Solutions for Quiz 4

Name ≔ Acain, Karl Roy M.

1. For the beam loaded as shown, determine:

a. Shear and Moment throughout the beam as a function of x
b. Shear and Moment at the midspan

ΣMa ＝ 0 ΣFv ＝ 0
1 L 1
By ⋅ L ＝ ―⋅ w ⋅ L ⋅ ― Ay + By ＝ ―⋅ w ⋅ L
2 3 2
1 1
Ay ＝ ―⋅ w ⋅ L - ―⋅ w ⋅ L
w⋅L 2 6
By ＝ ――
6
w⋅L
Ay ＝ ――
a. Shear and Moment as a function of x: 3
0≤x≤L

y w
――＝ ―
L-x L

w
y ＝ ―⋅ ((L - x))
L

ΣFv ＝ 0
1 1 w⋅L
V + ―⋅ x ⋅ y + ―⋅ w ⋅ x ＝ ――
2 2 3

wL 1 1 ⎛w ⎞
V ＝ ―― - ―⋅ wx - ―⋅ x ⋅ ⎜―⋅ ((L - x))⎟
3 2 2 ⎝L ⎠

wL wx wx
V ＝ ―― - ――- ―― ((L - x))
3 2 2L
 ΣM ＝ 0
1 ⎛x⎞ 1 ⎛2 x⎞ w⋅L
M + ―⋅ x ⋅ y ⋅ ⎜―⎟ + ―⋅ w ⋅ x ⋅ ⎜―― ⎟ ＝ ――⋅ x
2 ⎝3⎠ 2 ⎝ 3 ⎠ 3

wLx x 2 y wx 2
M ＝ ―― - ――- ――
3 6 3

wLx x 2 ⎛ w ⎞ wx 2
M ＝ ―― - ― ⋅ ⎜―⋅ ((L - x))⎟ - ――
3 6 ⎝L ⎠ 3

wLx wx 2 wx 2
M ＝ ―― - ――((L - x)) - ――
3 6L 3

b. Shear and Moment at the midspan (x=0.5L)

wL wx wx wL w ((0.5 L)) w ((0.5 L))
V ＝ ―― - ――- ―― ((L - x)) ＝ ―― - ――― - ――― ((L - 0.5 L))
3 2 2L 3 2 2L

wL wL wL
V ＝ ―― - ――- ――
3 4 8

wL
V ＝ -――
24
2 2
wLx wx 2 wx 2 wL ((0.5 L)) w ((0.5 L)) w ((0.5 L))
M ＝ ―― - ――((L - x)) - ―― ＝ ―――― - ―――― ((L - 0.5 L)) - ――――
3 6L 3 3 6L 3

wL 2 wL 2 wL 2
M ＝ ―― - ―― - ――
6 48 12

wL 2
M ＝ ――
16
Alternative Solution:

y w
――＝ ―
L-x L

w
y ＝ ―⋅ ((L - x))
L

ΣFv ＝ 0
1 w⋅L
V + ―⋅ x ⋅ ((w - y)) + x ⋅ y ＝ ――
2 3

wL 1 ⎛ w ⎞ ⎛w ⎞
V ＝ ―― - ―⋅ x ⋅ ⎜w - ―⋅ ((L - x))⎟ - x ⋅ ⎜―⋅ ((L - x))⎟
3 2 ⎝ L ⎠ ⎝L ⎠

wL wx 2 wx
＝
V ―― - ―― + ―― ((L - x))
3 2L L

ΣM ＝ 0
⎛x⎞ 1 ⎛ 2 x ⎞ wL
M + x ⋅ y ⋅ ⎜―⎟ + ―⋅ x ⋅ ((w - y)) ⋅ ⎜―― ⎟ ＝ ――⋅x
⎝2⎠ 2 ⎝ 3 ⎠ 3

wL x2 y x2
M ＝ ―― ⋅ x - ――- ― ((w - y))
3 2 3
2 ⎛w ⎞
x ⎜―⋅ ((L - x))⎟ 2
wL ⎝L ⎠ x ⎛ w ⎞
M ＝ ―― ⋅ x - ――――― - ― ⎜w - ―⋅ ((L - x))⎟
3 2 3 ⎝ L ⎠

wLx wx 2 wx 3
M ＝ ―― - ――((L - x)) - ――
3 2L 3L

@x=0.5L
2
wL wx 2 wx wL w ((0.5 L)) w ((0.5 L))
V ＝ ―― - ―― + ―― ((L - x)) ＝ ―― - ―――― + ――― ((L - x))
3 2L L 3 2L L

wL wL wL wL
V ＝ ―― - ――- ―― V ＝ -――
3 8 4 24
 2 3
wLx wx 2 wx 3 wL ((0.5 L)) w ((0.5 L)) w ((0.5 L))
M ＝ ―― - ――((L - x)) - ―― ＝ ―――― - ―――― ((L - 0.5 L)) - ――――
3 2L 3L 3 2L 3L

wL 2 wL 2 wL 2
M ＝ ―― - ―― - ――
6 16 24

wL 2
M ＝ ――
16

2. The beam is loaded as shown. Determine:

a. Internal shear force and bending moment at C
b. Internal shear force and bending moment 1m to the right from the pin support
c. Maximum moment in the beam and it's location from the pin support

P ≔ Name = 5 kN L1 ≔ Name =3 m
0,0 3,0

kN
w1 ≔ Name = 3 ―― L2 ≔ Name =1 m
1,0 m 4,0

ΣMB ＝ 0

ΣFv ＝ 0
0 ≤ x ≤ L2
ΣFv ＝ 0 ΣMx ＝ 0

V1 ((x)) + P ＝ 0 M1 ((x)) + P ⋅ x ＝ 0

V1 ((x)) ≔ -P M1 ((x)) ≔ -P ⋅ x

L2 ≤ x ≤ L2 + L1

w1 y
― ＝ ――
L1 x - L2

w1
y ＝ ― ⋅ ⎛⎝x - L2⎞⎠
L1

ΣFv ＝ 0
1
Ay ＝ P + ―⋅ y ⋅ ⎛⎝x - L2⎞⎠ + V ((x))
2

1 ⎛ w1 ⎞
Ay ＝ P + ―⋅ ⎜― ⋅ ⎛⎝x - L2⎞⎠⎟ ⋅ ⎛⎝x - L2⎞⎠ + V ((x))
2 ⎝ L1 ⎠

w1 2
V2 ((x)) ≔ Ay - P - ――⋅ ⎛⎝x - L2⎞⎠
2 ⋅ L1
ΣM ＝ 0
1 ⎛⎝x - L2⎞⎠
M ((x)) + ―⋅ y ⋅ ⎛⎝x - L2⎞⎠ ⋅ ――― + P ⋅ x ＝ Ay ⋅ ⎛⎝x - L2⎞⎠
2 3

1 ⎛ w1 ⎞ ⎛⎝x - L2⎞⎠
M ((x)) + ―⋅ ⎜― ⋅ ⎛⎝x - L2⎞⎠⎟ ⋅ ⎛⎝x - L2⎞⎠ ⋅ ――― + P ⋅ x ＝ Ay ⋅ ⎛⎝x - L2⎞⎠
2 ⎝ L1 ⎠ 3

w1 3
M2 ((x)) ≔ Ay ⋅ ⎛⎝x - L2⎞⎠ - ―― ⋅ ⎛⎝x - L2⎞⎠ - P ⋅ x
6 L1
V ((x)) ≔ if 0 m M ((x)) ≔ if 0 m ≤ x ≤ L2
‖ 0 kN ‖ -P ⋅ x
‖ ‖
else if 0.0001 m ≤ x ≤ L2 else
‖ -P ‖ w1 3
‖ ‖ Ay ⋅ ⎛⎝x - L2⎞⎠ - ―― ⋅ ⎛⎝x - L2⎞⎠ - P ⋅ x
else if L2 ≤ x ≤ ⎛⎝L1 + L2⎞⎠ ‖ 6 L1
‖
‖ w1 2
‖ Ay - P - ―― ⋅ ⎛⎝x - L2⎞⎠
‖ 2 ⋅ L1
‖
else x ≔ 0 m , 0.01 m ‥ ⎛⎝L1 + L2 + 0.01 m⎞⎠
‖ 0 kN
‖

3.4

2.55

1.7

0.85

0
0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4 4.4
-0.85
V ((x)) ((kN))
-1.7

-2.55

-3.4

-4.25

-5.1

x ((m))

0.55

0
0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4 4.4
-0.55

-1.1

-1.65

-2.2

-2.75 M ((x)) ((kN ⋅ m))

-3.3

-3.85

-4.4

-4.95

-5.5

x ((m))
a. Internal shear force and bending moment at C

xc ≔ L2 + 0.5 ⋅ L1 = 2.5 m

Vc ≔ V ⎛⎝xc⎞⎠ = 2.042 kN

Mc ≔ M ⎛⎝xc⎞⎠ = -0.813 kN ⋅ m

b. Internal shear force and bending moment 1m to the right from the pin support

x2m ≔ L2 + 1 m = 2 m

V2m ≔ V ⎛⎝x2m⎞⎠ = 2.667 kN

M2m ≔ M ⎛⎝x2m⎞⎠ = -2 m ⋅ kN

c. Maximum moment in the beam and it's location from the pin support

Maximum Negative Moment

xneg ≔ L2 = 1 m

Mmax.neg ≔ M ⎛⎝xneg⎞⎠ = -5 kN ⋅ m

Maximum Positive Moment (Shear is 0)

w1 2
V2 ((x)) ＝ Ay - P - ――⋅ ⎛⎝x - L2⎞⎠
2 ⋅ L1

Mmax.pos ≔ M ⎛⎝xpos⎞⎠ = 0.313 kN ⋅ m

3. The beam is loaded as shown. Determine:
a. Internal shear force and bending moment at B
b. The shear and moment at the fixed end
c. Draw the shear and moment diagram

P ≔ Name = 5 kN L1 ≔ Name =3 m
0,0 3,0

kN
w1 ≔ Name = 3 ―― L2 ≔ Name =1 m
1,0 m 4,0

kN
w2 ≔ Name = 10 ――
2,0 m

ΣFv ＝ 0
Dy ≔ P + w1 ⋅ ⎛⎝2 ⋅ L2 + L1⎞⎠ + w2 ⋅ L1 = 50 kN

ΣMd ＝ 0
⎛ ⎛⎝2 ⋅ L2 + L1⎞⎠ ⎞ ⎛ L1 ⎞
Md + P ⋅ ⎛⎝2 ⋅ L2 + L1⎞⎠ + w1 ⋅ ⎛⎝2 ⋅ L2 + L1⎞⎠ ⋅ ⎜―――― ⎟ + w2 ⋅ L1 ⋅ ⎜―+ L2⎟ ＝ 0
⎝ 2 ⎠ ⎝ 2 ⎠

w1 2 ⎛ L1 ⎞
Md ≔ -⎛⎝P ⋅ ⎛⎝2 ⋅ L2 + L1⎞⎠⎞⎠ - ― ⋅ ⎛⎝2 ⋅ L2 + L1⎞⎠ - w2 ⋅ L1 ⋅ ⎜―+ L2⎟ = -137.5 kN ⋅ m
2 ⎝ 2 ⎠

0 m ≤ x ≤ L2 ΣFv ＝ 0
P + V1 ((x)) + w1 ⋅ x ＝ 0

V1 ((x)) ≔ -P - w1 ⋅ x

ΣMx ＝ 0
x
M1 ((x)) + w1 ⋅ x ⋅ ―+ P ⋅ x ＝ 0
2

w1
M1 ((x)) ≔ -P ⋅ x - ― ⋅ x 2
2
L2 ≤ x ≤ ⎛⎝L2 + L1⎞⎠

ΣFv ＝ 0
V2 ((x)) + P + w1 ⋅ x + w2 ⋅ ⎛⎝x - L2⎞⎠ ＝ 0

V2 ((x)) ≔ -P - w1 ⋅ x - w2 ⋅ ⎛⎝x - L2⎞⎠

ΣMx ＝ 0 ⎛⎝x - L2⎞⎠

x
M2 ((x)) + P ⋅ x + w1 ⋅ x ⋅ ―+ w2 ⋅ ⎛⎝x - L2⎞⎠ ⋅ ――― ＝0
2 2

w1 w2 2
M2 ((x)) ≔ -((P ⋅ x)) - ― ⋅ x 2 - ― ⋅ ⎛⎝x - L2⎞⎠
2 2

⎛⎝L2 + L1⎞⎠ ≤ x ≤ ⎛⎝2 ⋅ L2 + L1⎞⎠

ΣFv ＝ 0
V3 ((x)) + P + w1 ⋅ x + w2 ⋅ L1 ＝ 0

V3 ((x)) ≔ -P - w1 ⋅ x - w2 ⋅ L1

ΣMx ＝ 0
x
M3 ((x)) + P ⋅ x + w1 ⋅ x ⋅ ―+ w2 ⋅ L1 ⋅ ⎛⎝x - L2 - 0.5 L1⎞⎠ ＝ 0
2
x2
M3 ((x)) ≔ -P ⋅ x - w1 ⋅ ― - w2 ⋅ L1 ⋅ ⎛⎝x - L2 - 0.5 L1⎞⎠
2
V ((x)) ≔ if 0 m ≤ x ≤ L2 M ((x)) ≔ if 0 m ≤ x ≤ L2
‖ -P - w ⋅ x ‖ w1 2
‖ 1 ‖ -P ⋅ x - ― ⋅x
else if L2 ≤ x ≤ ⎛⎝L2 + L1⎞⎠ ‖
‖ 2
‖ -P - w ⋅ x - w ⋅ ⎛x - L ⎞ else if L2 ≤ x ≤ ⎛⎝L2 + L1⎞⎠
‖ 1 2 ⎝ 2⎠
‖
else w1 2 w2
‖ -((P ⋅ x)) - ―
2

‖ -P - w ⋅ x - w ⋅ L ⋅ x - ― ⋅ ⎛⎝x - L2⎞⎠
‖
‖ 2 2
‖ 1 2 1
else
‖
‖ -P ⋅ x - w1 ⋅ ― x2
x ≔ 0 m , 0.001 m ‥ ⎛⎝2 ⋅ L2 + L1⎞⎠ - w2 ⋅ L1 ⋅ ⎛⎝x - L2 - 0.5 L1⎞⎠
‖
‖ 2

a. Internal shear force and bending moment at B

xB ≔ L2 = 1 m

VB ≔ V ⎛⎝xB⎞⎠ = -8 kN

MB ≔ M ⎛⎝xB⎞⎠ = -6.5 kN ⋅ m

b. The shear and moment at the fixed end

xfxd ≔ 2 ⋅ L2 + L1 = 5 m

Vfxd ≔ V ⎛⎝xfxd⎞⎠ = -50 kN

Mfxd ≔ M ⎛⎝xfxd⎞⎠ = -137.5 kN ⋅ m

c. Draw the shear and moment diagram

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-5

-9.5

-14

-18.5

-23

-27.5
V ((x)) ((kN))
-32

-36.5

-41

-45.5

-50

x ((m))

15

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
-15

-30

-45

-60

-75 M ((x)) ((kN ⋅ m))

-90

-105

-120

-135

-150

x ((m))
SUMMARY OF ANSWERS:

Item 1.0
wL wx wx wL wx 2 wx
V ＝ ―― - ――- ―― ((L - x)) OR V ＝ ―― - ―― + ―― ((L - x)) (5 Points)
3 2 2L 3 2L L

wLx wx 2 wx 2 wLx wx 2 wx 3
M ＝ ―― - ――((L - x)) - ―― OR M ＝ ―― - ――((L - x)) - ―― (5 Points)
3 6L 3 3 2L 3L

wL
Vmid ＝ -―― (5 Points)
24

wL 2
Mmid ＝ ―― (5 Points)
16

Item 2.0

Vc = 2.042 kN (5 Points) Mmax.neg = -5 kN ⋅ m (2.5 Points)

Mc = -0.813 kN ⋅ m (5 Points) xneg = 1 m (2.5 Points)

V2m = 2.667 kN (5 Points) Mmax.pos = 0.313 kN ⋅ m (2.5 Points)

M2m = -2 kN ⋅ m (5 Points) xpos = 3.517 m (2.5 Points)

Item 3.0

VB = -8 kN (5 Points) Vfxd = -50 kN (5 Points)

MB = -6.5 kN ⋅ m (5 Points) Mfxd = -137.5 kN ⋅ m (10 Points)

(Significant points of the shear and moment diagram)

V ((0 m)) = -5 kN M ((0 m)) = 0 kN ⋅ m

V ⎛⎝L2⎞⎠ = -8 kN M ⎛⎝L2⎞⎠ = -6.5 kN ⋅ m

V ⎛⎝L2 + L1⎞⎠ = -47 kN M ⎛⎝L2 + L1⎞⎠ = -89 kN ⋅ m

V ⎛⎝L2 + L1 + L2⎞⎠ = -50 kN M ⎛⎝L2 + L1 + L2⎞⎠ = -137.5 kN ⋅ m