relationships between shear force, bending moment, and ... diff relationship… · relationships...
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Relationships between Shear Force, Bending Moment, and Distributed Loads
StevenVukazichSanJoseStateUniversity
Consider a beam with general supports, and general loading
P
Mw
x
y
∆𝑥
Look at equilibrium of a small slice of the beam
FBD of slice of beam
𝑤∆𝑥
x
∆𝑥+
+
+
Sign convention
𝑀 + ∆𝑀
𝑉 + ∆𝑉𝑉
𝑀
a
Force Equilibrium
𝑤∆𝑥
x
∆𝑥
𝑀 + ∆𝑀
𝑉 + ∆𝑉𝑉
𝑀
a
'𝐹)
�
�
= 0+
𝑑𝑉𝑑𝑥
= 𝑤
Differential Relationship between V and w
Slopeoftangentlinetothesheardiagram at a point
𝑑𝑉𝑑𝑥
= 𝑤
Value of the magnitude of the distributed load intensity at that point
=
Moment Equilibrium
𝑤∆𝑥
x
∆𝑥
𝑀 + ∆𝑀
𝑉 + ∆𝑉𝑉
𝑀
a
'𝑀=
�
�
= 0+
𝑑𝑀𝑑𝑥
= 𝑉
Differential Relationship between M and V
Slopeoftangentlinetothemomentdiagram at a point
𝑑𝑀𝑑𝑥
= 𝑉
Value of the magnitude of the ordinate of the shear diagram at that point
=
𝑑𝑉𝑑𝑥
= 𝑤
𝑑𝑉 = 𝑤𝑑𝑥
? 𝑑𝑉@
A= ? 𝑤𝑑𝑥
@
A
𝑉@ − 𝑉A = ? 𝑤𝑑𝑥 =@
A
areaunderthedistributedloadbetweenpointsAandB
Can integrate the differential relationship between w and V between two points on the beam
Can integrate the differential relationship between V and M between two points on the beam
𝑑𝑀 = 𝑉𝑑𝑥
? 𝑑𝑀@
A= ? 𝑉𝑑𝑥
@
A
𝑀@ −𝑀A = ? 𝑉𝑑𝑥 =@
A
areaunderthesheardiagrambetweenpointsAandB
𝑑𝑀𝑑𝑥
= 𝑉
Consider the V and M diagrams for the beam below
+
+
9 ft 2 ft 3 ft 4 ft
3 k
E
32 k-ft
2 k/ft
DCBA
2 k 10 k
V
M
2 k
-7 k3 2� = 4.24ft
9 k-ft
𝑀M=N = 4 2� = 5.66k−ft0
-23 k-ft
9 k-ft
-12 k-ft
3 k
0
3 k
-7 k
9 ft 2 ft 3 ft 4 ft
3 k
E
32 k-ft
2 k/ft
DCBA
2 k 10 k
V
2 k
-7 k
3 k 3 k
-7 k
Slopeoftangentlinetothesheardiagram at a point
𝑑𝑉𝑑𝑥
= 𝑤
Value of the magnitude of the distributed load intensity at that point =
slope = 0slope = -2 k/ftslope = 0
9 ft 2 ft 3 ft 4 ft
3 k
E
32 k-ft
2 k/ft
DCBA
2 k 10 k
V
2 k
-7 k
3 k 3 k
-7 k
𝑉@ − 𝑉A = ? 𝑤𝑑𝑥 =@
A
areaunderthedistributedloadbetweenpointsAandB
12
−2k/ft 9ft = −9k
VB – VA = – 7 – 2 = – 9 k OK
VE – VC = 3 – (– 7) = 10 knot OK why?
VD- – VB = – 7 – 7 = 0 OK
9 ft 2 ft 3 ft 4 ft
V
M
2 k
-7 k3 2� = 4.24ft
-9 k-ft
𝑀M=N = 4 2� = 5.66k−ft0
-23 k-ft
9 k-ft
-12 k-ft
3 k
0
3 k
-7 k
Slopeoftangentlinetothemomentdiagram at a point
𝑑𝑀𝑑𝑥
= 𝑉
Value of the magnitude of the ordinate of the shear diagram at that point =
EDCBA
slope = 2 k-ft/ftslope = 0slope = -7 k-ft/ftslope = 3 k-ft/ft
9 ft 2 ft 3 ft 4 ft
V
M
2 k
-7 k3 2� = 4.24ft
-9 k-ft
𝑀M=N = 4 2� = 5.66k−ft0
-23 k-ft
9 k-ft
-12 k-ft
3 k
0
3 k
-7 k
EDCBA
𝑀@ −𝑀A = ? 𝑉𝑑𝑥 =@
A
areaunderthesheardiagrambetweenpointsAandB
−7k 2ft = −14k−ft−7k 3ft = −21k−ft
3k 4ft = 12k−ft
MC- - MB = – 23 – (– 9) = – 14 k-ft OK
MD – MC+ = – 12 – 9 = – 21 k-ft OKME - MD = 0 – (– 12) = 12 k-ft OK
ME – MC+ = 0 – 9 = – 9 k-ft OK
ME – MB = 0 – (– 9) = 9 k-ftnot OK why?