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]

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]

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]

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]

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, ----

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)=2 -0.2 *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]

∫ 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)=2 -0.2 *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

∫ ∫ 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)=2 -0.2 *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

∫ 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, ---

∫ ∫ 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 ∫ = 2 -0.2 *x * dx k k ft2 V(x)=2 *x-0.1 *x2+c ft

∫ ∫ 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 ∫ = 2 -0.2 *x * dx k k ft2 V(x)=2 *x-0.1 *x2+c ft

∫ 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-0.1 *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

∫ 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-0.1 *x2 Plugging in the shear force, V of x, the moment in beam AB equals, --- ft ft2 ∫ M(x)= V(x) * dx

∫ 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-0.1 *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 *x2- *x3+c ft ft2 30

∫ 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-0.1 *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 *x2- *x3+c ft ft2 30

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 *x2- *x3 ft ft2 30

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 *x2- *x3 ft ft2 30

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 *x2- *x3 ft ft2 30 M(x)=20.83 [k*ft]

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 *x2- *x3 ft ft2 30 M(x)=20.83 [k*ft]

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 *x2- *x3 ft ft2 30 M(x)=20.83 [k*ft] AnswerC

? 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