Question

2. (10 points) Consider the beam shown below: (a) draw the complete shear diagram with corresponding labels and units (b) draw the complete moment diagram with corresponding labels and units (c) find the magnitude of maximum compressive normal stress (d) find the magnitude of the maximum tensile normal stress Give the value and location for all quantities along the beam and on the cross section. NOTE: The moment of inertia is 12 = 99.75 in4 and the centroid is given. Clearly label all significant quantities in your shear and moment diagrams. 1 in 1 in 斗 6 in 2.5 in 1 in 9 in ī 15 k-ft 2.5 kip/ft 2.5 kip/ft B D A 5 ft 5 ft 5 ft *** *** 17.25 kip 7.75 kip

Answer 1

Draw the free body diagram of the beam.

2.5 kip/ft 2.5 kip/ft 15 k-ft B FBD A D 5 m 5 m 5 m 17.25 kip 7.75 kip

a) Shear force diagram.

Shear force at A is 0

Shear force at left of the A is -2.5x5 = -12.5 kip

Shear force at right of the A is -2.5x5+17.25 = 4.75 kip

Shear force at C is 4.75 kip

Shear force at left of the D is 4.75 - 2,5x5 = -7.75 kip

Shear force at right of the A is -7.75 + 7.5 = 0

2.5 kip/ft 2.5 kip/ft 15 k-ft B FBD D 5 m 5 m 5 m 17.25 kip 7.75 kip 4.75 kip + +ve SFD A B D с -ve -ve - 7.75 kip 12.5 kip

b) Moment diagram

Bending moment at A is 15 kip-ft

Bending moment at X, where Bending moment is zero, is. 15 -2.5xXx(X/2)=0

X = 3.464 ft

Bending moment at B is 15 - 2.5x5x2.5 = -16.25 kip-ft

Bending moment at X, where Bending moment is zero, is. 15 -2.5x5x(X -2.5)+ 17.25(X- 5)=0

X = 8.42 ft

Bending moment at C is 15 - 2.5x5x7.5 + 17.25x5= 7.5 kip-ft

Bending moment at x = 11.9 ft is 15 - 2.5x5x9.4 + 17.25x6.9-2.5x1.9x0.95 = 12 kipt-ft

Bending moment at D is 15 - 2.5x5x12.5 + 17.25x10 - 2.5x5x2.5= 0

2.5 kip/ft 2.5 kip/ft 15 k-ft B FBD D 5 m 5 m 5 m 17.25 kip 7.75 kip 4.75 kip + +ve SFD A B D с -ve -ve 7.75 kip 12.5 kip 8.42 ft 1.9 ft 15 kip-ft + 12 kip-ft + 7.5 kip-ft +ve BMD +ve B A D ve 3.464 ft 16.25 kip-ft

c) Maximum compressive normal stress.

The maximum bending moment is 16.25 kif-ft

The maximum compressive normal stress formula is

м ху compr

where M = maximum bending moment = 16.25 kip-ft

y = distance of toppest layer from centroid = 3.5"

I = moment of inertia

substitute the respective values in the above equation.

O compr 16.25 х 12 х 3.5 99.75 О сотрт 6.842 ksi

1 in 1 in →it 6.842 ksi maximum compressive stress 6 in - - 2.5 in 业不 1 in 9 in

d) Maximum tensile normal stress.

The maximum bending moment is 16.25 kif-ft

The maximum compressive normal stress formula is

$\sigma _{tensile}=\frac{M\times y}{I_{z}}$

where M = maximum bending moment = 16.25 kip-ft

y = distance of toppest layer from centroid = 2.5"

I_z= moment of inertia

substitute the respective values in the above equation.

$\\\sigma _{tensile}=\frac{16.25\times 2.5}{99.75}\\ \\ \sigma _{tensile}=4.887 \text{ ksi}$

1 in 习 1 in 到k 6 in 2.5 in 业 1 in 4.887 ksi maximum tensile stress 9 in 个