Problem 1

Determine the midspan deflection of the beam shown in Figure 1 if E = 10 GPa and I = 20 × 10 6 mm4.

2m 5 KN
3 KN.m

6m

From Case No. 7, midspan deflection is From Case No. 8, midspan deflection is

Pb 5 wo L
4
δ= ( 3 L2−4 b2 ) when a>b δ=
48 EI 384 EI

Midspan deflection of the given beam

EIδ = EIδ due to 5 kN concentrated load + EIδ due to 3 kN/m uniform loading
4
Pb ( 2 5 wo L
3 L −4 b ) +
2
δ=
48 EI 384 EI
4
5 (2) 5(2)6
EIδ= ( 3(6)2−4 (2)2 ) +
48 384

115 135
EIδ= +
6 4

635
EIδ=
12

635 4
(1000 )
12
δ= 6
10,000(20 X 10 )

δ=264.583 mm answer
Problem 2

Determine the midspan value of EIδ for the beam loaded as shown in Figure 2. Use the method of
superposition.

170 lb 100 lb

3 ft 6 ft 4 ft

From Case No. 7 of Summary of Beam Loadings, deflection at the center is

Pb
δ= ( 3 L2−4 b2 ) when a>b
48 EI

Thus,
EIδmidspan = EIδmidspan due to 170 lb force + EIδmidspan due to 100 lb force

Pb
EI δ midspan =∑ ( 3 L2−4 b2 )
48 EI

170(3) 100(4)
EI δ midspan = ( 3(a)2−4 (6)2 ) + ( 3(a)2−4 ( 4)2 )
48 48

EI δ midspan =6.9282+4.6188

3
EI δ midspan =11.5470lb . ft answer
Problem3

Find the value of EIδ midway between the supports for the beam shown in Fig. 3. (Hint: Combine Case
No. 11 and one half of Case No. 8.)

100 lb/ft 100 lb.ft

3 ft 3 ft 3 ft

The midspan deflection from Case No. 8 and Case No. 11 are respectively,

∑ MB=0 ∑ MA=0

R1 ( 6 )−100 ( 3 )( 4.5 ) +100 ( 3 ) (1.5 )=0 - R2 ( 6 ) +100 ( 3 ) ( 7.5 ) +100 ( 3 ) ( 1.5 )=0

R1=150 lb lb
R2=450
EIδ = ½ of EIδ due to uniform load over
the entire span - EIδ due to end moment

[] (
1 5 ( 100 ) ( 6 ) 450 (6 )
)
4 2
EI δ midspan = −
2 384 16

EI δ midspan =843.75−1012.5

EI δ midspan =−168.75

3
EI δ midspan =168.75 lb. ft upward answer
Problem 4

Determine the value of EIδ under each concentrated load in Figure below.

100 N 300 N

4m 4m 3m

From Case No. 7 of Summary of Beam Loadings, the deflection equations are

Pb
δ= ( 3 L2−4 b2 ) when a>b
48 EI
Pb
δ= ¿
6L

Deflection under the 100 N load EIδ = EIδ due to 100 N load + EIδ due to 300 N load

100 (4 )(7) 2 2 2 300 ( 3 ) ( 4 )

EIδ= ( 11 −4 −7 ) + (112−32−42 )
46(11) 6 (11 )

3
EIδ=7612.1212 N . m answer

Deflection under the 300 N load EIδ = EIδ due to 300 N load + EIδ due to 100 N load

100 (7)
EIδ=
6¿¿

3
EIδ=1745.4545 N .m answer
Problem 5

The beam shown in Figure below has a rectangular cross section 5 inches wide by 11 inches deep.
Compute the value of P that will limit the midspan deflection to 0.3 inch. Use E = 2 × 10 6 psi.

3000 lb/ft

3 ft 15ft 3 ft

The overhang is resolved into simple beam with Moment of inertia of beam section
end moments. The magnitude of end moment is,
3
M =3000 ( 3 ) ( 1.5 ) +2 P bd 3 5(9)
I= =
12 12

M =13500+2 P 1215 4
I= ¿
4
Type of loading

2 2
5 wo L
4 ML ML
δ= δ= δ=
384 EI 16 EI 16 EI

4 2
5 wo L ML
EI δ midspan = −2 [ ]
384 EI 16 EI
4 3 2
5(3000)(15 )(12 ) (3000+2 P)(15 )
0.3= −2[ ]
6 1215 6 1215
384 (2 ×10 )( ) 16 (2 ×10 )( )
4 4
45 3000+2 P
0.3= −
8 12500

P=31781.25 lb answer