1. Find: Mmax [lb*ft] in ABP 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 ---

2. Find: Mmax [lb*ft] in ABP 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.

3. Find: Mmax [lb*ft] in ABP 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]

4. Find: Mmax [lb*ft] in ABP 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 ---

5. Find: Mmax [lb*ft] in ABP 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]

6. Find: Mmax [lb*ft] in ABP 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]

7. Find: Mmax [lb*ft] in ABP 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]

8. Find: Mmax [lb*ft] in ABP 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]

9. Find: Mmax [lb*ft] in ABP 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]

10. Find: Mmax [lb*ft] in ABP 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]

11. Find: Mmax [lb*ft] in ABP 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]

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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.

22. 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

23. ∫ 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

24. 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

25. ∫ 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

26. ∫ 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

27. ∫ 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

28. ∫ 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]

29. ∫ 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

30. ∫ 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

31. ∫ 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]

32. ∫ ∫ 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

33. ∫ ∫ 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

34. ∫ ∫ 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

35. ∫ ∫ 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

36. ∫ ∫ Find: Mmax [lb*ft] in AB AnswerD 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]
AnswerD
M [lb*ft] C D
the answer is D.
19,200
14,400 A

37. ? 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