20 Find the reactions for the beams a x 6 ft b x 4 ft c x

20. Find the reactions for the beams: (a) x = 6 ft, (b) x = 4 ft, (c) x = 5 ft; (d) x = 7 ft 2 in. oo lb RI R2 Io\' o0 lb (2) TRI 2 Io\' - Figure C.7 Problem 20.

Solution

All the reaction forces can be determined from moment equilibrium equations

1) x=6 ft

R1= 100*6/16=37.5 lb(vertically upwards)

R2=100-37.5=62.5 lb(vertically upwards)

x=4 ft

R1=100*4/14=28.57 lb(vertically upwards)

R2=100-28.57=71.43 lb(vertically upwards)

x=5 ft

R1=100*5/15=33.33 lb(vertically upwards)

R2=100-33.33=67.67 lb(vertically upwards)

x=7 ft 2 in = 7.167 ft

R1=100*7.167/17.16741.75 lb(vertically upwards)

R2=58.25 lb(vertically upwards)

2) x= 6 ft

R1=100*6/10=60 lb (vertically downwards)

R2 = 100+60=160 lb (vertically upwards)

x= 4 ft

R1=100*4/10=40 lb (vertically downwards)

R2 = 100+40=140 lb (vertically upwards)

x= 5 ft

R1=100*5/10=50 lb (vertically downwards)

R2 = 100+50=150 lb (vertically upwards)

x= 7 ft 2in=7.167 ft

R1=100*7.167/10=71.67 lb (vertically downwards)

R2 = 100+71.67=171.67 lb (vertically upwards)