Using Beam Influence Lines ExampleStevenVukazich

SanJoseStateUniversity

Example Problem

x

24 ft 20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

18 ft

Q

The hinged concrete beam from our previous example is subjected to a uniform dead load of 1.5 k/ft and a uniform live load of 5.5 k/ft or a point live load of 90 k. In order to design the steel reinforcement at point Q of the concrete beam, it is desired to find the following:

1. The maximum positive bending moment at point Q due to the to the uniform live load and dead load;2. The maximum positive bending moment at point Q due to the to the point live load and dead load;3. The maximum negative bending moment at point Q due to the uniform live live load and dead load.4. The maximum negative bending moment at point Q due to the uniform point live load and dead load.

To place the live loads we need to construct the influence line for the bending moment at point Q. In a previous example, we constructed the MQ influence line.

MQ Influence Line

20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

MQ

0

18 ft

Q

6 ft

x

+

13.5 ft

4.5 ft

x = MQ

0 018– ft 4.5 ft18 + ft 4.5 ft24 ft 036 ft – 9.0 ft44 ft 056 ft 13.5 ft76 ft 084 ft – 5.4 ft

– 5.4 ft– 9.0 ft

Question 1

20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

MQ

0

18 ft

Q

6 ft

x

+

13.5 ft

4.5 ft

– 5.4 ft– 9.0 ft

wD = 1.5 k/ftwL = 5.5 k/ft wL = 5.5 k/ft

Place the distributed live load over all of the positive area of the MQ influence line to find 𝑀#max

'

Question 1

20 ft12 ft 8 ft12 ft 8 ft

BA D F

MQ

0

18 ft 6 ft

x

+

13.5 ft4.5 ft

– 5.4 ft– 9.0 ft

𝑀#(= 1.5k/ft 54.0ft4 − 90ft4 + 216ft4 − 21.6ft4

Q

C

E

G

Area<= =12 24ft 4.5ft = 54.0ft4

Area(>? =12 32ft 13.5ft = 216ft4

Area?A =12 8ft −5.4ft = −21.6ft4

Area=C( =12 20ft −9.0ft = −90ft4

𝑀#(= 1.5k/ft 158.4ft4 = 237.6k−ft

Question 1

20 ft12 ft 8 ft12 ft 8 ft

BA D F

MQ

0

18 ft 6 ft

x

+

13.5 ft4.5 ft

– 5.4 ft– 9.0 ft

𝑀#EFG'

H= 5.5k/ft 54.0ft4 + 216ft4 = 1485.0k−ft

Q

C

E

G

Area<= =12 24ft 4.5ft = 54.0ft4

Area(>? =12 32ft 13.5ft = 216ft4

Area?A =12 8ft −5.4ft = −21.6ft4

Area=C( =12 20ft −9.0ft = −90ft4

𝑀#EFG' = 237.6k−ft+ 1485.0 k−ft=1722.6 k−ft

𝑀#(= 237.6k−ft

Question 2

20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

MQ

0

18 ft

Q

6 ft

x

+

13.5 ft

4.5 ft

– 5.4 ft– 9.0 ft

wD = 1.5 k/ftPL = 90 k

Place the point live load over the maximum positive ordinate of the MQ influence line to find 𝑀#max

'

Question 2

20 ft12 ft 8 ft12 ft 8 ft

BA D F

MQ

0

18 ft 6 ft

x

+

13.5 ft4.5 ft

– 5.4 ft– 9.0 ft

𝑀#EFG'

H= 90k 13.5ft =1215 k−ft

Q

C

E

G

𝑀#(= 237.6k−ft

𝑀#EFG' = 237.6k−ft+ 1215 k−ft=1452.6 k−ft

Dead load is fixed, so MQD remains the same.

Question 3

20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

MQ

0

18 ft

Q

6 ft

x

+

13.5 ft

4.5 ft

– 5.4 ft– 9.0 ft

wD = 1.5 k/ftwL = 5.5 k/ft wL = 5.5 k/ft

Place the distributed live load over all of the negative area of the MQ influence line to find 𝑀#max

I

Question 3

20 ft12 ft 8 ft12 ft 8 ft

BA D F

MQ

0

18 ft 6 ft

x

+

13.5 ft4.5 ft

– 5.4 ft– 9.0 ft

𝑀#EFGI

H= 5.5k/ft −90ft4 − 21.6ft4 = −613.8k−ft

Q

C

E

G

Area<= =12 24ft 4.5ft = 54.0ft4

Area(>? =12 32ft 13.5ft = 216ft4

Area?A =12 8ft −5.4ft = −21.6ft4

Area=C( =12 20ft −9.0ft = −90ft4

𝑀#EFGI = 237.6k−ft−613.8 k−ft=−376.2 k−ft

𝑀#(= 237.6k−ft Dead load is fixed, so MQD remains the same.

Question 4

20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

MQ

0

18 ft

Q

6 ft

x

+

13.5 ft

4.5 ft

– 5.4 ft– 9.0 ft

wD = 1.5 k/ftPL = 90 k

Place the point live load over the maximum negative ordinate of the MQ influence line to find 𝑀#max

'

Question 4

20 ft12 ft 8 ft12 ft 8 ft

BA D F

MQ

0

18 ft 6 ft

x

+

13.5 ft4.5 ft

– 5.4 ft– 9.0 ft

𝑀#EFG'

H= 90k −9.0ft = −810 k−ft

Q

C

E

G

𝑀#(= 237.6k−ft

𝑀#EFG' = 237.6k−ft−810 k−ft=−572.4 k−ft

Dead load is fixed, so MQD remains the same.

Summary of Results for MQmax

x

24 ft 20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

18 ft

Q

L MQmax

5.5 k/ft 1722.6 k-ft

90 k 1452.6 k-ft

5.5 k/ft –376.2 k-ft

90 k –572.4 k-ft

wD = 1.5 k/ftPL = 90 k wL = 5.5 k/ftOR

Concrete BeamSection at Point Q

Another Example

x

24 ft 20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

The hinged concrete beam from our previous example is subjected to a uniform dead load of 1.5 k/ft and a uniform live load of 5.5 k/ft or a point live load of 90 k. In order to design the pin support at point A, it is desired to find the following:

1. The maximum positive vertical reaction at A due to the to the uniform live load and dead load;2. The maximum positive vertical reaction at A due to the point live load and dead load;3. The maximum negative vertical reaction at A due to the uniform live live load and dead load.4. The maximum negative vertical reaction at A due to the point live load and dead load.

To place the live loads we need to construct the influence line for the vertical reaction at point A. In a previous example, we constructed the Ay influence line.

Ay Influence Line

x

24 ft 20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

Ay

0+

– 0.5

1.0

– 0.3

0.75

x = Ay

0 1.024 ft 036 ft – 0.544 ft 056 ft 0.7576 ft 084 ft – 0.3

Question 1

20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

24 ft

x

wD = 1.5 k/ftwL = 5.5 k/ft wL = 5.5 k/ft

Place the distributed live load over all of the positive area of the Ay influence line to find 𝐴Kmax

'

Ay

0+

– 0.5

1.0

– 0.3

0.75

Question 1

20 ft12 ft 8 ft12 ft 8 ft24 ft

x

Area<= =12 24ft 1.0 = 12ft

Area(>? =12 32ft 0.75 = 12ft

Area?A =12 8ft −0.3 = −1.2ft

Area=C( =12 20ft −0.5 = −5ft

Ay

0+

– 0.5

1.0

– 0.3

0.75

𝐴K( = 1.5k/ft 12ft − 5ft + 12ft − 1.2ft

𝐴K( = 1.5k/ft 17.8ft = 26.7k

BA D F

C

E

G

Question 1

20 ft12 ft 8 ft12 ft 8 ft24 ft

x

𝐴KEFG'H= 5.5k/ft 12ft + 12ft =132 k

𝐴K( = 26.7k

𝐴KEFG' = 26.7k+132 k=158.7 k

Ay

0+

– 0.5

1.0

– 0.3

0.75

Area<= =12 24ft 1.0 = 12ft

Area(>? =12 32ft 0.75 = 12ft

Area?A =12 8ft −0.3 = −1.2ft

Area=C( =12 20ft −0.5 = −5ft

Question 2

20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

24 ft

x

wD = 1.5 k/ft

Place the point live load over the maximum positive ordinate of the Ay influence line to find 𝐴Kmax

'

Ay

0+

– 0.5

1.0

– 0.3

0.75

PL = 90 k

Question 2

20 ft12 ft 8 ft12 ft 8 ft24 ft

x

𝐴KEFG'H= 90k 1 =90 k

𝐴K( = 26.7k

𝐴KEFG' = 26.7k+90 k=116.7 k

Ay

0+

– 0.5

1.0

– 0.3

0.75

Area<= =12 24ft 1.0 = 12ft

Area(>? =12 32ft 0.75 = 12ft

Area?A =12 8ft −0.3 = −1.2ft

Area=C( =12 20ft −0.5 = −5ft

Dead load is fixed, so AyD remains the same.

Question 3

20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

24 ft

x

wD = 1.5 k/ft

wL = 5.5 k/ft wL = 5.5 k/ft

Place the distributed live load over all of the negative area of the Ay influence line to find 𝐴Kmax

I

Ay

0+

– 0.5

1.0

– 0.3

0.75

Question 3

20 ft12 ft 8 ft12 ft 8 ft24 ft

x

𝐴KEFGIH= 5.5k/ft −5ft − 1.2ft = −34.1 k

𝐴K( = 26.7k

𝐴KEFGI = 26.7k − 34.1 k= − 7.4k

Ay

0+

– 0.5

1.0

– 0.3

0.75

Area<= =12 24ft 1.0 = 12ft

Area(>? =12 32ft 0.75 = 12ft

Area?A =12 8ft −0.3 = −1.2ft

Area=C( =12 20ft −0.5 = −5ft

Dead load is fixed, so AyD remains the same.

Question 4

20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

24 ft

x

wD = 1.5 k/ft

Place the point live load over the maximum negative ordinate of the Ay influence line to find 𝐴Kmax

I

Ay

0+

– 0.5

1.0

– 0.3

0.75

PL = 90 k

Question 4

20 ft12 ft 8 ft12 ft 8 ft24 ft

x

𝐴KEFGIH= 90k −0.5 = −45 k

𝐴K( = 26.7k

Ay

0+

1.0

– 0.3

0.75

Area<= =12 24ft 1.0 = 12ft

Area(>? =12 32ft 0.75 = 12ft

Area?A =12 8ft −0.3 = −1.2ft

Area=C( =12 20ft −0.5 = −5ft

Dead load is fixed, so AyD remains the same.

– 0.5

𝐴KEFGI = 26.7k − 45 k= − 19.3 k

Summary of Results for Aymax

x

24 ft 20 ft12 ft 8 ft12 ft 8 ftBA

CD

EF

G

L Aymax

5.5 k/ft 158.7 k

90 k 116.7 k

5.5 k/ft –7.4 k

90 k –19.3 k-ft

wD = 1.5 k/ftPL = 90 k wL = 5.5 k/ftOR

Pin Support at Point A