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Problem 10.4-2 The propped cantilever beam shown in the figure supports a uniform load of intensity q on the left-hand half of the beam. Find the reactions RA , RB , and MA , and then draw the shear-force and bending-moment diagrams, labeling all critical ordinates.
Solution 10.4-2 Propped cantilever beam
SECTION 10.4 Method of Superposition 643
Select RB as redundant.
EQUILIBRIUM
RELEASED STRUCTURE AND FORCE-DISPLACEMENT
RELATIONS
COMPATIBILITY B ( B )
1
( B )
2
Substitute for ( B ) 1 and ( B ) 2 and solve for RB :
OTHER REACTIONS (FROM EQUILIBRIUM)
MA
9 qL^2 128
RA
57 qL 128
RB
7 qL 128
MA
qL^2 8
RA RBL
qL 2
RB
SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS
M max
945 qL^2 32,
x 1
57 L
A (^) B MA RA RB
q
—L 2 —^ L 2
A
B ( B ) 1 7 qL^4 384 EI
( B ) 2 RB L^3 3 EI A L B R (^) B
q
L 2
L 2
x 1
RA
RB
V
O
M O
M max
MA
Problem 10.4-3 The figure shows a propped cantilever beam ABC having span length L and an overhang of length a. A concentrated load P acts at the end of the overhang. Determine the reactions RA , RB , and MA for this beam. Also, draw the shear-force and bending-moment diagrams, labeling all critical ordinates.
A
L a
B C
MA
RA RB
P
Solution 10.4-3 Beam with an overhang
644 CHAPTER 10 Statically Indeterminate Beams
Select MA as redundant.
EQUILIBRIUM
RELEASED STRUCTURE AND FORCE-DISPL. EQS.
C OMPATIBILITY A ( A )
1
( A )
2
Substitute for ( A ) 1 and ( A ) 2 and solve for MA :
MA
Pa 2
RB
L
RA ( MA PL Pa )
L
( MA Pa )
OTHER REACTIONS (FROM EQUILIBRIUM)
SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS
RB
P
2 L
RA (2 L 3 a )
3 Pa 2 L
A B
( A )
MA ( A )
L
( A ) 2 MAL 3 EI
( A ) 1 PaL 6 EI
C
P
a
P
3 Pa 2 L
V O
Pa 2
Pa
M
O
L 3
Problem 10.4-4 Two flat beams AB and CD , lying in horizontal planes, cross at right angles and jointly support a vertical load P at their midpoints (see figure). Before the load P is applied, the beams just touch each other. Both beams are made of the same material and have the same widths. Also, the ends of both beams are simply supported. The lengths of beams AB and CD are LAB and LCD , respectively. What should be the ratio tAB / tCD of the thicknesses of the beams if all four reactions are to be the same?
Solution 10.4-4 Two beams supporting a load P
P
B
D
C
A
tAB tCD
For all four reactions to be the same, each beam must support one-half of the load P.
DEFLECTIONS
CD
( P 2) L^3 CD
48 EICD
AB
( P 2) L^3 AB
48 EIAB
COMPATIBILITY
AB CD or
MOMENT OF INERTIA
tAB tCD
LAB
LCD
L^3 AB
t AB^3
L^3 CD
t CD^3
ICD
IAB bt CD^3
bt^3 AB
L^3 AB
IAB
L^3 CD
ICD
Problem 10.4-6 A continuous beam ABC with two unequal spans, one of length L and one of length 2 L , supports a uniform load of intensity q (see figure). Determine the reactions RA , RB , and RC for this beam. Also, draw the shear-force and bending-moment diagrams, labeling all critical ordinates.
Solution 10.4-6 Continuous beam with two spans
646 CHAPTER 10 Statically Indeterminate Beams
A
RA RB RC
L
B C
2 L
q
Select RB as redundant.
EQUILIBRIUM
RELEASED STRUCTURE AND FORCE-DISPLACEMENT
RELATIONS
( B ) 2
4 RBL^3
9 EI
( B ) 1
11 qL^4 12 EI
RC
3 qL 2
RA RB
3 qL 2
RB
COMPATIBILITY
B ( B ) 1 ( B ) 2 0
OTHER REACTIONS (FROM EQUILIBRIUM)
SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS
RC
13 qL 16
RA
qL 8
RB
33 qL 16
11 qL^4 12 EI
4 RBL^3
9 EI
A
L
B
C
2 L
q
( B )
13 L 16
13 16 169 512
19 16
7 8
1 8
1 128
L 8
L 4
V qL
M qL^2
3 8
O
O 13 L 8
A
R (^) B
B
C ( B ) 2
SECTION 10.4 Method of Superposition 647
Problem 10.4-7 Beam ABC is fixed at support A and rests (at point B ) upon the midpoint of beam DE (see the first part of the figure). Thus, beam ABC may be represented as a propped cantilever beam with an overhang BC and a linearly elastic support of stiffness k at point B (see the second part of the figure). The distance from A to B is L 10 ft, the distance from B to C is L /2 5 ft, and the length of beam DE is L 10 ft. Both beams have the same flexural rigidity EI. A concentrated load P 1700 lb acts at the free end of beam ABC. Determine the reactions RA , RB , and MA for beam ABC. Also, draw the shear-force and bending-moment diagrams for beam ABC , labeling all critical ordinates.
Solution 10.4-7 Beam with spring support
D
E
A B^ C
P = 1700 lb
L = 10 ft = 5 ft
k
MA
RA RB
B (^) C
P
A
L 2 —
Select RB as redundant.
EQUILIBRIUM RA RB P MA RB L 3 PL 2
R ELEASED STRUCTURE AND FORCE-DISPL. EQS.
C OMPATIBILITY
Beam DE :
RB
28 P
7 PL^3
12 EI
RBL^3
3 EI
RB L^3
48 EI
k
48 EI
L^3
B ( B ) 1 ( B ) 2
RB
k
OTHER REACTIONS (FROM EQUILIBRIUM)
NUMERICAL VALUES
P 1700 lb L 10 ft 120 in. RA 1100 lb RB 2800 lb MA 30,000 lb-in.
SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS
27.27 in.
x 1
in.
MA
5 PL
RA
11 P
A
B ( B ) 1 7 PL^3 12 EI
( B ) 2 RB L^3 3 EI A
L L 2
R (^) B
C
P
1100
1700
V (lb) (^) O
x 1
M (lb - in.) O
102,
30,
Solution 10.4-9 Beam supported by a tie rod
SECTION 10.4 Method of Superposition 649
Select the force T in the tie rod as redundant.
RELEASED STRUCTURE AND FORCE-DISPLACEMENT
RELATIONS
COMPATIBILITY ( B ) 1 ( B ) 2 ( B ) 3
or
T
3 qAL^4 8 AL^3 24 IH
qL^4 8 EI
TL^3
3 EI
TH
EA
NUMERICAL VALUES
q 200 lb/ft L 6 ft H 3 ft E 30 106 psi Beam: S 6 12.5 I 22.1 in.^4 Tie Rod: d 0.25 in. A 0.04909 in.^2 Substitute: T 398 lb
SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS
RA qL T 802 lb
14,530 lb-in.
x 1 23.9 in. x 2 24.2 in.
MA
qL^2 2
TL
A
A
B
B
B
L
T
T
q
( B ) 1 qL^4 8 EI
( B ) 2 TL^3 3 EI
( B ) 3 TH H EA
C
A
B L
q MA
RA T
x 1
V (lb)
M (lb-in.)
O
O
x 2
398
802
4760
14,
Problem 10.4-10 The figure shows a nonprismatic, propped cantilever beam AB with flexural rigidity 2 EI from A to C and EI from C to B. Determine all reactions of the beam due to the uniform load of intensity q. ( Hint: Use the results of Problems 9.7-1 and 9.7-2.) A
B C MA RA RB
q
2 EI EI
L 2
L — 2 —
Solution 10.4-10 Nonprismatic beam
650 CHAPTER 10 Statically Indeterminate Beams
Select RB as redundant.
RELEASED STRUCTURE
( B )
1 downward deflection of end B due to load q
( B )
2 upward deflection due to reaction RB
FORCE - DISPLACEMENT RELATIONS
From Prob. 9.7-2:
From Prob. 9.7-1:
COMPATIBILITY
B ( B ) 1 ( B ) 2 0
or
EQUILIBRIUM
MA
qL^2 2
RBL
7 qL^2 48
RA qL RB
31 qL 48
RB
17 qL 48
17 qL^4 256 EI
3 RBL^3
16 EI
( B ) 2
3 RBL^3
16 EI
B
PL^3
24 EI 1
I 1
I 2
≤
( B ) 1
17 qL^4 256 EI
I 1 S I I 2 S 2 I
B
qL^4 128 EI 1
I 1
I 2
≤
A B
q
2 EI EI L 2
L — 2 —
Problem 10.4-11 A beam ABC is fixed at end A and supported by beam DE at point B (see figure). Both beams have the same cross section and are made of the same material. (a) Determine all reactions due to the load P. (b) What is the numerically largest bending moment in either beam?
A
D
B C
E
P
MA RA RD (^) RE L — 4
—^ L 4 —^ L 4 —^ L 4
A B
RB
2 EI EI
Problem 10.4-13 A beam AC rests on simple supports at points A and C (see figure). A small gap 0.4 in. exists between the unloaded beam and a support at point B , which is midway between the ends of the beam. The beam has total length 2 L 80 in. and flexural rigidity EI 0.4 109 lb-in.^2 Plot a graph of the bending moment MB at the midpoint of the beam as a function of the intensity q of the uniform load. Hints: Begin by determining the intensity q 0 of the load that will just close the gap. Then determine the corresponding bending moment ( MB ) 0. Next, determine the bending moment MB (in terms of q ) for the case where q q 0. Finally, make a statically indeterminate analysis and determine the moment MB (in terms of q ) for the case where q q 0. Plot MB (units of lb-in.) versus q (units of lb/in.) with q varying from 0 to 2500 lb/in.
652 CHAPTER 10 Statically Indeterminate Beams
A
L
D
L
q
L
RA RB (^) R C RD
B C
2 L 5
3 5
3 5
1 2
1 2
2 5
2 25
2 25 1 40
2 5
V qL
M qL^2
2 L 5
1 10 1 10
O
O
A (^) C B
RA RC
L = 40 in. L = 40 in.
q
RB
= 0.4 in.
LOADING, SHEAR-FORCE , AND BENDING-MOMENT DIAGRAMS
M max
2 qL^2 25
MB MC
qL^2 10
SECTION 10.4 Method of Superposition 653
q 0 load required to close the gap
magnitude of gap
( MB ) 0 bending moment when q q 0
CASE 1 q q 0
RA RC qL
CASE 2 q q 0
CASE 3 q q 0 (statically indeterminate)
Select RB as redundant.
RELEASED STRUCTURE
( B ) 2
RBL^3
6 EI
( B ) 1
5 qL^4 24 EI
( MB ) 0
q 0 L^2 2
12 EI ¢
5 L^2
q 0
24 EI ¢
5 L^4
B ¢
5 q 0 L^4 24 EI
MB
qL^2 2
B
5 qL^4 24 EI
COMPATIBILITY B ( B ) 1 ( B ) 2
or
EQUILIBRIUM
RA RC 2 RA 2 qL RB 0
NUMERICAL VALUES
0.4 in. L 40 in. EI 0.4 109 lb-in.^2 Units: lb, in. From Eqs. (1) and (2): q 0 300 lbin. ( MB ) 0 240,000 lb-in. For q q 0 : MB 800 q (3) For q q 0 : MB 300,000 200 q (4)
GRAPH OF BENDING MOMENT MB (EQS. 3 AND 4)
MB RAL
qL^2 2
3 EI ¢
L^2
qL^2 8
RA RC
3 qL 8
3 EI ¢
L^3
RB
5 qL 4
6 EI ¢
L^3
5 qL^4 24 EI
RB L^3
6 EI
Solution 10.4-13 Beam on a support with a gap
A (^) B C
RA (^) RC
L L
q
R (^) B
A
A
R (^) B
C
C
q
( B ) 2
( B ) 1
O
300,
200,
1000 2000
200,
100,
100,
q 0 300
EQ. (3) EQ. (4)
M (^) B (lb-in.) (MB ) 0 ^ 240,
q (lb/in.)
M (^) B 0 at q 1500
