Find: Mmax [lb*ft] in AB P P P P A B d d d d 1,200 3,200 14,400 19,200 P=800 [lb] d=12 [ft] Find the maximum moment in beam AB. [pause] In this problem, a 48 foot long beam AB ---
Find: Mmax [lb*ft] in AB P P P P A B d d d d 1,200 3,200 14,400 19,200 P=800 [lb] d=12 [ft] is connected with a pin, to vertical wall, at joint A, and with a roller, to a flat surface, at joint B.
Find: Mmax [lb*ft] in AB P P P P A B d d d d 1,200 3,200 14,400 19,200 P=800 [lb] d=12 [ft] At distances of 12, 24, 36 and 48 feet away from joint A, a downward point force, of 800 pounds, is applied to the beam. [pause]
Find: Mmax [lb*ft] in AB P P P P P=800 [lb] A B d=12 [ft] d d d d To find the maximum moment in the beam, it may be easiest to ---
Find: Mmax [lb*ft] in AB P P P P P=800 [lb] A B d=12 [ft] d d d d V [lb] M sketch out the shear and moment diagrams, for the beam. To sketch the shear diagram, we first need to determine, --- [lb*ft]
Find: Mmax [lb*ft] in AB P P P P P=800 [lb] A B d=12 [ft] d d d d FB,y FA,y M the vertical reaction forces at joints A and B. By summing the moments about joint A, --- [lb*ft]
Find: Mmax [lb*ft] in AB P P P P P=800 [lb] A B d=12 [ft] d d d d FB,y FA,y ΣMA = 0=-P*d-P*2*d-P*3*d -P*4*d+FB,y*4*d M and setting the value equal to zero, the vertical reaction force at joint B equals, --- [lb*ft]
Find: Mmax [lb*ft] in AB P P P P P=800 [lb] A B d=12 [ft] d d d d FB,y FA,y ΣMA = 0=-P*d-P*2*d-P*3*d -P*4*d+FB,y*4*d M 2.5 times P, or, 2,000 pounds. [pause] The vertical reaction force at point A can be determined by ---- [lb*ft] FB,y=2.5*P FB,y=2,000 [lb]
Find: Mmax [lb*ft] in AB P P P P P=800 [lb] A B d=12 [ft] d d d d FA,y FB,y=2,000 [lb] ΣFy = 0=FA,y+FB,y-4*P M summing the forces in the vertical direction, setting the value equal to zero, and then solving for --- [lb*ft]
Find: Mmax [lb*ft] in AB P P P P P=800 [lb] A B d=12 [ft] d d d d FA,y FB,y=2,000 [lb] ΣFy = 0=FA,y+FB,y-4*P M the vertical reaction at joint A, which is the sum of the loads, minus, the vertical reaction force at joint B, or --- FA,y=4*P-FB,y [lb*ft]
Find: Mmax [lb*ft] in AB P P P P P=800 [lb] A B d=12 [ft] d d d d FA,y FB,y=2,000 [lb] ΣFy = 0=FA,y+FB,y-4*P M 1,200 pounds. [pause] We’ll begin sketching the shear diagram, ---- FA,y=4*P-FB,y [lb*ft] FA,y=1,200 [lb]
Find: Mmax [lb*ft] in AB P=800 [lb] P d A B d=12 [ft] FA,y=1,200 [lb] FB,y=2,000 [lb] V [lb] from joint A, and work our way over towards joint B. Since the vertical reaction force at joint A equals 1,200 pounds, --- A B
Find: Mmax [lb*ft] in AB P=800 [lb] P d A B d=12 [ft] FA,y=1,200 [lb] FB,y=2,000 [lb] V [lb] 1,200 the shear in the beam, just right of joint A equals 1,200 pounds, and the shear along beam AB remains 1,200 pounds until --- A B
Find: Mmax [lb*ft] in AB P=800 [lb] P d A B d=12 [ft] FA,y=1,200 [lb] FB,y=2,000 [lb] V [lb] 1,200 12 feet in the direction of B, where a point force of 800 pounds downward acts on the beam. For convenience sake, ---- A B
Find: Mmax [lb*ft] in AB P=800 [lb] P d A B d=12 [ft] C D E FA,y=1,200 [lb] FB,y=2,000 [lb] V [lb] 1,200 we’ll label the 3 intermediate points between joints A and B, as C, D and E. So between points C and D, the shear in the beam equals, --- A B
Find: Mmax [lb*ft] in AB P=800 [lb] P d A B d=12 [ft] C D E FA,y=1,200 [lb] FB,y=2,000 [lb] VCD=VAC+P V [lb] =1,200 [lb]+(-800 [lb]) 1,200 the shear in the beam from A to C, 1,200 pounds, plus the point load, P, of negative 800 pounds, which equals --- A B
Find: Mmax [lb*ft] in AB P=800 [lb] P d A B d=12 [ft] C D E FA,y=1,200 [lb] FB,y=2,000 [lb] VCD=VAC+P V [lb] =1,200 [lb]+(-800 [lb]) 1,200 400 pounds. [pause] Similarly, the shear force betweens point D and E equals, --- =400 [lb] 400 A B
Find: Mmax [lb*ft] in AB P=800 [lb] P d A B d=12 [ft] C D E FA,y=1,200 [lb] FB,y=2,000 [lb] VDE=VCD+P V [lb] =400 [lb]+(-800 [lb]) 1,200 the shear force between C and D, 400 pounds, plus the point load, P, of negative 800 pounds, which equals, --- 400 A B
Find: Mmax [lb*ft] in AB P=800 [lb] P d A B d=12 [ft] C D E FA,y=1,200 [lb] FB,y=2,000 [lb] VDE=VCD+P V [lb] =400 [lb]+(-800 [lb]) 1,200 negative 400 pounds. Lastly, the shear between point E and joint B equals, --- =-400 [lb] 400 -400 B
Find: Mmax [lb*ft] in AB P=800 [lb] P d A B d=12 [ft] C D E FA,y=1,200 [lb] FB,y=2,000 [lb] VEB=VDE+P V [lb] =-400 [lb]+(-800 [lb]) 1,200 negative 1,200 pounds. [pause] Since the problem asks to find the maximum moment in beam AB, we can use --- =-1,200 [lb] 400 -400 B -1,200
Find: Mmax [lb*ft] in AB V [lb] 1,200 400 -400 -1,200 the shear diagram to construct the moment bending diagram, and identify the maximum moment.
Find: Mmax [lb*ft] in AB V [lb] 1,200 400 -400 -1,200 dM Since the shear in the beam is equal to the change in moment along the length of the beam, then the moment is equal to --- V= dx
∫ Find: Mmax [lb*ft] in AB dM V= M= V*dx dx V [lb] 1,200 400 -400 -1,200 dM the shear, times the length along the beam. Since there is no moment in the beam at the pin connection at point A, --- ∫ V= M= V*dx dx
Find: Mmax [lb*ft] in AB V [lb] 1,200 400 -400 -1,200 M [lb*ft] C D E B the moment at point A equals 0. Since the shear is constant between joint A and point C, --- A
∫ Find: Mmax [lb*ft] in AB A MC=MA+ V*dx V [lb] 1,200 400 -400 -1,200 M [lb*ft] C D E B the moment at joint C can be calculated as, --- A
∫ Find: Mmax [lb*ft] in AB A MC=MA+ V*dx V [lb] 1,200 400 -400 -1,200 M [lb*ft] C D E B the moment at joint A plus the integral of --- A
∫ Find: Mmax [lb*ft] in AB A MC=MA+ V*dx V [lb] 1,200 400 -400 -1,200 M [lb*ft] C D E B the shear force in the beam, over the length of the beam, from A to C. The moment at point C equals, --- A
∫ Find: Mmax [lb*ft] in AB A MC=MA+ V*dx V [lb] 1,200 400 -400 -1,200 M [lb*ft] C D E B 14,400 pound feet. [pause] The moment at point D is calculated the same way, --- MC=0 [lb*ft] 14,400 +1,200 [lb]*12 [ft] A MC=14,400 [lb*ft]
∫ Find: Mmax [lb*ft] in AB A MD=MC+ V*dx V [lb] 1,200 400 -400 -1,200 M [lb*ft] C D E B by adding the moment at point C to the integral of the shear force in the beam, --- 14,400 A
∫ Find: Mmax [lb*ft] in AB A MD=MC+ V*dx V [lb] 1,200 400 -400 -1,200 M [lb*ft] C D E B times the length of the beam, from point C and point D. The moment at point D equals, --- MD=14,400 [lb*ft] 14,400 +400 [lb]*12 [ft] A
∫ Find: Mmax [lb*ft] in AB A MD=MC+ V*dx V [lb] 1,200 400 -400 -1,200 M [lb*ft] C D E B 19,200 pounds feet. [pause] The moments at point E and Joint B are calculated the same way and equal, --- 19,200 MD=14,400 [lb*ft] 14,400 +400 [lb]*12 [ft] A MD=19,200 [lb*ft]
∫ ∫ Find: Mmax [lb*ft] in AB A ME=MD+ V*dx MB=ME+ V*dx ME=19,200 [lb*ft] MB=14,400 [lb*ft] -400 [lb]*12 [ft] -1,200 [lb]*12 [ft] ME=14,400 [lb*ft] MB=0 [lb*ft] M [lb*ft] C D E B 14,400 pounds feet, and 0 pounds feet respectively. 19,200 14,400 A
∫ ∫ Find: Mmax [lb*ft] in AB A ME=MD+ V*dx MB=ME+ V*dx ME=19,200 [lb*ft] MB=14,400 [lb*ft] -400 [lb]*12 [ft] -1,200 [lb]*12 [ft] ME=14,400 [lb*ft] MB=0 [lb*ft] M [lb*ft] C D E B The questions asks to find the maximum moment in beam AB, --- 19,200 14,400 A
∫ ∫ Find: Mmax [lb*ft] in AB A ME=MD+ V*dx MB=ME+ V*dx ME=19,200 [lb*ft] MB=14,400 [lb*ft] -400 [lb]*12 [ft] -1,200 [lb]*12 [ft] ME=14,400 [lb*ft] MB=0 [lb*ft] M [lb*ft] C D E B this occurs at point D, and equals 19,200 pounds feet. 19,200 14,400 A
∫ ∫ Find: Mmax [lb*ft] in AB A ME=MD+ V*dx MB=ME+ V*dx 1,200 3,200 14,400 19,200 ME=19,200 [lb*ft] MB=14,400 [lb*ft] -400 [lb]*12 [ft] -1,200 [lb]*12 [ft] ME=14,400 [lb*ft] MB=0 [lb*ft] M [lb*ft] C D E B When reviewing the possible solutions, --- 19,200 14,400 A
∫ ∫ Find: Mmax [lb*ft] in AB AnswerD A ME=MD+ V*dx MB=ME+ V*dx 1,200 3,200 14,400 19,200 ME=19,200 [lb*ft] MB=14,400 [lb*ft] -400 [lb]*12 [ft] -1,200 [lb]*12 [ft] ME=14,400 [lb*ft] MB=0 [lb*ft] AnswerD M [lb*ft] C D the answer is D. 19,200 14,400 A
? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1 Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4