Let $X\subset\mathbb{P}^n$ be a variety singular along a smooth subvariety $Z\subset X$ of positive dimension. Let us assume that $X$ has ordinary singularities along $Z$. Now, let $\pi:Y\rightarrow \mathbb{P}^n$ be the blow-up of a point $p\in Z$, and let $\widetilde{X}\subset Y$ be the strict transform of $X$ with exceptional divisor $E$. Then $\widetilde{X}$ is singular along the strict transform $\widetilde{Z}$ of $Z$. 


Here, ordinary means that when we blow-up just $Z$ the strict transform of $X$ is smooth and intersects transversally the exceptional divisor.


Could it happen that $\widetilde{X}$ is singular somewhere else in the intersection $\widetilde{X}\cap E$ ?



Blowing-up a point in the singular locus
Note that without any assumption on the singularities the answer to your question is yes. For instance, consider the hypersurface $Y = \{x\_0^2+x\_1^3+x\_2^4\}\subset\mathbb{P}^3$. Then $Sing(Y) = [0:0:0:1]$. The projective tangent cone of $Y$ in the singular point is not reduced. Therefore, the singularity is not ordinary. Let $\pi:X\rightarrow\mathbb{P}^3$ be the blow-up of $[0:0:0:1]$. It is easy to check that the singular locus of the strict transform $\widetilde{Y}\subset X$ contains a curve. 


On the other hand, if the singularities are ordinary this can not happen. 


More precisely:


*Let $W\subset Z\subset X$ be smooth projective varieties, and let $Y\subset X$ be a projective variety such that $Sing(Y) = Z$ and $Y$ has ordinary singularities along $Z$. Let $\pi\_W:X\_W\rightarrow X$ be the blow-up of $W$, and $Z\_W$, $Y\_W$ the strict transforms of $Z$ and $Y$ respectively. Then $Sing(Y\_W) = Z\_W$ and $Y\_W$ has ordinary singularities along $Z\_W$.*


Here is the proof:


The inverse image via $\pi\_W$ of $Z$ is given by $Z\_W\cup E\_W$, where $E\_W$ is the exceptional divisor of $\pi\_W$. Now, we may consider the blow-up $\pi\_{Z\_w}:X\_{Z\_W}\rightarrow X\_W$ of $Z\_W\cup E\_W$ in $X\_W$, and the blow-up $\pi\_Z:X\_Z\rightarrow X$ of $Z$ in $X$. Note that since $E\_W$ is a Cartier divisor in $X\_W$ its blow-up does not produce any effect. By Corollary 7.15 in Harshorne there exists a morphism $f:X\_{Z\_W}\rightarrow X\_Z$ such that $\pi\_Z\circ f = \pi\_W\circ \pi\_{Z\_{W}}$. Therefore, the morphism $f$ must map the exceptional divisor $E\_{Z\_W}$ of $\pi\_{Z\_W}$ onto the exceptional divisor $E\_{Z}$ of $\pi\_Z$. Furthermore, $f$ contracts the strict transform $\tilde{E}\_W$ of $E\_W$ in $X\_{Z\_W}$ onto $\pi\_{Z}^{-1}(W)\subset E\_Z$.


Now, let $g:\widetilde{X}\_Z\rightarrow X\_Z$ be the blow-up of $\pi\_{Z}^{-1}(W)$. By the universal property of the blow-up, see Proposition 7.14 Hartshorne, there exits a unique morphism $\phi:X\_{Z\_W}\rightarrow \widetilde{X}\_Z$ such that $g\circ\phi = f$.


Note that since we just blew-up smooth subvarieties both $X\_{Z\_W}$ and $\widetilde{X}\_Z$ are smooth with Picard number $\rho(X\_{Z\_W}) = \rho(\widetilde{X}\_Z) = \rho(X)+2$. In particular, the birational morphism $\phi:X\_{Z\_W}\rightarrow \widetilde{X}\_Z$ can not be a divisorial contraction. Notice that $\phi$ can not be a small contraction either because otherwise $\widetilde{X}\_Z$ would be singular. We conclude that $\phi$ is a bijective morphism, and since both $X\_{Z\_W}$ and $\widetilde{X}\_Z$ are smooth $\phi$ is an isomorphism.


Now, let $Y\_W$ be the strict transform of $Y$ in $X\_W$, and assume that $Z\_W\subsetneqq Sing(Y\_W)$. Therefore, the strict transform $Y\_{Z\_W}$ of $Y\_W$ in $X\_{Z\_W}$ is singular, and since $g$ is just the blow-up of a smooth variety the image $(g\circ \phi)(Y\_{Z\_W}) = f(Y\_{Z\_W})\subset X\_Z$ is singular as well. Note that $f(Y\_{Z\_W})\subset X\_Z$ is nothing but the strict transform of $Y$ via $\pi\_Z$, and since $Sing(Y) = Z$ and $Y$ has ordinary singularities along $Z$ the blow-up $\pi\_Z$ resolves the singularities of $Y$. A contradiction. We conclude that $Sing(Y\_W) = Z\_W$. Finally, if the intersection $Y\_{Z\_W}\cap E\_{Z\_W}$ is not transversal then the intersection $Y\_Z\cap E\_Z$ is not transversal as well. But this can not happen because the singularities of $Y$ are ordinary. Therefore, $Y\_W$ has ordinary singularities along $Z\_W$. 


