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2.001 - MECHANICS AND MATERIALS I

Lecture #

9/27/

Prof. Carol Livermore

Recall:

Sign Convention

Positive internal forces and moments shown.

Distributed Loads

[

[

q(x) = q 0

∫ L ]L

FTotal Loading = q(x)dx = q 0 x = q 0 L (^0 )

∫ (^) L 2 ]L

MADistributed Load = q(x)xdx =

q 0

2

x

q 0

2

L

2

(^0 )

Lump: (Equivalent Forces)

FBD

Lump:

Fx = 0

N = 0

Fy = 0

q 0 L L − q 0 x − Vy = 0 ⇒ Vy q 0 − x 2 2

M∗ = 0

q 0 L x + q 0 x

x +Mz = 0 2 2

L x

2

Mz = q 0 x − 2 2

Plot

Sanity Check

Pinned and Pinned on rollers at end

Cannot support moment M at ends

Can support x and y loads ⇒ OK

Fy = 0

Vy − (Vy + δVy ) + qδx = 0

δVy = qδx

Limit as δx 0

qdVy → ⇒ dx

M 0 = 0

δx δx −Mz + Mz + δMz − Vy −(Vy + δVy ) = 0 2 2

δx δx δMz − 2 Vy − δVy 2 2

δx δMz = Vy δx − δVy 2

δMz δVy Vy = + δx 2

Take limit δx → 0

dMz Vy = dx

dVy q = dx

But what if load is not uniformly distributed?

This is a 2-force member.

RAy = tan θ =

L 2 =

RAx L 2

RAx = 2RAy ⇒

RAx = 2 RAy

RAx = − 1 RDx

RAy = − 1 RDy

RAx

RAy

RDx RDy

RAx

RDx

RAy RDy

So:

RAx = 2 ⇒ RDx = 2RDy RDy

So:

M

RDy = L

M RAy = − L

2 M RDx = L

− 2 M RAx = L

Fx = 0

RAx − RCx = 0

− 2 M

RCx = L

Fy = 0

RAy − RCy = 0

−M

RCy = L

Take cut

FBD

FBD of Cut 2

Fx = 0

2 M

N − = 0

L

2 M

N =

L

Fy = 0

M

− − Vy = 0 L

−M Vy = L

M∗ = 0

M

x + Mz = 0 L

−M Mz = x L

2 M

Vy = L

M∗ = 0

2 M

Mz + x = 0 L

2 M Mz = x L

FBD of Cut 4

Fx = 0

M

− + N = 0

L

M

N =

L

Fy = 0

2 M

− Vy = 0 L

2 M Vy = L

M∗ = 0

2 M

M − x + Mz = 0 L

2 x Mz = −M 1 − L