w 120 lbft W 151 lb I1 4 ft I2 4 ft b 05 ft h 08 ft M

w = 120 lb/ft W = 151 lb I_1 = 4 ft I_2 = 4 ft b = 0.5 ft h = 0.8 ft M = 150 Ib.ft Calculate the vertical reaction at A: lb Calculate the horizontal reaction at A: lb Calculate the vertical reaction at B: lb Develop the shear and bending moment equations: Section 1: Range: ft lessthanorequalto x lessthanorequalto ft Calculate the maximum shear force in the beam: lb Calculate the location of the maximum bending moment: ft Calculate the maximum bending moment: Ib.ft Calculate the maximum shear stress: psf Calculate the maximum tensile stress: psf

Solution

Data available,

w = 120 lb/ft, W = 151 lb, l1 = l2 = 4 ft, M = 150 lb/ft

Applying static equlibrium conditions,

sigma Fh = 0

Va = 0 ------- horizontal force at A

sigma Fv = Ra + Rb - wL - W = 0

Ra + Rb = wL + W = 120 * 4 + 151 = 631 ---- eqn1

Taking moment about A,

sigma Ma = 0

Rb * L1 - wL1^2 / 2 - W*(L1+L2/2) = 0

Rb = [ wL1^2 / 2 + W*(L1+L2/2) ] / L1

Rb = (960 + 906)/4 = 466.5 lb -------- vertical force at B

Ra = 631 - 466.5 = 164.5 lb -------- vertical force at A

Bending moment: (+CCW)

Ma = 0

Mb = wL1^2 / 2 - Ra * L1 = 120*16 / 2 - 164.5 * 4 = 960 - 658 = 302 lb-ft

Mc =  wL1( L1/2 + L2/2) - Ra (L1+L2/2) - Rb * L2/2

Mc = 120*4 (2+2) - 164.5 * (4+2) - 466.5 * 2 =1920 - 987 - 933 = 0

Md = -150 lb-ft

Me range is 0 to -150 lb-ft (CW)

Shear force:(+ upward)

Ra = 164.5 lb

R_L/2 = 164.5 - 120*4*2 = 164.5 - 960 = - 800 lb

Rb = -800 + 466.5 = -333.5 lb

Rc = -333.5 + 150 = -183.5 lb

Rd = -183.5 lb

Maximum shear force at L1/2, 800 lf (downward)

Location of max bending moment at B i.e L1/2 = 2 ft from A

Max bending moment at B, 302 lb-ft

Max shear stress at L1 / 2 or 2 ft from A, which is 800 / 2 = 400 lb/ft2

Using the Max shear stress theory, tau max = sigma max / 2,

so sigma max = tau max * 2 = 400 * 2 = 800 lb / ft2