section through members, CD, DG and FG, --- 60 [kN]

Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEG o +FCD * dDG * cos(90-θ) 7[m] FDG FFG G F E summing the moments at point G, setting the value to zero, and solving for the for the force --- 5 [m] 5 [m] 60 [kN] FE,y

Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEG o +FCD * dDG * cos(90-θ) 7[m] FDG 60 [kN] * dFG-FE,y*dEG FCD = FFG F E dDG * cos(90-θ) o in member CD. [pause] This equation assumes, ---- 5 [m] 60 [kN] FE,y

Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEG o +FCD * dDG * cos(90-θ) 7[m] FDG 60 [kN] * dFG-FE,y*dEG FCD = FFG F E dDG * cos(90-θ) o member CD experiences a tensile force. Currently, the unknown variables in this equation include --- 5 [m] tensile 60 [kN] FE,y

Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEG o +FCD * dDG * cos(90-θ) 7[m] FDG 60 [kN] * dFG-FE,y*dEG FCD = FFG F E dDG * cos(90-θ) o the length of member DG, the vertical force at joint E, and the angle theta, made by members CD and DG. 5 [m] tensile 60 [kN] FE,y

Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEG o +FCD * dDG * cos(90-θ) 7[m] FDG 60 [kN] * dFG-FE,y*dEG FCD = FFG F E dDG * cos(90-θ) o To solve for the vertical force at joint E, --- 5 [m] tensile 60 [kN] FE,y

Find: FCD [kN] ΣMA = 0 C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] 5 [m] We’ll sum the moments about joint A, which includes --- FE,y 60 [kN]

Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] the 20 kiloNewton force at joint B, --- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE

Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] the 60 kiloNewton force at joint F, --- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE

Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] and the vertical reaction force, at the roller connection, at point E. Solving for the vertical force at joint E, --- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE

Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] we’re left without knowing the angle phi, ---- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= dAE

Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] 20 [kN] 7[m] A H G F E 5 [m] or the distances between the joints, A and B, A and E, and A and F. FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= dAE

Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] dAB= (5[m])2+(7[m])2 7[m] 7[m] A H G F E 5 [m] 5 [m] 5 [m] 5 [m] ΣMA = 0 The length of member AB equals the square root of 74 --- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= dAE

Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] dAB= (5[m])2+(7[m])2 7[m] 7[m] H G F E 5 [m] 5 [m] 5 [m] 5 [m] ΣMA = 0 meters. The length from joint A to joint E, equals, --- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= dAE

Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] dAB= (5[m])2+(7[m])2 7[m] H G F E dAE = 20 [m] 5 [m] 5 [m] 5 [m] 5 [m] ΣMA = 0 20 meters, and the length from joint A to joint F, equals, ---- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= dAE

Find: FCD [kN] ΣMA = 0 φ C B D 2 [m] dAB= (5[m])2+(7[m])2 7[m] H G F E dAE = 20 [m] dAF = 15 [m] 5 [m] 5 [m] 5 [m] 5 [m] ΣMA = 0 15 meters. [pause] The angle phi is calculated --- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= dAE

Find: FCD [kN] ΣMA = 0 φ=tan-1 φ C 5 [m] B D 2 [m] 7 [m] 7[m] 7[m] A H G F dAB= 74 [m] E dAE = 20 [m] dAF = 15 [m] 5 [m] 5 [m] 5 [m] 5 [m] ΣMA = 0 by taking the arctangent of 5 meters, divided by 7 meters, which equals, --- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= dAE

Find: FCD [kN] ΣMA = 0 φ=tan-1 φ=35.5 φ C 5 [m] B D 2 [m] 7 [m] 7[m] o φ 7[m] 7[m] A H G F dAB= 74 [m] E dAE = 20 [m] dAF = 15 [m] 5 [m] 5 [m] 5 [m] 5 [m] ΣMA = 0 35.5 degrees. [pause] Plugging in the angle, the lengths, and the forces, --- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= dAE

Find: FCD [kN] ΣMA = 0 φ=35.5 φ C B D 2 [m] 20 [kN] 7[m] A H G F o φ 7[m] A H G F dAB= 74 [m] E dAE = 20 [m] dAF = 15 [m] 5 [m] 5 [m] 5 [m] 5 [m] ΣMA = 0 the vertical reaction force at joint E equals --- FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= dAE

Find: FCD [kN] ΣMA = 0 φ=35.5 φ C B D 2 [m] 20 [kN] 7[m] A H G F o φ 7[m] A H G F dAB= 74 [m] E dAE = 20 [m] dAF = 15 [m] 5 [m] 5 [m] 5 [m] 5 [m] ΣMA = 0 52 kiloNewtons. And we can add this value to the figure. FE,y 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= =52 [kN] dAE

Find: FCD [kN] ΣMA = 0 φ=35.5 φ C B D 2 [m] 20 [kN] 7[m] A H G F o φ 7[m] A H G F dAB= 74 [m] E dAE = 20 [m] dAF = 15 [m] 5 [m] 5 [m] 5 [m] 5 [m] ΣMA = 0 [pause] Returning to the section we cut earlier, --- 52 [kN] 60 [kN] 0 = -20 [kN]*dAB*cos(φ) - 60[kN]*dAF + FE,y*dAE 20 [kN]*dAB*cos(φ) + 60[kN]*dAF FE,y= =52 [kN] dAE

Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEG o +FCD * dDG * cos(90-θ) 7[m] FDG 60 [kN] * dFG-FE,y*dEG FCD = FFG F E dDG * cos(90-θ) o we can plug in our value of 52 kiloNewtons for the vertical reaction at joint E. 5 [m] 60 [kN] FE,y

Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEG o +FCD * dDG * cos(90-θ) 7[m] FDG 60 [kN] * dFG-FE,y*dEG FCD = FFG F E dDG * cos(90-θ) o The other two unknown variables needed to solve for the force --- 5 [m] 60 [kN] 52 [kN]

Find: FCD [kN] ΣMG = 0 θ =-60 [kN] * dFG+FE,y * dEG o +FCD * dDG * cos(90-θ) 7[m] FDG 60 [kN] * dFG-FE,y*dEG FCD = FFG F E dDG * cos(90-θ) o in member CD are, the length of member DG, and the angle, theta. [pause] 5 [m] 60 [kN] 52 [kN]

Find: FCD [kN] dDG θ 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ) C 2 [m] dDG= (5[m])2+(7[m])2 FCD D θ dDG 7[m] FDG FFG G F E The length of member DG, is the same hypotenuse we previously calculated, --- 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ)

Find: FCD [kN] dDG θ 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ) C 2 [m] dDG= (5[m])2+(7[m])2 FCD D θ dDG= 74 [m] dDG 7[m] FDG FFG G F E the square root of 74, meters. [pause] To find the angle theta, --- 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ)

Find: FCD [kN] θ 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ) C C 2 [m] FCD D θ D 7[m] FDG FFG G F E we’ll extract the triangle CDG from the truss, and define the angles, ---- 5 [m] 5 [m] 60 [kN] 52 [kN] G 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ1 θ θ2 60 [kN] * dFG-FE,y*dEG FCD = dDG * cos(90-θ) C 2 [m] FCD D θ1 θ D 7[m] θ2 FDG FFG G F E theta 1 and theta 2. Which makes theta equal to --- 5 [m] 5 [m] 60 [kN] 52 [kN] G 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ1 θ θ2 60 [kN] * dFG-FE,y*dEG FCD = 2 [m] θ = θ1 + θ2 FCD D θ1 θ D 7[m] θ2 FDG FFG G F E theta 1 plus theta 2. From the given dimensions of the truss, --- 5 [m] 5 [m] 60 [kN] 52 [kN] G 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ1 θ θ1=tan-1 60 [kN] * dFG-FE,y*dEG FCD = 2 [m] θ = θ1 + θ2 FCD D 2 [m] θ1 θ 5 [m] 7[m] FDG 2 [m] θ1=tan-1 FFG G F E 5 [m] Theta 1 equals the arctangent of 2 meters divided by 5 meters, or --- 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ1 θ θ1=tan-1 θ1=21.8 2 [m] θ = θ1 + θ2 FCD D 2 [m] θ1 θ 5 [m] 7[m] FDG 2 [m] θ1=tan-1 FFG G F E 5 [m] o θ1=21.8 21.8 degrees. [pause] Likewise, theta 2 equals, --- 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ2 θ2=tan-1 60 [kN] * dFG-FE,y*dEG FCD = 2 [m] θ = θ1 + θ2 FCD D θ 5 [m] 7[m] θ2 FDG 7 [m] 7 [m] FFG θ2=tan-1 F E 5 [m] the arctangent of 7 meters divided by 5 meters, which equals, --- 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ2 θ2=tan-1 θ2=54.5 2 [m] θ = θ1 + θ2 FCD D θ 5 [m] 7[m] θ2 FDG 7 [m] 7 [m] FFG θ2=tan-1 F E 5 [m] 54.5 degrees. [pause] Adding the values of theta 1, --- 5 [m] 5 [m] o θ2=54.5 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 60 [kN] * dFG-FE,y*dEG 2 [m] θ = θ1 + θ2 FCD D θ o θ1=21.8 7[m] o θ2=54.5 FDG FFG G F E and theta 2, theta equals, --- 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 2 [m] θ = θ1 + θ2 FCD D θ o θ1=21.8 7[m] o θ2=54.5 o FDG θ=76.3 FFG G F E 76.3 degrees. [pause] From observation, the length of --- 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 2 [m] θ = θ1 + θ2 FCD D θ o θ1=21.8 7[m] o θ2=54.5 o FDG θ=76.3 FFG dFG=5 [m] G F E member FG equals 5 meters, and the length between joints E and G equals --- 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 2 [m] θ = θ1 + θ2 FCD D θ o θ1=21.8 7[m] o θ2=54.5 o FDG θ=76.3 FFG dFG=5 [m] G F E dEG=10 [m] 10 meters. Plugging these values into our equation, 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 2 [m] θ = θ1 + θ2 FCD D θ o θ1=21.8 7[m] o θ2=54.5 o FDG θ=76.3 FFG dFG=5 [m] G F E dEG=10 [m] the tensile force in member CD equals, --- 5 [m] 5 [m] 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 FCD =-26.3 [kN] 2 [m] θ = θ1 + θ2 FCD D θ o θ1=21.8 7[m] o θ2=54.5 o FDG θ=76.3 FFG G F E FCD =-26.3 [kN] negative 26.3 kiloNewtons, in tension, which is equivalent a compressive force of, -- 5 [m] 5 [m] (tension) 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 FCD =26.3 [kN] 2 [m] θ = θ1 + θ2 FCD D θ o θ1=21.8 7[m] o θ2=54.5 o FDG θ=76.3 FFG G F E FCD =26.3 [kN] 26.3 kiloNewtons. 5 [m] 5 [m] (compression) 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 25.6 (tension) 2 [m] θ = θ1 + θ2 FCD D θ o θ1=21.8 7[m] o θ2=54.5 o FDG θ=76.3 25.6 (tension) 25.6 (compression) 26.3 (tension) 26.3 (compression) FFG G F E FCD =26.3 [kN] When reviewing the possible solutions, --- 5 [m] 5 [m] (compression) 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

Find: FCD [kN] θ = θ1 + θ2 θ θ1=21.8 θ2=54.5 θ=76.3 AnswerD 2 [m] θ = θ1 + θ2 FCD D θ o θ1=21.8 7[m] o θ2=54.5 o FDG θ=76.3 25.6 (tension) 25.6 (compression) 26.3 (tension) 26.3 (compression) FFG AnswerD G F E FCD =26.3 [kN] the answer is D. 5 [m] 5 [m] (compression) 60 [kN] 52 [kN] 60 [kN] * dFG-FE,y*dEG FCD = dDG= 74 [m] dDG * cos(90-θ)

( ) ? τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal Find: σ’v ρc d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ φ γ Δ d ˚ d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] φ=α1-α2 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] τ [lb/ft2] (5 [cm])2 * π/4 ( ) H*C σfinal ρcn= 1+e σinitial log ‘ φ c=0 400 1,400 σ3 Sand σ1