CEE/CNE 210 Statics
Arizona State University
Subject: Recitation 03e
Name:
Date:
Distributed Loads
Problem 1
The rigid bent plate ABC forms a gate in water channel (shown in
cross-section), resisting the force of the water on the left. The
gate is hinged at A, has a roller support at C and is w = 4 ft wide
perpendicular to the cross-section (into the page).
B
C
L2
The uniform gate has a self-weight of m(s)g along it’s entire
cross-sectional length.
Determine the support reactions at A and C.
Answers: 𝑁𝐶 = 2858.8 lb; 𝐑𝐴 = − 5993 lb 𝐞1 + 2551 lb 𝐞2
q
L1
A
L1 = 8 ft; L2 = 5 ft; q = 60; m(s)g = 150 lb/ft
w = 62.43 lb/ft3
A. Modeling & Techniques
This problem has no forces in the e3 direction and no moments in the e2, or e1 directions this problem
can therefore model as 2-D. There is a hinge at point A that acts as a pin. The hinge at point A has an
unknown magnitude and an unknown direction. The roller support has an unknown magnitude, but
the direction of the roller acts perpendicular to the gate. There are two particles needed to solve this
situation. One particle will be drawn between points A to B. In order to model the change of
geometry one particle will need to be drawn between B and C. The self weight of m(s)g must be
integrated along with the hydrostatic force in order to find the net forces and moments.
B. Solution Strategy
Only one free body is required to model this problem. There are a total of three unknowns. There are
two unknowns at hinge A, the magnitude and direction. There is an unknown magnitude at roller C.
We can sum the moments of point A, to find the normal force at C. Then we can use force
equilibrium to solve for the force support reaction at point A.
CEE/CNE 210 Statics
Arizona State University
Subject:
C. Setup and Geometry
.
Name:
Date:
M1(g)=m(s1)gds1
L2
M2(g)=m(s2)gds2
S2
P(z)=γz
Z=L1sin θ-s1sin θ
P(z)=(L1-s1)sin θ γ
P1s1=γw(L1-s1)sin θ M × wds1
X2
X1
M
θ
S1
L1
N
M= cosθe1+sinθe2
X1=S1M
X2=L1M+S2e1
N= -sinθe1+cosθe2
D. FBD
.
P1(s1)M
-m(s2)ge2ds2
-m(s1)ge2ds1
Ra
Rbe2
CEE/CNE 210 Statics
Arizona State University
Subject: Recitation 03d
Distributed Loads
Name:
Date:
Problem 1
The cross-section of a flood gate is shown. The quarter cylinder
gate AB has a horizontal width of 6 ft (normal to the paper) and
controls the flow of fresh water over the ledge at B. The gate is
hinged about its upper edge A and rests on the frictionless ledge at
B. Accounting for the 6 ft cross-sectional width, the gate has a
weight of 481 lb/ft along its curved length.
Determine the minimum force P required to keep the gate closed.
water = 62.43 lb/ft3
Answer:
P = 10.8 kip
A. Modeling & Techniques
This situation has no force in the e3 direction and no moments in the e2, or e1 directions and can therefore
be analyzed as a 2-d problem. There is a pin at point A with two unknowns. Point A has an unknown
magnitude of force along with an unknown direction. The gate rests on the ledge at point B and has a
known magnitude a direction. We know the magnitude is perpendicular to the ledge and the force is zero
since we are solving for the minimum. The third unknown is the force at P. The direction of B is going to
be perpendicular to the wall it rests on. One particle will need to be solved for. The particle is in between
point A and point B towards the center of the circular arc. The gate has 3 unknown support components
and 3 equation components of equilibrium and is therefore stable and statically determinate.
B. Solution Strategy
Only need one free body diagram to represent this situation. The free body diagram will be drawn of the
gate with one particle. There are three scalar equations with three independent unknowns. The first step
would be summing the forces about point A. This eliminates the force at point A from the moment
equations. This is because the moment arm would become zero. This results in only one unknown at point
A. The moment at P can now be directly solved for.
CEE/CNE 210 Statics
Arizona State University
Subject:
Name:
Date:
E. Force equilibrium
F = 0=Ra+
𝑠
P1(s1)M −
𝑠
−m(s1)ge2ds1 −
𝑠
m(s2)ge2ds2+Rbe2
1
=Ra+(2 γwL12wsin θ )M-W1e2-W2e2Rbe2
F. Moment equilibrium
Ma = 0=(r x F)
M = 0=(0 × Ra)+
𝑎
(𝑠1M ×u1( s1)N)+
𝑠1
(𝑠1M× − m(s1)ge2ds1) +
L1M+s2e1 × -n(s2)ge2ds2+ (L1M+L2e1 × Rbe2)
𝑠2
1
1
1
= - 6 γWL13sin θe3 − 2 W1L1cos θe3 -W2(2 L2+L1cos θ) e3+Rbe2(L1cos+L2) e3
G. Distributed effects
Net force of incline AB hydrostatic load
P1(s1)M =
𝑠
1
𝑠1 2
γw(L1−s1) sinθnwds1= γWsinθ[L1s1-
𝑠1
2
1
]= (2 γwL12wsin θ )M
Net force of AB self weight
𝐿2
𝐿2
−𝑚𝑔𝐞2𝑑𝑠1=- 0 −[𝑚𝑔𝐞2𝑑𝑠1]=-mgL1e2=-W1e2
0
Net force of horizontal segment BC caused by self-weight
𝐿2
-
𝐿2
m(s2)ge2ds2 = -
−mge2 ds2 =-mgL2e2=-W2e2
Net moment of point A, caused by the self-weight of segment AB is
L M+s2e1 × -n(s2)ge2ds2=
𝑠2 1
1
1
L
)=W
(
2
2
6
6
𝐿2
(−mg L1(M ×e2)ds2)-mgL1L2cos θ e3=-mgL2(L1cos θ +
L2-L1cos θ)
Net moment of point A, caused by the hydrostatic load on segment Ab
0
(𝑠1M ×u1( s1)N)=γWsin θ
𝐿1
(
0
1
L1 −𝑠12)(−𝑒3)=- 2 γWL12sin θe3
H. Solving Equilibrium
1
1
1
(- 6 γWL13sin θe3 − 2 W1L1cos θe3 -W2(2 L2+L1cos θ) e3+(L1cos+L2) e3) · e3
Rbe2(9ft)=18454.5lb·ft+2400lb·ft+4875lb·ft
Rbe2=2858.8lb
1
Ra+(2 γwL12wsin θ )M-W1e2-W2e2Rbe2=0
1
Ra=-(2 γwL12wsin θ )M+W1e2+W2e2Rbe2
1
Ra=[2 *62.43 lb/ft3 )(8ft)2(4ft)sin60] (sin60e1-cos60e2)+1200e2+750e2-(2858.8lbe2)-(6920.4lb)
(sin60e1-cos60e2)+1200e2+750e2- 2858.8lbe2 𝐑𝐴 = − 5993 lb 𝐞1 + 2551 lb 𝐞2
CEE/CNE 210 Statics
Arizona State University
Subject:
Name:
Date:
C. Setup and Geometry
.
S
M
P(z)=γz
N
P(θ)=(H-Hsin θ) γ
Z
L1
P(θ)dA=γ(H-Hsin θ)wnds
X
Y
θ
L2
M= -sinθe1+cosθe2
Z=R-Rsinθ R=Z+RSinθ
N= cosθe1+sinθe2
D. FBD
.
Ra
P(S)nds
-mge2ds
-Pe1
CEE/CNE 210 Statics
Arizona State University
Subject:
E. Force equilibrium
𝐹 = 0 = Ra +
𝑠
P(S)nds −
𝑠
mge2ds−Pe1
F. Moment equilibrium
M = 0 = r×F = (0×Ra)+ −𝑠𝑚× P(S)nds - (𝑠𝑚×mge2ds
Name:
Date:
Subject: Recitation 03d
CEE/CNE 210 Statics
Arizona State University
Distributed Loads
Name:
Date:
Problem 2
The cross-section of a rigid flood gate is shown. The gate resists the
hydrostatic force of the water, which has a unit weight, w = 62.43 lb/ft3. The
gate ACB resists the force of the water on the left. The gate is 6 ft wide (into
the page) and is supported by a hinge along the top edge A (modeled as a pin
support). Friction can be neglected where the gate rests on the floor at B.
(a) If the self-weight of the gate is neglected, determine the support reactions
at A and B.
C
(b) If the gate weighs 50 lb/ft2 of face area (already accounting for the
thickness of the gate), how do the reactions change?
Answers: (a) NB = 43846 lb;
RA = (-18720e1 - 40476e2) lb
(b) NB = 46052 lb;
RA = (-18720e1 - 39470e2) lb
A. Modeling & Techniques
This problem has no forces in the e3 direction and no moments in the e2, or e1 directions this
problem can therefore model as 2-D. There is a pin at the top of edge A. The pin has an
unknown magnitude and an unknown direction. The gates rests on the floor at point B. B has a
known direction that is perpendicular to the floor and an unknown magnitude. This problem
will also require two particles. One particle will be needed for each segment of the gate.
Friction will be neglected at the floor at point b. The gate is stable and determinate with 3
unknown support components along with 3 equation components of equilibrium.
B. Solution strategy
Only one free body diagram will be required for this situation. The free body diagram will be
drawn of the gate with the two particles, one for each segment of the gate. The first step to
solving this situation is summing the moments at A. This eliminates the effect of force at that
point from the moment equation because the moment arm r(M )=( r×F) is zero. There is
therefore only one unknown in the moment equation. It is then possible to use the force
equilibrium equation force equilibrium equation to solve for the force reaction at A.
CEE/CNE 210 Statics
Arizona State University
Subject:
C. Setup and Geometry
.
N
M1(g)=-m(s1)gds1
M
M2(g)=-m(s2)gds2
S2
P(z)=γz
Z=L3sin θ-s3sin θ
L1
P(z)=(L3-s3)sin θ γ
P1s1=γw(L3-s3)sin θ M × wds1
L4
X2
L3
X1
S1
L2
M= cosθe1+sinθe2
X1=-S1M
N= sinθe1-cosθe2
D. FBD
.
Ra
P2(s2)M
-m(s2)ge2ds2
P1(s1)e1
-m(s1)ge2ds1
Rbe2
X2=L1M+S2e1
Name:
Date:
CEE/CNE 210 Statics
Arizona State University
Subject:
Name:
Date:
E. Force equilibrium
F = 0=Ra+
𝑠
P1(s1)M −
𝑠
−m(s1)ge2ds1 −
𝑠
m(s2)ge2ds2+Rbe2
1
=Ra+(2 γwL12wsin θ )M-W1e2-W2e2Rbe2
F. Moment equilibrium
Ma = 0=(r × F)
M = 0=(0 x Ra)+(x2 ×Rbe2)+
4𝑓𝑡
𝑠1
0
×P1(s1)e1+