1. Find: M [k*ft] at L/2 A B w 5 w=2 [k/ft] 8 21 L=10 [ft] 33 L
Find the moment in kip feet, half way along the length of the beam AB. [pause] In this problem, ---
L=10 [ft]

2. Find: M [k*ft] at L/2 A B w 5 w=2 [k/ft] 8 21 L=10 [ft] 33 L
a cantilever beam is 10 feet long, and is fixed to a wall, at point B. The loading along the beam, ---
L=10 [ft]

3. Find: M [k*ft] at L/2 A B w 5 w=2 [k/ft] 8 21 L=10 [ft] 33 L
is a continuous downward force, which is 2 kips per foot at the end of the bream, and decreases linearly, to 0 kips per foot, at the wall.
L=10 [ft]

4. Find: M [k*ft] at L/2 A B w 5 w=2 [k/ft] 8 21 L=10 [ft] 33 L
To solve this problem, it would be helpful to assign a Cartesian coordinate system to the diagram, with the origin at point A, ---
L=10 [ft]

5. Find: M [k*ft] at L/2 B A x y w=2 [k/ft] 0 [k/ft] L=10 [ft]
the x-axis in the horizontal direction, and the y-axis in the vertical direction. That way, we can define the continuous loading, ----

6. Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x w=2 [k/ft] 0 [k/ft]
L=10 [ft] k k
w(x)= *x
w of x, as, a 2 kips per foot – 2 / 10 ths kips per foot squared, times x, where x ranges from 0 to L. ft
ft2
0 [ft] ≤ x ≤ 10 [ft]

7. ∫ Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x M(x)= V(x) * dx
w=2 [k/ft]
0 [k/ft] B A x
L=10 [ft] k k
w(x)= *x
The moment at any point in the beam AB, is the integral of the shear force, V of x, over the length of the beam, dx, and the shear force --- ft
ft2 ∫
M(x)= V(x) * dx

8. ∫ ∫ Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x M(x)= V(x) * dx
w=2 [k/ft]
0 [k/ft] B A x
L=10 [ft] k k
w(x)= *x
at any point in the beam AB, is the integral of the loading, W of x, over the length of the beam, dx. ft
ft2 ∫
M(x)= V(x) * dx ∫
V(x)= w(x) * dx

9. ∫ Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x V(x)= w(x) * dx k k ft
L=10 [ft] ∫
V(x)= w(x) * dx
Plugging in our equation for the loading, w of x, the shear force in beam AB equals, ---

10. ∫ ∫ Find: M [k*ft] at L/2 = B A x y w(x)=2 -0.2 *x V(x)= w(x) * dx
L=10 [ft]
V(x)= w(x) * dx ∫
2 kips per foot, times x, minus 1 / 10 th kips per foot squared, times x squared, plus a constant of integration, --- k ft k
ft2 ∫ = *x
* dx k k
ft2
V(x)=2 *x *x2+c ft

11. ∫ ∫ Find: M [k*ft] at L/2 = B A x y w(x)=2 -0.2 *x V(x)= w(x) * dx
L=10 [ft]
V(x)= w(x) * dx ∫
which equals zero, since there is no shear force at the end of the beam, where x equals 0. k ft k
ft2 ∫ = *x
* dx k k
ft2
V(x)=2 *x *x2+c ft

12. ∫ Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x V(x)=2 *x-0.1 *x2
L=10 [ft] k k
V(x)=2 *x *x2
Next, the moment along beam AB is calculated, as the integral of the shear force, V of x, along length the length of the beam, dx. ft
ft2 ∫
M(x)= V(x) * dx

13. ∫ Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x V(x)=2 *x-0.1 *x2
L=10 [ft] k k
V(x)=2 *x *x2
Plugging in the shear force, V of x, the moment in beam AB equals, --- ft
ft2 ∫
M(x)= V(x) * dx

14. ∫ Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x V(x)=2 *x-0.1 *x2
L=10 [ft] k
V(x)=2 *x *x2
1 kip per foot, times x squared, minus, 1 / 30 th, kip per foot squared, times x cubed, plus a constant of integration, ---
ft2 ∫
M(x)= V(x) * dx k k 1
M(x)=1 *x *x3+c ft
ft2 30

15. ∫ Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x V(x)=2 *x-0.1 *x2
L=10 [ft] k
V(x)=2 *x *x2
which equal 0, because there is no moment at the end of the beam, where x equals zero. [pause]
ft2 ∫
M(x)= V(x) * dx k k 1
M(x)=1 *x *x3+c ft
ft2 30

16. Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x M(x)=1 *x2- *x3 k k ft
L=10 [ft]
Since the problem asks to find the moment at a length of L / 2, and the beam is 10 feet long, --- k k 1
M(x)=1 *x *x3 ft
ft2 30

17. Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x M(x)=1 *x2- *x3 k k ft
L=10 [ft] L
x= =5 [ft] 2
then we’ll plug the value of 5 feet in for x, and the moment at halfway along the beam, equals, --- k k 1
M(x)=1 *x *x3 ft
ft2 30

18. Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x M(x)=1 *x2- *x3
L=10 [ft] L
x= =5 [ft] 2
20.83, kip feet. k k 1
M(x)=1 *x *x3 ft
ft2 30
M(x)=20.83 [k*ft]

19. Find: M [k*ft] at L/2 B A x y w(x)=2 -0.2 *x 5 8 21 33 M(x)=1 *x2- *x3
L=10 [ft] 5 8 21 33 L
x= =5 [ft] 2
When reviewing the possible solutions, --- k k 1
M(x)=1 *x *x3 ft
ft2 30
M(x)=20.83 [k*ft]

20. Find: M [k*ft] at L/2 AnswerC B A x y w(x)=2 -0.2 *x 5 8 21 33
L=10 [ft] 5 8 21 33 L
x= =5 [ft] 2
the answer is C. k k 1
M(x)=1 *x *x3 ft
ft2 30
M(x)=20.83 [k*ft]
AnswerC

21. ? 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